Physics Book - User contributions [en]

Loop Rule

2016-04-16T22:22:42Z

Ldarian3: /* References */

'''CLAIMED BY LAYLA DARIAN'''

The Loop Rule is a fundamental principle of electric circuits that states when you travel around an entire circuit loop, you will return to the starting voltage.
==The Energy Principle and Loop Rule==

The loop rule simply states that in any round trip path in a circuit,
[[Electric Potential]] equals zero. Keep in mind that this applies through ANY round trip path; there can be
multiple round trip paths through more complex circuits. This principle deals with the conservation of energy within a circuit. Loop Rule and [[Node Rule]] are the two
fundamental principle of electric circuits and are used to determine the behaviors of electric circuits.
This principle is often used to solve for resistance of the light bulbs or other types of resistors or the
current passing through these resistors.

===A Mathematical Model===
A mathematical representation is:
<math> \Sigma \Delta {V}_{i} = 0 </math>
AND
<math> \Delta {V}_{1} + \Delta {V}_{2} + \space.... = 0 </math> along any closed path in a circuit.

In circuits this Energy Principle is often called Kirchoff's Loop Rule and can be represented as: <math> emf = \Delta {V}_{1} + \Delta {V}_{2} + \space..... </math>

===A Visual Model===

[[File:loop rule.jpg]]

To figure out the sign of the voltages, act as an observer walking along the path. Start at the negative end of the emf and continue walking along the path. The emf will be positive in the loop rule because you are moving from low to high voltage. Once you reach a resistor or capacitor, this will be negative in the loop rule equation because it is high to low voltage. Continue along the path until you return to the starting position.

[[File:loop 1.jpg]]

[[File:loop 2.jpg]]

[[File:loop 3.jpg]]

==Examples==

For further instruction, check out these YouTube videos
[https://www.youtube.com/watch?v=Z2QDXjG2ynU]
[https://www.youtube.com/watch?v=paDs-Hnmklo]

===Simple===
[[File:SimpleLoopRule.jpg]]

The circuit shown above consists of a single battery and a single resistor. The resistance of the wires is negligible for this problem.

Problem:

If the <math> emf </math> is 5 V and the resistance of the resistor is 10 ohms, what is the current passing through the resistor.

Solution:

Although we can solve this using the V = IR equation for the whole loop, lets examine this problem using the loop rule equation.

The loop rule equation would be <math> {V}_{battery} - {V}_{resistor} = 0 </math>

Since we know the <math> emf </math> of the battery we just need to find the potential difference through the resistor. For this we can use the equation of V = IR.

Thus we now have an loop rule equation of <math> emf - IR = 0 </math>
From here it is a relatively simple process to find the current. We can rewrite the loop rule equation as <math> emf = IR </math> and then plug in 5 for the emf and
10 for the resistance, leaving us with I = .5 amperes.

===Middling===

[[File:MediumExampleLoopRule.jpg]]

Problem:

The circuit shown above consists of a single battery, whose <math> emf </math> is 1.3 V, and three wires made of the same material, but having different cross-sectional areas.
Let the length of the thin wires be <math> {L}_{thick} </math> and the length of the thin wire be <math> {L}_{thin} </math>
Find a loop rule equation that starts at the negative end of the battery and goes counterclockwise through the circuit.

Solution:

When beginning this problem, you must notice that the difference in cross-sectional areas affects the electric field in each wire. Because of this
we will denote the electric field at D. as <math> {E}_{D} </math> and the electric field everywhere else as <math> {E}_{A} </math>.
To begin we will go around the circuit clockwise and add up each component. First, we know that the <math> emf </math> of the battery is 1.3 V. Then, we will add up the potential
voltage of each of the wires.

Remember that the electric potential of a wire is equal to the electric field * length of the wire. From his we can now find the potential difference
of each section of the wires. The electric potential of location A - C is <math> {E}_{A} * {L}_{thick} </math>. This is the same for the electric potential of location E - G of the wire.
For the thin section of the wire, the electric potential is <math> {E}_{D} * {L}_{thin} </math>. From here we just go around the circuit counterclockwise and add each potential difference to the loop rule equation.

Thus we can find that a loop rule equation is: <math> emf - 2 ({E}_{A} * {L}_{thick}) - {E}_{D} * {L}_{thin} = 0 </math>

This can also be rewritten as: <math> emf = 2 ({E}_{A} * {L}_{thick}) + {E}_{D} * {L}_{thin} = 0 </math>

===Difficult===

[[File:HardLoopRule.jpg]]

Problem:

For the circuit above, imagine a situation where the switch has been closed for a long time.
Calculate the current at a,b,c,d,e and charge Q of the capacitor. Answer these using <math> emf, {R}_{1}, {R}_{2},
and \space C </math>

Solution:

First, write loop rule equations for each of the possible loops in the circuit. There are 3 loop equations that are possible.

<math> emf - {I}_{1}{R}_{1} - {I}_{2}{R}_{2} = 0 </math>

<math> emf - {I}_{1}{R}_{1} - Q/C = 0 </math>

<math> {I}_{2}{R}_{2} - Q/C = 0 </math>

From here, we can then solve for the current passing through a,b,d and e. We also know that the current passing through
these points must be the same so <math> {I}_{1} = {I}_{2} </math>

<math> emf - {I}_{1}{R}_{1} - {I}_{1}{R}_{2} = 0 </math>

<math> emf = {I}_{1}({R}_{1} + {R}_{2}) = 0 </math>

<math> emf/({R}_{1} + {R}_{2}) = {I}_{1} </math>

So the current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

You must also know that once a capacitor is charging for a long time, current no longer flows through the capacitor. We can then
easily solve for c because since current is no longer flowing through the capacitor, the current at c = 0.

Lastly, we will use the loop rule equation of <math> {I}_{2}{R}_{2} - Q/C = 0 </math> to solve for Q.

<math> {I}_{2}{R}_{2} = Q/C </math>

<math> C*({I}_{2}{R}_{2}) = Q </math> Since <math> {I}_{1} = {I}_{2} </math>

<math> C*({I}_{1}{R}_{2}) = Q </math> Lastly, we will plug in what we found {I}_{1} equals from before.

<math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}) = Q </math>

Lastly, Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math> and we have now solved the problem.

Answers:

Current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

Current at c = 0

Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math>

==History==

The Loop Rule is formally known as the Kirchhoff Circuit Law, named after Gustav Kirchhoff discovered and defined this fundamental concept of electric circuits. He discovered this during his time as a student at Albertus University of Königsberg in 1845. Kirchoff went on to explore the topics of spectroscopy and black body radiation after his graduation from Albertus. Nowadays, it is used very often in electrical engineering. For more in depth information about Gustav Kirchhoff,
visit the full wiki page on him: [[Gustav Kirchhoff]]

[[File:KirchoffLoopRule.jpg|thumb|right||Gustav Kirchhoff]]

==Connectedness==
Because I am a Computer Science majors, while circuits are not directly connected to programming, computers and other electronics are made of different
circuits. Loop Rule is used in the electrical engineering that is required to create the circuits for the computers I use to program.

One interesting note about Loop Rule is that is does not apply universally to all circuits. In particular, AC (alternating current) circuits, have a fluctuating
electric charge that changes direction. This causes the electric potential of a round trip path around the circuit to no longer be zero. However, DC (direct current)
circuits still follow the loop rule.

== See also ==
Other Circuit Concepts you can check out :

[[Node Rule]]

[[Components]]

[[Resistors and Conductivity]]

[[Current]]

===More Information and External links===

[https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws Kirchoff's Circuit Laws - Wikipedia]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/circuits-and-direct-currents-20/kirchhoff-s-rules-152/the-loop-rule-540-5636/ Loop Rule - Boundless.com Textbook]

[https://www.youtube.com/watch?v=IlyUtYRqMLs Loop Rule - Doc Physics Video Lecture]

==References==

* Dr. Nicholas Darnton's lecture notes
* https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff.27s_voltage_law
* Chabay, Ruth W. Matter and Interactions: Electric and Magnetic Interactions. John Wiley, 2015. Print.
* http://www.ux1.eiu.edu/~cfadd/1360/28DC/Loop.html
* Mason, J. (Producer). (2012). How to Solve a Kirchhoff's Rules Problem [Motion picture]. United States of America: YouTube.
*Anderson, P. (2015, March 09). Kirchoff's Loop Rule. Retrieved April 16, 2016, from https://www.youtube.com/watch?v=paDs-Hnmklo

Claimed by Bmock7
[[Category: Simple Circuits]]

Loop Rule

2016-04-16T17:55:38Z

Ldarian3: /* References */

'''CLAIMED BY LAYLA DARIAN'''

The Loop Rule is a fundamental principle of electric circuits that states when you travel around an entire circuit loop, you will return to the starting voltage.
==The Energy Principle and Loop Rule==

The loop rule simply states that in any round trip path in a circuit,
[[Electric Potential]] equals zero. Keep in mind that this applies through ANY round trip path; there can be
multiple round trip paths through more complex circuits. This principle deals with the conservation of energy within a circuit. Loop Rule and [[Node Rule]] are the two
fundamental principle of electric circuits and are used to determine the behaviors of electric circuits.
This principle is often used to solve for resistance of the light bulbs or other types of resistors or the
current passing through these resistors.

===A Mathematical Model===
A mathematical representation is:
<math> \Sigma \Delta {V}_{i} = 0 </math>
AND
<math> \Delta {V}_{1} + \Delta {V}_{2} + \space.... = 0 </math> along any closed path in a circuit.

In circuits this Energy Principle is often called Kirchoff's Loop Rule and can be represented as: <math> emf = \Delta {V}_{1} + \Delta {V}_{2} + \space..... </math>

===A Visual Model===

[[File:loop rule.jpg]]

To figure out the sign of the voltages, act as an observer walking along the path. Start at the negative end of the emf and continue walking along the path. The emf will be positive in the loop rule because you are moving from low to high voltage. Once you reach a resistor or capacitor, this will be negative in the loop rule equation because it is high to low voltage. Continue along the path until you return to the starting position.

[[File:loop 1.jpg]]

[[File:loop 2.jpg]]

[[File:loop 3.jpg]]

==Examples==

For further instruction, check out these YouTube videos
[https://www.youtube.com/watch?v=Z2QDXjG2ynU]
[https://www.youtube.com/watch?v=paDs-Hnmklo]

===Simple===
[[File:SimpleLoopRule.jpg]]

The circuit shown above consists of a single battery and a single resistor. The resistance of the wires is negligible for this problem.

Problem:

If the <math> emf </math> is 5 V and the resistance of the resistor is 10 ohms, what is the current passing through the resistor.

Solution:

Although we can solve this using the V = IR equation for the whole loop, lets examine this problem using the loop rule equation.

The loop rule equation would be <math> {V}_{battery} - {V}_{resistor} = 0 </math>

Since we know the <math> emf </math> of the battery we just need to find the potential difference through the resistor. For this we can use the equation of V = IR.

Thus we now have an loop rule equation of <math> emf - IR = 0 </math>
From here it is a relatively simple process to find the current. We can rewrite the loop rule equation as <math> emf = IR </math> and then plug in 5 for the emf and
10 for the resistance, leaving us with I = .5 amperes.

===Middling===

[[File:MediumExampleLoopRule.jpg]]

Problem:

The circuit shown above consists of a single battery, whose <math> emf </math> is 1.3 V, and three wires made of the same material, but having different cross-sectional areas.
Let the length of the thin wires be <math> {L}_{thick} </math> and the length of the thin wire be <math> {L}_{thin} </math>
Find a loop rule equation that starts at the negative end of the battery and goes counterclockwise through the circuit.

Solution:

When beginning this problem, you must notice that the difference in cross-sectional areas affects the electric field in each wire. Because of this
we will denote the electric field at D. as <math> {E}_{D} </math> and the electric field everywhere else as <math> {E}_{A} </math>.
To begin we will go around the circuit clockwise and add up each component. First, we know that the <math> emf </math> of the battery is 1.3 V. Then, we will add up the potential
voltage of each of the wires.

Remember that the electric potential of a wire is equal to the electric field * length of the wire. From his we can now find the potential difference
of each section of the wires. The electric potential of location A - C is <math> {E}_{A} * {L}_{thick} </math>. This is the same for the electric potential of location E - G of the wire.
For the thin section of the wire, the electric potential is <math> {E}_{D} * {L}_{thin} </math>. From here we just go around the circuit counterclockwise and add each potential difference to the loop rule equation.

Thus we can find that a loop rule equation is: <math> emf - 2 ({E}_{A} * {L}_{thick}) - {E}_{D} * {L}_{thin} = 0 </math>

This can also be rewritten as: <math> emf = 2 ({E}_{A} * {L}_{thick}) + {E}_{D} * {L}_{thin} = 0 </math>

===Difficult===

[[File:HardLoopRule.jpg]]

Problem:

For the circuit above, imagine a situation where the switch has been closed for a long time.
Calculate the current at a,b,c,d,e and charge Q of the capacitor. Answer these using <math> emf, {R}_{1}, {R}_{2},
and \space C </math>

Solution:

First, write loop rule equations for each of the possible loops in the circuit. There are 3 loop equations that are possible.

<math> emf - {I}_{1}{R}_{1} - {I}_{2}{R}_{2} = 0 </math>

<math> emf - {I}_{1}{R}_{1} - Q/C = 0 </math>

<math> {I}_{2}{R}_{2} - Q/C = 0 </math>

From here, we can then solve for the current passing through a,b,d and e. We also know that the current passing through
these points must be the same so <math> {I}_{1} = {I}_{2} </math>

<math> emf - {I}_{1}{R}_{1} - {I}_{1}{R}_{2} = 0 </math>

<math> emf = {I}_{1}({R}_{1} + {R}_{2}) = 0 </math>

<math> emf/({R}_{1} + {R}_{2}) = {I}_{1} </math>

So the current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

You must also know that once a capacitor is charging for a long time, current no longer flows through the capacitor. We can then
easily solve for c because since current is no longer flowing through the capacitor, the current at c = 0.

Lastly, we will use the loop rule equation of <math> {I}_{2}{R}_{2} - Q/C = 0 </math> to solve for Q.

<math> {I}_{2}{R}_{2} = Q/C </math>

<math> C*({I}_{2}{R}_{2}) = Q </math> Since <math> {I}_{1} = {I}_{2} </math>

<math> C*({I}_{1}{R}_{2}) = Q </math> Lastly, we will plug in what we found {I}_{1} equals from before.

<math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}) = Q </math>

Lastly, Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math> and we have now solved the problem.

Answers:

Current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

Current at c = 0

Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math>

==History==

The Loop Rule is formally known as the Kirchhoff Circuit Law, named after Gustav Kirchhoff discovered and defined this fundamental concept of electric circuits. He discovered this during his time as a student at Albertus University of Königsberg in 1845. Kirchoff went on to explore the topics of spectroscopy and black body radiation after his graduation from Albertus. Nowadays, it is used very often in electrical engineering. For more in depth information about Gustav Kirchhoff,
visit the full wiki page on him: [[Gustav Kirchhoff]]

[[File:KirchoffLoopRule.jpg|thumb|right||Gustav Kirchhoff]]

==Connectedness==
Because I am a Computer Science majors, while circuits are not directly connected to programming, computers and other electronics are made of different
circuits. Loop Rule is used in the electrical engineering that is required to create the circuits for the computers I use to program.

One interesting note about Loop Rule is that is does not apply universally to all circuits. In particular, AC (alternating current) circuits, have a fluctuating
electric charge that changes direction. This causes the electric potential of a round trip path around the circuit to no longer be zero. However, DC (direct current)
circuits still follow the loop rule.

== See also ==
Other Circuit Concepts you can check out :

[[Node Rule]]

[[Components]]

[[Resistors and Conductivity]]

[[Current]]

===More Information and External links===

[https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws Kirchoff's Circuit Laws - Wikipedia]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/circuits-and-direct-currents-20/kirchhoff-s-rules-152/the-loop-rule-540-5636/ Loop Rule - Boundless.com Textbook]

[https://www.youtube.com/watch?v=IlyUtYRqMLs Loop Rule - Doc Physics Video Lecture]

==References==

* Dr. Nicholas Darnton's lecture notes
* https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff.27s_voltage_law
* Chabay, Ruth W. Matter and Interactions: Electric and Magnetic Interactions. John Wiley, 2015. Print.
* http://www.ux1.eiu.edu/~cfadd/1360/28DC/Loop.html
* https://www.youtube.com/watch?v=paDs-Hnmklo
*https://www.youtube.com/watch?v=Z2QDXjG2ynU

Claimed by Bmock7
[[Category: Simple Circuits]]

Loop Rule

2016-04-16T17:54:31Z

Ldarian3: /* References */

'''CLAIMED BY LAYLA DARIAN'''

The Loop Rule is a fundamental principle of electric circuits that states when you travel around an entire circuit loop, you will return to the starting voltage.
==The Energy Principle and Loop Rule==

The loop rule simply states that in any round trip path in a circuit,
[[Electric Potential]] equals zero. Keep in mind that this applies through ANY round trip path; there can be
multiple round trip paths through more complex circuits. This principle deals with the conservation of energy within a circuit. Loop Rule and [[Node Rule]] are the two
fundamental principle of electric circuits and are used to determine the behaviors of electric circuits.
This principle is often used to solve for resistance of the light bulbs or other types of resistors or the
current passing through these resistors.

===A Mathematical Model===
A mathematical representation is:
<math> \Sigma \Delta {V}_{i} = 0 </math>
AND
<math> \Delta {V}_{1} + \Delta {V}_{2} + \space.... = 0 </math> along any closed path in a circuit.

In circuits this Energy Principle is often called Kirchoff's Loop Rule and can be represented as: <math> emf = \Delta {V}_{1} + \Delta {V}_{2} + \space..... </math>

===A Visual Model===

[[File:loop rule.jpg]]

To figure out the sign of the voltages, act as an observer walking along the path. Start at the negative end of the emf and continue walking along the path. The emf will be positive in the loop rule because you are moving from low to high voltage. Once you reach a resistor or capacitor, this will be negative in the loop rule equation because it is high to low voltage. Continue along the path until you return to the starting position.

[[File:loop 1.jpg]]

[[File:loop 2.jpg]]

[[File:loop 3.jpg]]

==Examples==

For further instruction, check out these YouTube videos
[https://www.youtube.com/watch?v=Z2QDXjG2ynU]
[https://www.youtube.com/watch?v=paDs-Hnmklo]

===Simple===
[[File:SimpleLoopRule.jpg]]

The circuit shown above consists of a single battery and a single resistor. The resistance of the wires is negligible for this problem.

Problem:

If the <math> emf </math> is 5 V and the resistance of the resistor is 10 ohms, what is the current passing through the resistor.

Solution:

Although we can solve this using the V = IR equation for the whole loop, lets examine this problem using the loop rule equation.

The loop rule equation would be <math> {V}_{battery} - {V}_{resistor} = 0 </math>

Since we know the <math> emf </math> of the battery we just need to find the potential difference through the resistor. For this we can use the equation of V = IR.

Thus we now have an loop rule equation of <math> emf - IR = 0 </math>
From here it is a relatively simple process to find the current. We can rewrite the loop rule equation as <math> emf = IR </math> and then plug in 5 for the emf and
10 for the resistance, leaving us with I = .5 amperes.

===Middling===

[[File:MediumExampleLoopRule.jpg]]

Problem:

The circuit shown above consists of a single battery, whose <math> emf </math> is 1.3 V, and three wires made of the same material, but having different cross-sectional areas.
Let the length of the thin wires be <math> {L}_{thick} </math> and the length of the thin wire be <math> {L}_{thin} </math>
Find a loop rule equation that starts at the negative end of the battery and goes counterclockwise through the circuit.

Solution:

When beginning this problem, you must notice that the difference in cross-sectional areas affects the electric field in each wire. Because of this
we will denote the electric field at D. as <math> {E}_{D} </math> and the electric field everywhere else as <math> {E}_{A} </math>.
To begin we will go around the circuit clockwise and add up each component. First, we know that the <math> emf </math> of the battery is 1.3 V. Then, we will add up the potential
voltage of each of the wires.

Remember that the electric potential of a wire is equal to the electric field * length of the wire. From his we can now find the potential difference
of each section of the wires. The electric potential of location A - C is <math> {E}_{A} * {L}_{thick} </math>. This is the same for the electric potential of location E - G of the wire.
For the thin section of the wire, the electric potential is <math> {E}_{D} * {L}_{thin} </math>. From here we just go around the circuit counterclockwise and add each potential difference to the loop rule equation.

Thus we can find that a loop rule equation is: <math> emf - 2 ({E}_{A} * {L}_{thick}) - {E}_{D} * {L}_{thin} = 0 </math>

This can also be rewritten as: <math> emf = 2 ({E}_{A} * {L}_{thick}) + {E}_{D} * {L}_{thin} = 0 </math>

===Difficult===

[[File:HardLoopRule.jpg]]

Problem:

For the circuit above, imagine a situation where the switch has been closed for a long time.
Calculate the current at a,b,c,d,e and charge Q of the capacitor. Answer these using <math> emf, {R}_{1}, {R}_{2},
and \space C </math>

Solution:

First, write loop rule equations for each of the possible loops in the circuit. There are 3 loop equations that are possible.

<math> emf - {I}_{1}{R}_{1} - {I}_{2}{R}_{2} = 0 </math>

<math> emf - {I}_{1}{R}_{1} - Q/C = 0 </math>

<math> {I}_{2}{R}_{2} - Q/C = 0 </math>

From here, we can then solve for the current passing through a,b,d and e. We also know that the current passing through
these points must be the same so <math> {I}_{1} = {I}_{2} </math>

<math> emf - {I}_{1}{R}_{1} - {I}_{1}{R}_{2} = 0 </math>

<math> emf = {I}_{1}({R}_{1} + {R}_{2}) = 0 </math>

<math> emf/({R}_{1} + {R}_{2}) = {I}_{1} </math>

So the current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

You must also know that once a capacitor is charging for a long time, current no longer flows through the capacitor. We can then
easily solve for c because since current is no longer flowing through the capacitor, the current at c = 0.

Lastly, we will use the loop rule equation of <math> {I}_{2}{R}_{2} - Q/C = 0 </math> to solve for Q.

<math> {I}_{2}{R}_{2} = Q/C </math>

<math> C*({I}_{2}{R}_{2}) = Q </math> Since <math> {I}_{1} = {I}_{2} </math>

<math> C*({I}_{1}{R}_{2}) = Q </math> Lastly, we will plug in what we found {I}_{1} equals from before.

<math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}) = Q </math>

Lastly, Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math> and we have now solved the problem.

Answers:

Current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

Current at c = 0

Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math>

==History==

The Loop Rule is formally known as the Kirchhoff Circuit Law, named after Gustav Kirchhoff discovered and defined this fundamental concept of electric circuits. He discovered this during his time as a student at Albertus University of Königsberg in 1845. Kirchoff went on to explore the topics of spectroscopy and black body radiation after his graduation from Albertus. Nowadays, it is used very often in electrical engineering. For more in depth information about Gustav Kirchhoff,
visit the full wiki page on him: [[Gustav Kirchhoff]]

[[File:KirchoffLoopRule.jpg|thumb|right||Gustav Kirchhoff]]

==Connectedness==
Because I am a Computer Science majors, while circuits are not directly connected to programming, computers and other electronics are made of different
circuits. Loop Rule is used in the electrical engineering that is required to create the circuits for the computers I use to program.

One interesting note about Loop Rule is that is does not apply universally to all circuits. In particular, AC (alternating current) circuits, have a fluctuating
electric charge that changes direction. This causes the electric potential of a round trip path around the circuit to no longer be zero. However, DC (direct current)
circuits still follow the loop rule.

== See also ==
Other Circuit Concepts you can check out :

[[Node Rule]]

[[Components]]

[[Resistors and Conductivity]]

[[Current]]

===More Information and External links===

[https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws Kirchoff's Circuit Laws - Wikipedia]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/circuits-and-direct-currents-20/kirchhoff-s-rules-152/the-loop-rule-540-5636/ Loop Rule - Boundless.com Textbook]

[https://www.youtube.com/watch?v=IlyUtYRqMLs Loop Rule - Doc Physics Video Lecture]

==References==

* https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff.27s_voltage_law
* Chabay, Ruth W. Matter and Interactions: Electric and Magnetic Interactions. John Wiley, 2015. Print.
* http://www.ux1.eiu.edu/~cfadd/1360/28DC/Loop.html
* https://www.youtube.com/watch?v=paDs-Hnmklo
*

Claimed by Bmock7
[[Category: Simple Circuits]]

Loop Rule

2016-04-16T17:43:05Z

Ldarian3: /* Examples */

'''CLAIMED BY LAYLA DARIAN'''

The Loop Rule is a fundamental principle of electric circuits that states when you travel around an entire circuit loop, you will return to the starting voltage.
==The Energy Principle and Loop Rule==

The loop rule simply states that in any round trip path in a circuit,
[[Electric Potential]] equals zero. Keep in mind that this applies through ANY round trip path; there can be
multiple round trip paths through more complex circuits. This principle deals with the conservation of energy within a circuit. Loop Rule and [[Node Rule]] are the two
fundamental principle of electric circuits and are used to determine the behaviors of electric circuits.
This principle is often used to solve for resistance of the light bulbs or other types of resistors or the
current passing through these resistors.

===A Mathematical Model===
A mathematical representation is:
<math> \Sigma \Delta {V}_{i} = 0 </math>
AND
<math> \Delta {V}_{1} + \Delta {V}_{2} + \space.... = 0 </math> along any closed path in a circuit.

In circuits this Energy Principle is often called Kirchoff's Loop Rule and can be represented as: <math> emf = \Delta {V}_{1} + \Delta {V}_{2} + \space..... </math>

===A Visual Model===

[[File:loop rule.jpg]]

To figure out the sign of the voltages, act as an observer walking along the path. Start at the negative end of the emf and continue walking along the path. The emf will be positive in the loop rule because you are moving from low to high voltage. Once you reach a resistor or capacitor, this will be negative in the loop rule equation because it is high to low voltage. Continue along the path until you return to the starting position.

[[File:loop 1.jpg]]

[[File:loop 2.jpg]]

[[File:loop 3.jpg]]

==Examples==

For further instruction, check out these YouTube videos
[https://www.youtube.com/watch?v=Z2QDXjG2ynU]
[https://www.youtube.com/watch?v=paDs-Hnmklo]

===Simple===
[[File:SimpleLoopRule.jpg]]

The circuit shown above consists of a single battery and a single resistor. The resistance of the wires is negligible for this problem.

Problem:

If the <math> emf </math> is 5 V and the resistance of the resistor is 10 ohms, what is the current passing through the resistor.

Solution:

Although we can solve this using the V = IR equation for the whole loop, lets examine this problem using the loop rule equation.

The loop rule equation would be <math> {V}_{battery} - {V}_{resistor} = 0 </math>

Since we know the <math> emf </math> of the battery we just need to find the potential difference through the resistor. For this we can use the equation of V = IR.

Thus we now have an loop rule equation of <math> emf - IR = 0 </math>
From here it is a relatively simple process to find the current. We can rewrite the loop rule equation as <math> emf = IR </math> and then plug in 5 for the emf and
10 for the resistance, leaving us with I = .5 amperes.

===Middling===

[[File:MediumExampleLoopRule.jpg]]

Problem:

The circuit shown above consists of a single battery, whose <math> emf </math> is 1.3 V, and three wires made of the same material, but having different cross-sectional areas.
Let the length of the thin wires be <math> {L}_{thick} </math> and the length of the thin wire be <math> {L}_{thin} </math>
Find a loop rule equation that starts at the negative end of the battery and goes counterclockwise through the circuit.

Solution:

When beginning this problem, you must notice that the difference in cross-sectional areas affects the electric field in each wire. Because of this
we will denote the electric field at D. as <math> {E}_{D} </math> and the electric field everywhere else as <math> {E}_{A} </math>.
To begin we will go around the circuit clockwise and add up each component. First, we know that the <math> emf </math> of the battery is 1.3 V. Then, we will add up the potential
voltage of each of the wires.

Remember that the electric potential of a wire is equal to the electric field * length of the wire. From his we can now find the potential difference
of each section of the wires. The electric potential of location A - C is <math> {E}_{A} * {L}_{thick} </math>. This is the same for the electric potential of location E - G of the wire.
For the thin section of the wire, the electric potential is <math> {E}_{D} * {L}_{thin} </math>. From here we just go around the circuit counterclockwise and add each potential difference to the loop rule equation.

Thus we can find that a loop rule equation is: <math> emf - 2 ({E}_{A} * {L}_{thick}) - {E}_{D} * {L}_{thin} = 0 </math>

This can also be rewritten as: <math> emf = 2 ({E}_{A} * {L}_{thick}) + {E}_{D} * {L}_{thin} = 0 </math>

===Difficult===

[[File:HardLoopRule.jpg]]

Problem:

For the circuit above, imagine a situation where the switch has been closed for a long time.
Calculate the current at a,b,c,d,e and charge Q of the capacitor. Answer these using <math> emf, {R}_{1}, {R}_{2},
and \space C </math>

Solution:

First, write loop rule equations for each of the possible loops in the circuit. There are 3 loop equations that are possible.

<math> emf - {I}_{1}{R}_{1} - {I}_{2}{R}_{2} = 0 </math>

<math> emf - {I}_{1}{R}_{1} - Q/C = 0 </math>

<math> {I}_{2}{R}_{2} - Q/C = 0 </math>

From here, we can then solve for the current passing through a,b,d and e. We also know that the current passing through
these points must be the same so <math> {I}_{1} = {I}_{2} </math>

<math> emf - {I}_{1}{R}_{1} - {I}_{1}{R}_{2} = 0 </math>

<math> emf = {I}_{1}({R}_{1} + {R}_{2}) = 0 </math>

<math> emf/({R}_{1} + {R}_{2}) = {I}_{1} </math>

So the current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

You must also know that once a capacitor is charging for a long time, current no longer flows through the capacitor. We can then
easily solve for c because since current is no longer flowing through the capacitor, the current at c = 0.

Lastly, we will use the loop rule equation of <math> {I}_{2}{R}_{2} - Q/C = 0 </math> to solve for Q.

<math> {I}_{2}{R}_{2} = Q/C </math>

<math> C*({I}_{2}{R}_{2}) = Q </math> Since <math> {I}_{1} = {I}_{2} </math>

<math> C*({I}_{1}{R}_{2}) = Q </math> Lastly, we will plug in what we found {I}_{1} equals from before.

<math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}) = Q </math>

Lastly, Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math> and we have now solved the problem.

Answers:

Current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

Current at c = 0

Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math>

==History==

The Loop Rule is formally known as the Kirchhoff Circuit Law, named after Gustav Kirchhoff discovered and defined this fundamental concept of electric circuits. He discovered this during his time as a student at Albertus University of Königsberg in 1845. Kirchoff went on to explore the topics of spectroscopy and black body radiation after his graduation from Albertus. Nowadays, it is used very often in electrical engineering. For more in depth information about Gustav Kirchhoff,
visit the full wiki page on him: [[Gustav Kirchhoff]]

[[File:KirchoffLoopRule.jpg|thumb|right||Gustav Kirchhoff]]

==Connectedness==
Because I am a Computer Science majors, while circuits are not directly connected to programming, computers and other electronics are made of different
circuits. Loop Rule is used in the electrical engineering that is required to create the circuits for the computers I use to program.

One interesting note about Loop Rule is that is does not apply universally to all circuits. In particular, AC (alternating current) circuits, have a fluctuating
electric charge that changes direction. This causes the electric potential of a round trip path around the circuit to no longer be zero. However, DC (direct current)
circuits still follow the loop rule.

== See also ==
Other Circuit Concepts you can check out :

[[Node Rule]]

[[Components]]

[[Resistors and Conductivity]]

[[Current]]

===More Information and External links===

[https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws Kirchoff's Circuit Laws - Wikipedia]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/circuits-and-direct-currents-20/kirchhoff-s-rules-152/the-loop-rule-540-5636/ Loop Rule - Boundless.com Textbook]

[https://www.youtube.com/watch?v=IlyUtYRqMLs Loop Rule - Doc Physics Video Lecture]

==References==

* https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff.27s_voltage_law
* Chabay, Ruth W. Matter and Interactions: Electric and Magnetic Interactions. John Wiley, 2015. Print.
* http://www.ux1.eiu.edu/~cfadd/1360/28DC/Loop.html

Claimed by Bmock7
[[Category: Simple Circuits]]

Loop Rule

2016-04-16T17:41:03Z

Ldarian3: /* Examples */

'''CLAIMED BY LAYLA DARIAN'''

The Loop Rule is a fundamental principle of electric circuits that states when you travel around an entire circuit loop, you will return to the starting voltage.
==The Energy Principle and Loop Rule==

The loop rule simply states that in any round trip path in a circuit,
[[Electric Potential]] equals zero. Keep in mind that this applies through ANY round trip path; there can be
multiple round trip paths through more complex circuits. This principle deals with the conservation of energy within a circuit. Loop Rule and [[Node Rule]] are the two
fundamental principle of electric circuits and are used to determine the behaviors of electric circuits.
This principle is often used to solve for resistance of the light bulbs or other types of resistors or the
current passing through these resistors.

===A Mathematical Model===
A mathematical representation is:
<math> \Sigma \Delta {V}_{i} = 0 </math>
AND
<math> \Delta {V}_{1} + \Delta {V}_{2} + \space.... = 0 </math> along any closed path in a circuit.

In circuits this Energy Principle is often called Kirchoff's Loop Rule and can be represented as: <math> emf = \Delta {V}_{1} + \Delta {V}_{2} + \space..... </math>

===A Visual Model===

[[File:loop rule.jpg]]

To figure out the sign of the voltages, act as an observer walking along the path. Start at the negative end of the emf and continue walking along the path. The emf will be positive in the loop rule because you are moving from low to high voltage. Once you reach a resistor or capacitor, this will be negative in the loop rule equation because it is high to low voltage. Continue along the path until you return to the starting position.

[[File:loop 1.jpg]]

[[File:loop 2.jpg]]

[[File:loop 3.jpg]]

==Examples==

For further instruction, check out these YouTube videos
[How to Solve a Kirchhoff's Rules Problem - Simple Example]
[Kirchoff's Loop Rule]
[https://www.youtube.com/watch?v=Z2QDXjG2ynU]
[https://www.youtube.com/watch?v=paDs-Hnmklo]

===Simple===
[[File:SimpleLoopRule.jpg]]

The circuit shown above consists of a single battery and a single resistor. The resistance of the wires is negligible for this problem.

Problem:

If the <math> emf </math> is 5 V and the resistance of the resistor is 10 ohms, what is the current passing through the resistor.

Solution:

Although we can solve this using the V = IR equation for the whole loop, lets examine this problem using the loop rule equation.

The loop rule equation would be <math> {V}_{battery} - {V}_{resistor} = 0 </math>

Since we know the <math> emf </math> of the battery we just need to find the potential difference through the resistor. For this we can use the equation of V = IR.

Thus we now have an loop rule equation of <math> emf - IR = 0 </math>
From here it is a relatively simple process to find the current. We can rewrite the loop rule equation as <math> emf = IR </math> and then plug in 5 for the emf and
10 for the resistance, leaving us with I = .5 amperes.

===Middling===

[[File:MediumExampleLoopRule.jpg]]

Problem:

The circuit shown above consists of a single battery, whose <math> emf </math> is 1.3 V, and three wires made of the same material, but having different cross-sectional areas.
Let the length of the thin wires be <math> {L}_{thick} </math> and the length of the thin wire be <math> {L}_{thin} </math>
Find a loop rule equation that starts at the negative end of the battery and goes counterclockwise through the circuit.

Solution:

When beginning this problem, you must notice that the difference in cross-sectional areas affects the electric field in each wire. Because of this
we will denote the electric field at D. as <math> {E}_{D} </math> and the electric field everywhere else as <math> {E}_{A} </math>.
To begin we will go around the circuit clockwise and add up each component. First, we know that the <math> emf </math> of the battery is 1.3 V. Then, we will add up the potential
voltage of each of the wires.

Remember that the electric potential of a wire is equal to the electric field * length of the wire. From his we can now find the potential difference
of each section of the wires. The electric potential of location A - C is <math> {E}_{A} * {L}_{thick} </math>. This is the same for the electric potential of location E - G of the wire.
For the thin section of the wire, the electric potential is <math> {E}_{D} * {L}_{thin} </math>. From here we just go around the circuit counterclockwise and add each potential difference to the loop rule equation.

Thus we can find that a loop rule equation is: <math> emf - 2 ({E}_{A} * {L}_{thick}) - {E}_{D} * {L}_{thin} = 0 </math>

This can also be rewritten as: <math> emf = 2 ({E}_{A} * {L}_{thick}) + {E}_{D} * {L}_{thin} = 0 </math>

===Difficult===

[[File:HardLoopRule.jpg]]

Problem:

For the circuit above, imagine a situation where the switch has been closed for a long time.
Calculate the current at a,b,c,d,e and charge Q of the capacitor. Answer these using <math> emf, {R}_{1}, {R}_{2},
and \space C </math>

Solution:

First, write loop rule equations for each of the possible loops in the circuit. There are 3 loop equations that are possible.

<math> emf - {I}_{1}{R}_{1} - {I}_{2}{R}_{2} = 0 </math>

<math> emf - {I}_{1}{R}_{1} - Q/C = 0 </math>

<math> {I}_{2}{R}_{2} - Q/C = 0 </math>

From here, we can then solve for the current passing through a,b,d and e. We also know that the current passing through
these points must be the same so <math> {I}_{1} = {I}_{2} </math>

<math> emf - {I}_{1}{R}_{1} - {I}_{1}{R}_{2} = 0 </math>

<math> emf = {I}_{1}({R}_{1} + {R}_{2}) = 0 </math>

<math> emf/({R}_{1} + {R}_{2}) = {I}_{1} </math>

So the current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

You must also know that once a capacitor is charging for a long time, current no longer flows through the capacitor. We can then
easily solve for c because since current is no longer flowing through the capacitor, the current at c = 0.

Lastly, we will use the loop rule equation of <math> {I}_{2}{R}_{2} - Q/C = 0 </math> to solve for Q.

<math> {I}_{2}{R}_{2} = Q/C </math>

<math> C*({I}_{2}{R}_{2}) = Q </math> Since <math> {I}_{1} = {I}_{2} </math>

<math> C*({I}_{1}{R}_{2}) = Q </math> Lastly, we will plug in what we found {I}_{1} equals from before.

<math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}) = Q </math>

Lastly, Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math> and we have now solved the problem.

Answers:

Current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

Current at c = 0

Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math>

==History==

The Loop Rule is formally known as the Kirchhoff Circuit Law, named after Gustav Kirchhoff discovered and defined this fundamental concept of electric circuits. He discovered this during his time as a student at Albertus University of Königsberg in 1845. Kirchoff went on to explore the topics of spectroscopy and black body radiation after his graduation from Albertus. Nowadays, it is used very often in electrical engineering. For more in depth information about Gustav Kirchhoff,
visit the full wiki page on him: [[Gustav Kirchhoff]]

[[File:KirchoffLoopRule.jpg|thumb|right||Gustav Kirchhoff]]

==Connectedness==
Because I am a Computer Science majors, while circuits are not directly connected to programming, computers and other electronics are made of different
circuits. Loop Rule is used in the electrical engineering that is required to create the circuits for the computers I use to program.

One interesting note about Loop Rule is that is does not apply universally to all circuits. In particular, AC (alternating current) circuits, have a fluctuating
electric charge that changes direction. This causes the electric potential of a round trip path around the circuit to no longer be zero. However, DC (direct current)
circuits still follow the loop rule.

== See also ==
Other Circuit Concepts you can check out :

[[Node Rule]]

[[Components]]

[[Resistors and Conductivity]]

[[Current]]

===More Information and External links===

[https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws Kirchoff's Circuit Laws - Wikipedia]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/circuits-and-direct-currents-20/kirchhoff-s-rules-152/the-loop-rule-540-5636/ Loop Rule - Boundless.com Textbook]

[https://www.youtube.com/watch?v=IlyUtYRqMLs Loop Rule - Doc Physics Video Lecture]

==References==

* https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff.27s_voltage_law
* Chabay, Ruth W. Matter and Interactions: Electric and Magnetic Interactions. John Wiley, 2015. Print.
* http://www.ux1.eiu.edu/~cfadd/1360/28DC/Loop.html

Claimed by Bmock7
[[Category: Simple Circuits]]

Loop Rule

2016-04-16T17:39:26Z

Ldarian3: /* Examples */

'''CLAIMED BY LAYLA DARIAN'''

The Loop Rule is a fundamental principle of electric circuits that states when you travel around an entire circuit loop, you will return to the starting voltage.
==The Energy Principle and Loop Rule==

The loop rule simply states that in any round trip path in a circuit,
[[Electric Potential]] equals zero. Keep in mind that this applies through ANY round trip path; there can be
multiple round trip paths through more complex circuits. This principle deals with the conservation of energy within a circuit. Loop Rule and [[Node Rule]] are the two
fundamental principle of electric circuits and are used to determine the behaviors of electric circuits.
This principle is often used to solve for resistance of the light bulbs or other types of resistors or the
current passing through these resistors.

===A Mathematical Model===
A mathematical representation is:
<math> \Sigma \Delta {V}_{i} = 0 </math>
AND
<math> \Delta {V}_{1} + \Delta {V}_{2} + \space.... = 0 </math> along any closed path in a circuit.

In circuits this Energy Principle is often called Kirchoff's Loop Rule and can be represented as: <math> emf = \Delta {V}_{1} + \Delta {V}_{2} + \space..... </math>

===A Visual Model===

[[File:loop rule.jpg]]

To figure out the sign of the voltages, act as an observer walking along the path. Start at the negative end of the emf and continue walking along the path. The emf will be positive in the loop rule because you are moving from low to high voltage. Once you reach a resistor or capacitor, this will be negative in the loop rule equation because it is high to low voltage. Continue along the path until you return to the starting position.

[[File:loop 1.jpg]]

[[File:loop 2.jpg]]

[[File:loop 3.jpg]]

==Examples==

For further instruction, check out these YouTube videos
[https://www.youtube.com/watch?v=Z2QDXjG2ynU]
[https://www.youtube.com/watch?v=paDs-Hnmklo]

===Simple===
[[File:SimpleLoopRule.jpg]]

The circuit shown above consists of a single battery and a single resistor. The resistance of the wires is negligible for this problem.

Problem:

If the <math> emf </math> is 5 V and the resistance of the resistor is 10 ohms, what is the current passing through the resistor.

Solution:

Although we can solve this using the V = IR equation for the whole loop, lets examine this problem using the loop rule equation.

The loop rule equation would be <math> {V}_{battery} - {V}_{resistor} = 0 </math>

Since we know the <math> emf </math> of the battery we just need to find the potential difference through the resistor. For this we can use the equation of V = IR.

Thus we now have an loop rule equation of <math> emf - IR = 0 </math>
From here it is a relatively simple process to find the current. We can rewrite the loop rule equation as <math> emf = IR </math> and then plug in 5 for the emf and
10 for the resistance, leaving us with I = .5 amperes.

===Middling===

[[File:MediumExampleLoopRule.jpg]]

Problem:

The circuit shown above consists of a single battery, whose <math> emf </math> is 1.3 V, and three wires made of the same material, but having different cross-sectional areas.
Let the length of the thin wires be <math> {L}_{thick} </math> and the length of the thin wire be <math> {L}_{thin} </math>
Find a loop rule equation that starts at the negative end of the battery and goes counterclockwise through the circuit.

Solution:

When beginning this problem, you must notice that the difference in cross-sectional areas affects the electric field in each wire. Because of this
we will denote the electric field at D. as <math> {E}_{D} </math> and the electric field everywhere else as <math> {E}_{A} </math>.
To begin we will go around the circuit clockwise and add up each component. First, we know that the <math> emf </math> of the battery is 1.3 V. Then, we will add up the potential
voltage of each of the wires.

Remember that the electric potential of a wire is equal to the electric field * length of the wire. From his we can now find the potential difference
of each section of the wires. The electric potential of location A - C is <math> {E}_{A} * {L}_{thick} </math>. This is the same for the electric potential of location E - G of the wire.
For the thin section of the wire, the electric potential is <math> {E}_{D} * {L}_{thin} </math>. From here we just go around the circuit counterclockwise and add each potential difference to the loop rule equation.

Thus we can find that a loop rule equation is: <math> emf - 2 ({E}_{A} * {L}_{thick}) - {E}_{D} * {L}_{thin} = 0 </math>

This can also be rewritten as: <math> emf = 2 ({E}_{A} * {L}_{thick}) + {E}_{D} * {L}_{thin} = 0 </math>

===Difficult===

[[File:HardLoopRule.jpg]]

Problem:

For the circuit above, imagine a situation where the switch has been closed for a long time.
Calculate the current at a,b,c,d,e and charge Q of the capacitor. Answer these using <math> emf, {R}_{1}, {R}_{2},
and \space C </math>

Solution:

First, write loop rule equations for each of the possible loops in the circuit. There are 3 loop equations that are possible.

<math> emf - {I}_{1}{R}_{1} - {I}_{2}{R}_{2} = 0 </math>

<math> emf - {I}_{1}{R}_{1} - Q/C = 0 </math>

<math> {I}_{2}{R}_{2} - Q/C = 0 </math>

From here, we can then solve for the current passing through a,b,d and e. We also know that the current passing through
these points must be the same so <math> {I}_{1} = {I}_{2} </math>

<math> emf - {I}_{1}{R}_{1} - {I}_{1}{R}_{2} = 0 </math>

<math> emf = {I}_{1}({R}_{1} + {R}_{2}) = 0 </math>

<math> emf/({R}_{1} + {R}_{2}) = {I}_{1} </math>

So the current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

You must also know that once a capacitor is charging for a long time, current no longer flows through the capacitor. We can then
easily solve for c because since current is no longer flowing through the capacitor, the current at c = 0.

Lastly, we will use the loop rule equation of <math> {I}_{2}{R}_{2} - Q/C = 0 </math> to solve for Q.

<math> {I}_{2}{R}_{2} = Q/C </math>

<math> C*({I}_{2}{R}_{2}) = Q </math> Since <math> {I}_{1} = {I}_{2} </math>

<math> C*({I}_{1}{R}_{2}) = Q </math> Lastly, we will plug in what we found {I}_{1} equals from before.

<math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}) = Q </math>

Lastly, Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math> and we have now solved the problem.

Answers:

Current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

Current at c = 0

Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math>

==History==

The Loop Rule is formally known as the Kirchhoff Circuit Law, named after Gustav Kirchhoff discovered and defined this fundamental concept of electric circuits. He discovered this during his time as a student at Albertus University of Königsberg in 1845. Kirchoff went on to explore the topics of spectroscopy and black body radiation after his graduation from Albertus. Nowadays, it is used very often in electrical engineering. For more in depth information about Gustav Kirchhoff,
visit the full wiki page on him: [[Gustav Kirchhoff]]

[[File:KirchoffLoopRule.jpg|thumb|right||Gustav Kirchhoff]]

==Connectedness==
Because I am a Computer Science majors, while circuits are not directly connected to programming, computers and other electronics are made of different
circuits. Loop Rule is used in the electrical engineering that is required to create the circuits for the computers I use to program.

One interesting note about Loop Rule is that is does not apply universally to all circuits. In particular, AC (alternating current) circuits, have a fluctuating
electric charge that changes direction. This causes the electric potential of a round trip path around the circuit to no longer be zero. However, DC (direct current)
circuits still follow the loop rule.

== See also ==
Other Circuit Concepts you can check out :

[[Node Rule]]

[[Components]]

[[Resistors and Conductivity]]

[[Current]]

===More Information and External links===

[https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws Kirchoff's Circuit Laws - Wikipedia]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/circuits-and-direct-currents-20/kirchhoff-s-rules-152/the-loop-rule-540-5636/ Loop Rule - Boundless.com Textbook]

[https://www.youtube.com/watch?v=IlyUtYRqMLs Loop Rule - Doc Physics Video Lecture]

==References==

* https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff.27s_voltage_law
* Chabay, Ruth W. Matter and Interactions: Electric and Magnetic Interactions. John Wiley, 2015. Print.
* http://www.ux1.eiu.edu/~cfadd/1360/28DC/Loop.html

Claimed by Bmock7
[[Category: Simple Circuits]]

File:Loop 3.jpg

2016-04-16T17:33:02Z

Ldarian3:

File:Loop 2.jpg

2016-04-16T17:31:48Z

Ldarian3:

File:Loop 1.jpg

2016-04-16T17:29:20Z

Ldarian3:

Loop Rule

2016-04-16T17:28:56Z

Ldarian3: /* A Visual Model */

'''CLAIMED BY LAYLA DARIAN'''

The Loop Rule is a fundamental principle of electric circuits that states when you travel around an entire circuit loop, you will return to the starting voltage.
==The Energy Principle and Loop Rule==

The loop rule simply states that in any round trip path in a circuit,
[[Electric Potential]] equals zero. Keep in mind that this applies through ANY round trip path; there can be
multiple round trip paths through more complex circuits. This principle deals with the conservation of energy within a circuit. Loop Rule and [[Node Rule]] are the two
fundamental principle of electric circuits and are used to determine the behaviors of electric circuits.
This principle is often used to solve for resistance of the light bulbs or other types of resistors or the
current passing through these resistors.

===A Mathematical Model===
A mathematical representation is:
<math> \Sigma \Delta {V}_{i} = 0 </math>
AND
<math> \Delta {V}_{1} + \Delta {V}_{2} + \space.... = 0 </math> along any closed path in a circuit.

In circuits this Energy Principle is often called Kirchoff's Loop Rule and can be represented as: <math> emf = \Delta {V}_{1} + \Delta {V}_{2} + \space..... </math>

===A Visual Model===

[[File:loop rule.jpg]]

To figure out the sign of the voltages, act as an observer walking along the path. Start at the negative end of the emf and continue walking along the path. The emf will be positive in the loop rule because you are moving from low to high voltage. Once you reach a resistor or capacitor, this will be negative in the loop rule equation because it is high to low voltage. Continue along the path until you return to the starting position.

[[File:loop 1.jpg]]

[[File:loop 2.jpg]]

[[File:loop 3.jpg]]

==Examples==

===Simple===
[[File:SimpleLoopRule.jpg]]

The circuit shown above consists of a single battery and a single resistor. The resistance of the wires is negligible for this problem.

Problem:

If the <math> emf </math> is 5 V and the resistance of the resistor is 10 ohms, what is the current passing through the resistor.

Solution:

Although we can solve this using the V = IR equation for the whole loop, lets examine this problem using the loop rule equation.

The loop rule equation would be <math> {V}_{battery} - {V}_{resistor} = 0 </math>

Since we know the <math> emf </math> of the battery we just need to find the potential difference through the resistor. For this we can use the equation of V = IR.

Thus we now have an loop rule equation of <math> emf - IR = 0 </math>
From here it is a relatively simple process to find the current. We can rewrite the loop rule equation as <math> emf = IR </math> and then plug in 5 for the emf and
10 for the resistance, leaving us with I = .5 amperes.

===Middling===

[[File:MediumExampleLoopRule.jpg]]

Problem:

The circuit shown above consists of a single battery, whose <math> emf </math> is 1.3 V, and three wires made of the same material, but having different cross-sectional areas.
Let the length of the thin wires be <math> {L}_{thick} </math> and the length of the thin wire be <math> {L}_{thin} </math>
Find a loop rule equation that starts at the negative end of the battery and goes counterclockwise through the circuit.

Solution:

When beginning this problem, you must notice that the difference in cross-sectional areas affects the electric field in each wire. Because of this
we will denote the electric field at D. as <math> {E}_{D} </math> and the electric field everywhere else as <math> {E}_{A} </math>.
To begin we will go around the circuit clockwise and add up each component. First, we know that the <math> emf </math> of the battery is 1.3 V. Then, we will add up the potential
voltage of each of the wires.

Remember that the electric potential of a wire is equal to the electric field * length of the wire. From his we can now find the potential difference
of each section of the wires. The electric potential of location A - C is <math> {E}_{A} * {L}_{thick} </math>. This is the same for the electric potential of location E - G of the wire.
For the thin section of the wire, the electric potential is <math> {E}_{D} * {L}_{thin} </math>. From here we just go around the circuit counterclockwise and add each potential difference to the loop rule equation.

Thus we can find that a loop rule equation is: <math> emf - 2 ({E}_{A} * {L}_{thick}) - {E}_{D} * {L}_{thin} = 0 </math>

This can also be rewritten as: <math> emf = 2 ({E}_{A} * {L}_{thick}) + {E}_{D} * {L}_{thin} = 0 </math>

===Difficult===

[[File:HardLoopRule.jpg]]

Problem:

For the circuit above, imagine a situation where the switch has been closed for a long time.
Calculate the current at a,b,c,d,e and charge Q of the capacitor. Answer these using <math> emf, {R}_{1}, {R}_{2},
and \space C </math>

Solution:

First, write loop rule equations for each of the possible loops in the circuit. There are 3 loop equations that are possible.

<math> emf - {I}_{1}{R}_{1} - {I}_{2}{R}_{2} = 0 </math>

<math> emf - {I}_{1}{R}_{1} - Q/C = 0 </math>

<math> {I}_{2}{R}_{2} - Q/C = 0 </math>

From here, we can then solve for the current passing through a,b,d and e. We also know that the current passing through
these points must be the same so <math> {I}_{1} = {I}_{2} </math>

<math> emf - {I}_{1}{R}_{1} - {I}_{1}{R}_{2} = 0 </math>

<math> emf = {I}_{1}({R}_{1} + {R}_{2}) = 0 </math>

<math> emf/({R}_{1} + {R}_{2}) = {I}_{1} </math>

So the current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

You must also know that once a capacitor is charging for a long time, current no longer flows through the capacitor. We can then
easily solve for c because since current is no longer flowing through the capacitor, the current at c = 0.

Lastly, we will use the loop rule equation of <math> {I}_{2}{R}_{2} - Q/C = 0 </math> to solve for Q.

<math> {I}_{2}{R}_{2} = Q/C </math>

<math> C*({I}_{2}{R}_{2}) = Q </math> Since <math> {I}_{1} = {I}_{2} </math>

<math> C*({I}_{1}{R}_{2}) = Q </math> Lastly, we will plug in what we found {I}_{1} equals from before.

<math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}) = Q </math>

Lastly, Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math> and we have now solved the problem.

Answers:

Current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

Current at c = 0

Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math>

==History==

The Loop Rule is formally known as the Kirchhoff Circuit Law, named after Gustav Kirchhoff discovered and defined this fundamental concept of electric circuits. He discovered this during his time as a student at Albertus University of Königsberg in 1845. Kirchoff went on to explore the topics of spectroscopy and black body radiation after his graduation from Albertus. Nowadays, it is used very often in electrical engineering. For more in depth information about Gustav Kirchhoff,
visit the full wiki page on him: [[Gustav Kirchhoff]]

[[File:KirchoffLoopRule.jpg|thumb|right||Gustav Kirchhoff]]

==Connectedness==
Because I am a Computer Science majors, while circuits are not directly connected to programming, computers and other electronics are made of different
circuits. Loop Rule is used in the electrical engineering that is required to create the circuits for the computers I use to program.

One interesting note about Loop Rule is that is does not apply universally to all circuits. In particular, AC (alternating current) circuits, have a fluctuating
electric charge that changes direction. This causes the electric potential of a round trip path around the circuit to no longer be zero. However, DC (direct current)
circuits still follow the loop rule.

== See also ==
Other Circuit Concepts you can check out :

[[Node Rule]]

[[Components]]

[[Resistors and Conductivity]]

[[Current]]

===More Information and External links===

[https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws Kirchoff's Circuit Laws - Wikipedia]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/circuits-and-direct-currents-20/kirchhoff-s-rules-152/the-loop-rule-540-5636/ Loop Rule - Boundless.com Textbook]

[https://www.youtube.com/watch?v=IlyUtYRqMLs Loop Rule - Doc Physics Video Lecture]

==References==

* https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff.27s_voltage_law
* Chabay, Ruth W. Matter and Interactions: Electric and Magnetic Interactions. John Wiley, 2015. Print.
* http://www.ux1.eiu.edu/~cfadd/1360/28DC/Loop.html

Claimed by Bmock7
[[Category: Simple Circuits]]

File:Loop rule.jpg

2016-04-16T16:41:09Z

Ldarian3:

Loop Rule

2016-04-16T16:36:37Z

Ldarian3: /* A Visual Model */

'''CLAIMED BY LAYLA DARIAN'''

The Loop Rule is a fundamental principle of electric circuits that states when you travel around an entire circuit loop, you will return to the starting voltage.
==The Energy Principle and Loop Rule==

The loop rule simply states that in any round trip path in a circuit,
[[Electric Potential]] equals zero. Keep in mind that this applies through ANY round trip path; there can be
multiple round trip paths through more complex circuits. This principle deals with the conservation of energy within a circuit. Loop Rule and [[Node Rule]] are the two
fundamental principle of electric circuits and are used to determine the behaviors of electric circuits.
This principle is often used to solve for resistance of the light bulbs or other types of resistors or the
current passing through these resistors.

===A Mathematical Model===
A mathematical representation is:
<math> \Sigma \Delta {V}_{i} = 0 </math>
AND
<math> \Delta {V}_{1} + \Delta {V}_{2} + \space.... = 0 </math> along any closed path in a circuit.

In circuits this Energy Principle is often called Kirchoff's Loop Rule and can be represented as: <math> emf = \Delta {V}_{1} + \Delta {V}_{2} + \space..... </math>

===A Visual Model===

[[File:loop rule.jpg]]

To figure out the sign of the voltages, act as an observer walking along the path

==Examples==

===Simple===
[[File:SimpleLoopRule.jpg]]

The circuit shown above consists of a single battery and a single resistor. The resistance of the wires is negligible for this problem.

Problem:

If the <math> emf </math> is 5 V and the resistance of the resistor is 10 ohms, what is the current passing through the resistor.

Solution:

Although we can solve this using the V = IR equation for the whole loop, lets examine this problem using the loop rule equation.

The loop rule equation would be <math> {V}_{battery} - {V}_{resistor} = 0 </math>

Since we know the <math> emf </math> of the battery we just need to find the potential difference through the resistor. For this we can use the equation of V = IR.

Thus we now have an loop rule equation of <math> emf - IR = 0 </math>
From here it is a relatively simple process to find the current. We can rewrite the loop rule equation as <math> emf = IR </math> and then plug in 5 for the emf and
10 for the resistance, leaving us with I = .5 amperes.

===Middling===

[[File:MediumExampleLoopRule.jpg]]

Problem:

The circuit shown above consists of a single battery, whose <math> emf </math> is 1.3 V, and three wires made of the same material, but having different cross-sectional areas.
Let the length of the thin wires be <math> {L}_{thick} </math> and the length of the thin wire be <math> {L}_{thin} </math>
Find a loop rule equation that starts at the negative end of the battery and goes counterclockwise through the circuit.

Solution:

When beginning this problem, you must notice that the difference in cross-sectional areas affects the electric field in each wire. Because of this
we will denote the electric field at D. as <math> {E}_{D} </math> and the electric field everywhere else as <math> {E}_{A} </math>.
To begin we will go around the circuit clockwise and add up each component. First, we know that the <math> emf </math> of the battery is 1.3 V. Then, we will add up the potential
voltage of each of the wires.

Remember that the electric potential of a wire is equal to the electric field * length of the wire. From his we can now find the potential difference
of each section of the wires. The electric potential of location A - C is <math> {E}_{A} * {L}_{thick} </math>. This is the same for the electric potential of location E - G of the wire.
For the thin section of the wire, the electric potential is <math> {E}_{D} * {L}_{thin} </math>. From here we just go around the circuit counterclockwise and add each potential difference to the loop rule equation.

Thus we can find that a loop rule equation is: <math> emf - 2 ({E}_{A} * {L}_{thick}) - {E}_{D} * {L}_{thin} = 0 </math>

This can also be rewritten as: <math> emf = 2 ({E}_{A} * {L}_{thick}) + {E}_{D} * {L}_{thin} = 0 </math>

===Difficult===

[[File:HardLoopRule.jpg]]

Problem:

For the circuit above, imagine a situation where the switch has been closed for a long time.
Calculate the current at a,b,c,d,e and charge Q of the capacitor. Answer these using <math> emf, {R}_{1}, {R}_{2},
and \space C </math>

Solution:

First, write loop rule equations for each of the possible loops in the circuit. There are 3 loop equations that are possible.

<math> emf - {I}_{1}{R}_{1} - {I}_{2}{R}_{2} = 0 </math>

<math> emf - {I}_{1}{R}_{1} - Q/C = 0 </math>

<math> {I}_{2}{R}_{2} - Q/C = 0 </math>

From here, we can then solve for the current passing through a,b,d and e. We also know that the current passing through
these points must be the same so <math> {I}_{1} = {I}_{2} </math>

<math> emf - {I}_{1}{R}_{1} - {I}_{1}{R}_{2} = 0 </math>

<math> emf = {I}_{1}({R}_{1} + {R}_{2}) = 0 </math>

<math> emf/({R}_{1} + {R}_{2}) = {I}_{1} </math>

So the current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

You must also know that once a capacitor is charging for a long time, current no longer flows through the capacitor. We can then
easily solve for c because since current is no longer flowing through the capacitor, the current at c = 0.

Lastly, we will use the loop rule equation of <math> {I}_{2}{R}_{2} - Q/C = 0 </math> to solve for Q.

<math> {I}_{2}{R}_{2} = Q/C </math>

<math> C*({I}_{2}{R}_{2}) = Q </math> Since <math> {I}_{1} = {I}_{2} </math>

<math> C*({I}_{1}{R}_{2}) = Q </math> Lastly, we will plug in what we found {I}_{1} equals from before.

<math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}) = Q </math>

Lastly, Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math> and we have now solved the problem.

Answers:

Current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

Current at c = 0

Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math>

==History==

The Loop Rule is formally known as the Kirchhoff Circuit Law, named after Gustav Kirchhoff discovered and defined this fundamental concept of electric circuits. He discovered this during his time as a student at Albertus University of Königsberg in 1845. Kirchoff went on to explore the topics of spectroscopy and black body radiation after his graduation from Albertus. Nowadays, it is used very often in electrical engineering. For more in depth information about Gustav Kirchhoff,
visit the full wiki page on him: [[Gustav Kirchhoff]]

[[File:KirchoffLoopRule.jpg|thumb|right||Gustav Kirchhoff]]

==Connectedness==
Because I am a Computer Science majors, while circuits are not directly connected to programming, computers and other electronics are made of different
circuits. Loop Rule is used in the electrical engineering that is required to create the circuits for the computers I use to program.

One interesting note about Loop Rule is that is does not apply universally to all circuits. In particular, AC (alternating current) circuits, have a fluctuating
electric charge that changes direction. This causes the electric potential of a round trip path around the circuit to no longer be zero. However, DC (direct current)
circuits still follow the loop rule.

== See also ==
Other Circuit Concepts you can check out :

[[Node Rule]]

[[Components]]

[[Resistors and Conductivity]]

[[Current]]

===More Information and External links===

[https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws Kirchoff's Circuit Laws - Wikipedia]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/circuits-and-direct-currents-20/kirchhoff-s-rules-152/the-loop-rule-540-5636/ Loop Rule - Boundless.com Textbook]

[https://www.youtube.com/watch?v=IlyUtYRqMLs Loop Rule - Doc Physics Video Lecture]

==References==

* https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff.27s_voltage_law
* Chabay, Ruth W. Matter and Interactions: Electric and Magnetic Interactions. John Wiley, 2015. Print.
* http://www.ux1.eiu.edu/~cfadd/1360/28DC/Loop.html

Claimed by Bmock7
[[Category: Simple Circuits]]

Loop Rule

2016-04-16T16:08:55Z

Ldarian3:

'''CLAIMED BY LAYLA DARIAN'''

The Loop Rule is a fundamental principle of electric circuits that states when you travel around an entire circuit loop, you will return to the starting voltage.
==The Energy Principle and Loop Rule==

The loop rule simply states that in any round trip path in a circuit,
[[Electric Potential]] equals zero. Keep in mind that this applies through ANY round trip path; there can be
multiple round trip paths through more complex circuits. This principle deals with the conservation of energy within a circuit. Loop Rule and [[Node Rule]] are the two
fundamental principle of electric circuits and are used to determine the behaviors of electric circuits.
This principle is often used to solve for resistance of the light bulbs or other types of resistors or the
current passing through these resistors.

===A Mathematical Model===
A mathematical representation is:
<math> \Sigma \Delta {V}_{i} = 0 </math>
AND
<math> \Delta {V}_{1} + \Delta {V}_{2} + \space.... = 0 </math> along any closed path in a circuit.

In circuits this Energy Principle is often called Kirchoff's Loop Rule and can be represented as: <math> emf = \Delta {V}_{1} + \Delta {V}_{2} + \space..... </math>

===A Visual Model===

[[File:LoopRule.jpg]]

The total voltage in the circuit is equal to all the individual voltages in the circuit added together.

Another way to think about it is <math> \Delta {V} - (\Delta {V}_{1} + \Delta {V}_{2} + \Delta {V}_{3})= 0 </math>

==Examples==

===Simple===
[[File:SimpleLoopRule.jpg]]

The circuit shown above consists of a single battery and a single resistor. The resistance of the wires is negligible for this problem.

Problem:

If the <math> emf </math> is 5 V and the resistance of the resistor is 10 ohms, what is the current passing through the resistor.

Solution:

Although we can solve this using the V = IR equation for the whole loop, lets examine this problem using the loop rule equation.

The loop rule equation would be <math> {V}_{battery} - {V}_{resistor} = 0 </math>

Since we know the <math> emf </math> of the battery we just need to find the potential difference through the resistor. For this we can use the equation of V = IR.

Thus we now have an loop rule equation of <math> emf - IR = 0 </math>
From here it is a relatively simple process to find the current. We can rewrite the loop rule equation as <math> emf = IR </math> and then plug in 5 for the emf and
10 for the resistance, leaving us with I = .5 amperes.

===Middling===

[[File:MediumExampleLoopRule.jpg]]

Problem:

The circuit shown above consists of a single battery, whose <math> emf </math> is 1.3 V, and three wires made of the same material, but having different cross-sectional areas.
Let the length of the thin wires be <math> {L}_{thick} </math> and the length of the thin wire be <math> {L}_{thin} </math>
Find a loop rule equation that starts at the negative end of the battery and goes counterclockwise through the circuit.

Solution:

When beginning this problem, you must notice that the difference in cross-sectional areas affects the electric field in each wire. Because of this
we will denote the electric field at D. as <math> {E}_{D} </math> and the electric field everywhere else as <math> {E}_{A} </math>.
To begin we will go around the circuit clockwise and add up each component. First, we know that the <math> emf </math> of the battery is 1.3 V. Then, we will add up the potential
voltage of each of the wires.

Remember that the electric potential of a wire is equal to the electric field * length of the wire. From his we can now find the potential difference
of each section of the wires. The electric potential of location A - C is <math> {E}_{A} * {L}_{thick} </math>. This is the same for the electric potential of location E - G of the wire.
For the thin section of the wire, the electric potential is <math> {E}_{D} * {L}_{thin} </math>. From here we just go around the circuit counterclockwise and add each potential difference to the loop rule equation.

Thus we can find that a loop rule equation is: <math> emf - 2 ({E}_{A} * {L}_{thick}) - {E}_{D} * {L}_{thin} = 0 </math>

This can also be rewritten as: <math> emf = 2 ({E}_{A} * {L}_{thick}) + {E}_{D} * {L}_{thin} = 0 </math>

===Difficult===

[[File:HardLoopRule.jpg]]

Problem:

For the circuit above, imagine a situation where the switch has been closed for a long time.
Calculate the current at a,b,c,d,e and charge Q of the capacitor. Answer these using <math> emf, {R}_{1}, {R}_{2},
and \space C </math>

Solution:

First, write loop rule equations for each of the possible loops in the circuit. There are 3 loop equations that are possible.

<math> emf - {I}_{1}{R}_{1} - {I}_{2}{R}_{2} = 0 </math>

<math> emf - {I}_{1}{R}_{1} - Q/C = 0 </math>

<math> {I}_{2}{R}_{2} - Q/C = 0 </math>

From here, we can then solve for the current passing through a,b,d and e. We also know that the current passing through
these points must be the same so <math> {I}_{1} = {I}_{2} </math>

<math> emf - {I}_{1}{R}_{1} - {I}_{1}{R}_{2} = 0 </math>

<math> emf = {I}_{1}({R}_{1} + {R}_{2}) = 0 </math>

<math> emf/({R}_{1} + {R}_{2}) = {I}_{1} </math>

So the current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

You must also know that once a capacitor is charging for a long time, current no longer flows through the capacitor. We can then
easily solve for c because since current is no longer flowing through the capacitor, the current at c = 0.

Lastly, we will use the loop rule equation of <math> {I}_{2}{R}_{2} - Q/C = 0 </math> to solve for Q.

<math> {I}_{2}{R}_{2} = Q/C </math>

<math> C*({I}_{2}{R}_{2}) = Q </math> Since <math> {I}_{1} = {I}_{2} </math>

<math> C*({I}_{1}{R}_{2}) = Q </math> Lastly, we will plug in what we found {I}_{1} equals from before.

<math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}) = Q </math>

Lastly, Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math> and we have now solved the problem.

Answers:

Current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

Current at c = 0

Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math>

==History==

The Loop Rule is formally known as the Kirchhoff Circuit Law, named after Gustav Kirchhoff discovered and defined this fundamental concept of electric circuits. He discovered this during his time as a student at Albertus University of Königsberg in 1845. Kirchoff went on to explore the topics of spectroscopy and black body radiation after his graduation from Albertus. Nowadays, it is used very often in electrical engineering. For more in depth information about Gustav Kirchhoff,
visit the full wiki page on him: [[Gustav Kirchhoff]]

[[File:KirchoffLoopRule.jpg|thumb|right||Gustav Kirchhoff]]

==Connectedness==
Because I am a Computer Science majors, while circuits are not directly connected to programming, computers and other electronics are made of different
circuits. Loop Rule is used in the electrical engineering that is required to create the circuits for the computers I use to program.

One interesting note about Loop Rule is that is does not apply universally to all circuits. In particular, AC (alternating current) circuits, have a fluctuating
electric charge that changes direction. This causes the electric potential of a round trip path around the circuit to no longer be zero. However, DC (direct current)
circuits still follow the loop rule.

== See also ==
Other Circuit Concepts you can check out :

[[Node Rule]]

[[Components]]

[[Resistors and Conductivity]]

[[Current]]

===More Information and External links===

[https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws Kirchoff's Circuit Laws - Wikipedia]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/circuits-and-direct-currents-20/kirchhoff-s-rules-152/the-loop-rule-540-5636/ Loop Rule - Boundless.com Textbook]

[https://www.youtube.com/watch?v=IlyUtYRqMLs Loop Rule - Doc Physics Video Lecture]

==References==

* https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff.27s_voltage_law
* Chabay, Ruth W. Matter and Interactions: Electric and Magnetic Interactions. John Wiley, 2015. Print.
* http://www.ux1.eiu.edu/~cfadd/1360/28DC/Loop.html

Claimed by Bmock7
[[Category: Simple Circuits]]

Loop Rule

2016-04-16T16:06:33Z

Ldarian3: /* A Mathematical Model */

'''CLAIMED BY LAYLA DARIAN'''

The Loop Rule is a fundamental principle of electric circuits that states when you travel around an entire circuit loop, you will return to the starting voltage.
==Loop Rule==

The loop rule simply states that in any round trip path in a circuit,
[[Electric Potential]] equals zero. Keep in mind that this applies through ANY round trip path; there can be
multiple round trip paths through more complex circuits. This principle deals with the conservation of energy within a circuit. Loop Rule and [[Node Rule]] are the two
fundamental principle of electric circuits and are used to determine the behaviors of electric circuits.
This principle is often used to solve for resistance of the light bulbs or other types of resistors or the
current passing through these resistors.

===A Mathematical Model===
A mathematical representation is:
<math> \Sigma \Delta {V}_{i} = 0 </math>
AND
<math> \Delta {V}_{1} + \Delta {V}_{2} + \space.... = 0 </math> along any closed path in a circuit.

In circuits this Energy Principle is often called Kirchoff's Loop Rule and can be represented as: <math> emf = \Delta {V}_{1} + \Delta {V}_{2} + \space..... </math>

===A Visual Model===

[[File:LoopRule.jpg]]

The total voltage in the circuit is equal to all the individual voltages in the circuit added together.

Another way to think about it is <math> \Delta {V} - (\Delta {V}_{1} + \Delta {V}_{2} + \Delta {V}_{3})= 0 </math>

==Examples==

===Simple===
[[File:SimpleLoopRule.jpg]]

The circuit shown above consists of a single battery and a single resistor. The resistance of the wires is negligible for this problem.

Problem:

If the <math> emf </math> is 5 V and the resistance of the resistor is 10 ohms, what is the current passing through the resistor.

Solution:

Although we can solve this using the V = IR equation for the whole loop, lets examine this problem using the loop rule equation.

The loop rule equation would be <math> {V}_{battery} - {V}_{resistor} = 0 </math>

Since we know the <math> emf </math> of the battery we just need to find the potential difference through the resistor. For this we can use the equation of V = IR.

Thus we now have an loop rule equation of <math> emf - IR = 0 </math>
From here it is a relatively simple process to find the current. We can rewrite the loop rule equation as <math> emf = IR </math> and then plug in 5 for the emf and
10 for the resistance, leaving us with I = .5 amperes.

===Middling===

[[File:MediumExampleLoopRule.jpg]]

Problem:

The circuit shown above consists of a single battery, whose <math> emf </math> is 1.3 V, and three wires made of the same material, but having different cross-sectional areas.
Let the length of the thin wires be <math> {L}_{thick} </math> and the length of the thin wire be <math> {L}_{thin} </math>
Find a loop rule equation that starts at the negative end of the battery and goes counterclockwise through the circuit.

Solution:

When beginning this problem, you must notice that the difference in cross-sectional areas affects the electric field in each wire. Because of this
we will denote the electric field at D. as <math> {E}_{D} </math> and the electric field everywhere else as <math> {E}_{A} </math>.
To begin we will go around the circuit clockwise and add up each component. First, we know that the <math> emf </math> of the battery is 1.3 V. Then, we will add up the potential
voltage of each of the wires.

Remember that the electric potential of a wire is equal to the electric field * length of the wire. From his we can now find the potential difference
of each section of the wires. The electric potential of location A - C is <math> {E}_{A} * {L}_{thick} </math>. This is the same for the electric potential of location E - G of the wire.
For the thin section of the wire, the electric potential is <math> {E}_{D} * {L}_{thin} </math>. From here we just go around the circuit counterclockwise and add each potential difference to the loop rule equation.

Thus we can find that a loop rule equation is: <math> emf - 2 ({E}_{A} * {L}_{thick}) - {E}_{D} * {L}_{thin} = 0 </math>

This can also be rewritten as: <math> emf = 2 ({E}_{A} * {L}_{thick}) + {E}_{D} * {L}_{thin} = 0 </math>

===Difficult===

[[File:HardLoopRule.jpg]]

Problem:

For the circuit above, imagine a situation where the switch has been closed for a long time.
Calculate the current at a,b,c,d,e and charge Q of the capacitor. Answer these using <math> emf, {R}_{1}, {R}_{2},
and \space C </math>

Solution:

First, write loop rule equations for each of the possible loops in the circuit. There are 3 loop equations that are possible.

<math> emf - {I}_{1}{R}_{1} - {I}_{2}{R}_{2} = 0 </math>

<math> emf - {I}_{1}{R}_{1} - Q/C = 0 </math>

<math> {I}_{2}{R}_{2} - Q/C = 0 </math>

From here, we can then solve for the current passing through a,b,d and e. We also know that the current passing through
these points must be the same so <math> {I}_{1} = {I}_{2} </math>

<math> emf - {I}_{1}{R}_{1} - {I}_{1}{R}_{2} = 0 </math>

<math> emf = {I}_{1}({R}_{1} + {R}_{2}) = 0 </math>

<math> emf/({R}_{1} + {R}_{2}) = {I}_{1} </math>

So the current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

You must also know that once a capacitor is charging for a long time, current no longer flows through the capacitor. We can then
easily solve for c because since current is no longer flowing through the capacitor, the current at c = 0.

Lastly, we will use the loop rule equation of <math> {I}_{2}{R}_{2} - Q/C = 0 </math> to solve for Q.

<math> {I}_{2}{R}_{2} = Q/C </math>

<math> C*({I}_{2}{R}_{2}) = Q </math> Since <math> {I}_{1} = {I}_{2} </math>

<math> C*({I}_{1}{R}_{2}) = Q </math> Lastly, we will plug in what we found {I}_{1} equals from before.

<math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}) = Q </math>

Lastly, Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math> and we have now solved the problem.

Answers:

Current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

Current at c = 0

Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math>

==History==

The Loop Rule is formally known as the Kirchhoff Circuit Law, named after Gustav Kirchhoff discovered and defined this fundamental concept of electric circuits. He discovered this during his time as a student at Albertus University of Königsberg in 1845. Kirchoff went on to explore the topics of spectroscopy and black body radiation after his graduation from Albertus. Nowadays, it is used very often in electrical engineering. For more in depth information about Gustav Kirchhoff,
visit the full wiki page on him: [[Gustav Kirchhoff]]

[[File:KirchoffLoopRule.jpg|thumb|right||Gustav Kirchhoff]]

==Connectedness==
Because I am a Computer Science majors, while circuits are not directly connected to programming, computers and other electronics are made of different
circuits. Loop Rule is used in the electrical engineering that is required to create the circuits for the computers I use to program.

One interesting note about Loop Rule is that is does not apply universally to all circuits. In particular, AC (alternating current) circuits, have a fluctuating
electric charge that changes direction. This causes the electric potential of a round trip path around the circuit to no longer be zero. However, DC (direct current)
circuits still follow the loop rule.

== See also ==
Other Circuit Concepts you can check out :

[[Node Rule]]

[[Components]]

[[Resistors and Conductivity]]

[[Current]]

===More Information and External links===

[https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws Kirchoff's Circuit Laws - Wikipedia]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/circuits-and-direct-currents-20/kirchhoff-s-rules-152/the-loop-rule-540-5636/ Loop Rule - Boundless.com Textbook]

[https://www.youtube.com/watch?v=IlyUtYRqMLs Loop Rule - Doc Physics Video Lecture]

==References==

* https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff.27s_voltage_law
* Chabay, Ruth W. Matter and Interactions: Electric and Magnetic Interactions. John Wiley, 2015. Print.
* http://www.ux1.eiu.edu/~cfadd/1360/28DC/Loop.html

Claimed by Bmock7
[[Category: Simple Circuits]]

Loop Rule

2016-04-16T16:05:25Z

Ldarian3: /* A Mathematical Model */

'''CLAIMED BY LAYLA DARIAN'''

The Loop Rule is a fundamental principle of electric circuits that states when you travel around an entire circuit loop, you will return to the starting voltage.
==Loop Rule==

The loop rule simply states that in any round trip path in a circuit,
[[Electric Potential]] equals zero. Keep in mind that this applies through ANY round trip path; there can be
multiple round trip paths through more complex circuits. This principle deals with the conservation of energy within a circuit. Loop Rule and [[Node Rule]] are the two
fundamental principle of electric circuits and are used to determine the behaviors of electric circuits.
This principle is often used to solve for resistance of the light bulbs or other types of resistors or the
current passing through these resistors.

===A Mathematical Model===
A mathematical representation is:
<math> \Sigma \Delta {V}_{i} = 0 </math>
AND
<math> \Delta {V}_{1} + \Delta {V}_{2} + \space.... = 0 </math> along any closed path in a circuit.

This can also be represented in a circuit as <math> emf = \Delta {V}_{1} + \Delta {V}_{2} + \space..... </math>

===A Visual Model===

[[File:LoopRule.jpg]]

The total voltage in the circuit is equal to all the individual voltages in the circuit added together.

Another way to think about it is <math> \Delta {V} - (\Delta {V}_{1} + \Delta {V}_{2} + \Delta {V}_{3})= 0 </math>

==Examples==

===Simple===
[[File:SimpleLoopRule.jpg]]

The circuit shown above consists of a single battery and a single resistor. The resistance of the wires is negligible for this problem.

Problem:

If the <math> emf </math> is 5 V and the resistance of the resistor is 10 ohms, what is the current passing through the resistor.

Solution:

Although we can solve this using the V = IR equation for the whole loop, lets examine this problem using the loop rule equation.

The loop rule equation would be <math> {V}_{battery} - {V}_{resistor} = 0 </math>

Since we know the <math> emf </math> of the battery we just need to find the potential difference through the resistor. For this we can use the equation of V = IR.

Thus we now have an loop rule equation of <math> emf - IR = 0 </math>
From here it is a relatively simple process to find the current. We can rewrite the loop rule equation as <math> emf = IR </math> and then plug in 5 for the emf and
10 for the resistance, leaving us with I = .5 amperes.

===Middling===

[[File:MediumExampleLoopRule.jpg]]

Problem:

The circuit shown above consists of a single battery, whose <math> emf </math> is 1.3 V, and three wires made of the same material, but having different cross-sectional areas.
Let the length of the thin wires be <math> {L}_{thick} </math> and the length of the thin wire be <math> {L}_{thin} </math>
Find a loop rule equation that starts at the negative end of the battery and goes counterclockwise through the circuit.

Solution:

When beginning this problem, you must notice that the difference in cross-sectional areas affects the electric field in each wire. Because of this
we will denote the electric field at D. as <math> {E}_{D} </math> and the electric field everywhere else as <math> {E}_{A} </math>.
To begin we will go around the circuit clockwise and add up each component. First, we know that the <math> emf </math> of the battery is 1.3 V. Then, we will add up the potential
voltage of each of the wires.

Remember that the electric potential of a wire is equal to the electric field * length of the wire. From his we can now find the potential difference
of each section of the wires. The electric potential of location A - C is <math> {E}_{A} * {L}_{thick} </math>. This is the same for the electric potential of location E - G of the wire.
For the thin section of the wire, the electric potential is <math> {E}_{D} * {L}_{thin} </math>. From here we just go around the circuit counterclockwise and add each potential difference to the loop rule equation.

Thus we can find that a loop rule equation is: <math> emf - 2 ({E}_{A} * {L}_{thick}) - {E}_{D} * {L}_{thin} = 0 </math>

This can also be rewritten as: <math> emf = 2 ({E}_{A} * {L}_{thick}) + {E}_{D} * {L}_{thin} = 0 </math>

===Difficult===

[[File:HardLoopRule.jpg]]

Problem:

For the circuit above, imagine a situation where the switch has been closed for a long time.
Calculate the current at a,b,c,d,e and charge Q of the capacitor. Answer these using <math> emf, {R}_{1}, {R}_{2},
and \space C </math>

Solution:

First, write loop rule equations for each of the possible loops in the circuit. There are 3 loop equations that are possible.

<math> emf - {I}_{1}{R}_{1} - {I}_{2}{R}_{2} = 0 </math>

<math> emf - {I}_{1}{R}_{1} - Q/C = 0 </math>

<math> {I}_{2}{R}_{2} - Q/C = 0 </math>

From here, we can then solve for the current passing through a,b,d and e. We also know that the current passing through
these points must be the same so <math> {I}_{1} = {I}_{2} </math>

<math> emf - {I}_{1}{R}_{1} - {I}_{1}{R}_{2} = 0 </math>

<math> emf = {I}_{1}({R}_{1} + {R}_{2}) = 0 </math>

<math> emf/({R}_{1} + {R}_{2}) = {I}_{1} </math>

So the current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

You must also know that once a capacitor is charging for a long time, current no longer flows through the capacitor. We can then
easily solve for c because since current is no longer flowing through the capacitor, the current at c = 0.

Lastly, we will use the loop rule equation of <math> {I}_{2}{R}_{2} - Q/C = 0 </math> to solve for Q.

<math> {I}_{2}{R}_{2} = Q/C </math>

<math> C*({I}_{2}{R}_{2}) = Q </math> Since <math> {I}_{1} = {I}_{2} </math>

<math> C*({I}_{1}{R}_{2}) = Q </math> Lastly, we will plug in what we found {I}_{1} equals from before.

<math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}) = Q </math>

Lastly, Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math> and we have now solved the problem.

Answers:

Current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

Current at c = 0

Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math>

==History==

The Loop Rule is formally known as the Kirchhoff Circuit Law, named after Gustav Kirchhoff discovered and defined this fundamental concept of electric circuits. He discovered this during his time as a student at Albertus University of Königsberg in 1845. Kirchoff went on to explore the topics of spectroscopy and black body radiation after his graduation from Albertus. Nowadays, it is used very often in electrical engineering. For more in depth information about Gustav Kirchhoff,
visit the full wiki page on him: [[Gustav Kirchhoff]]

[[File:KirchoffLoopRule.jpg|thumb|right||Gustav Kirchhoff]]

==Connectedness==
Because I am a Computer Science majors, while circuits are not directly connected to programming, computers and other electronics are made of different
circuits. Loop Rule is used in the electrical engineering that is required to create the circuits for the computers I use to program.

One interesting note about Loop Rule is that is does not apply universally to all circuits. In particular, AC (alternating current) circuits, have a fluctuating
electric charge that changes direction. This causes the electric potential of a round trip path around the circuit to no longer be zero. However, DC (direct current)
circuits still follow the loop rule.

== See also ==
Other Circuit Concepts you can check out :

[[Node Rule]]

[[Components]]

[[Resistors and Conductivity]]

[[Current]]

===More Information and External links===

[https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws Kirchoff's Circuit Laws - Wikipedia]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/circuits-and-direct-currents-20/kirchhoff-s-rules-152/the-loop-rule-540-5636/ Loop Rule - Boundless.com Textbook]

[https://www.youtube.com/watch?v=IlyUtYRqMLs Loop Rule - Doc Physics Video Lecture]

==References==

* https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff.27s_voltage_law
* Chabay, Ruth W. Matter and Interactions: Electric and Magnetic Interactions. John Wiley, 2015. Print.
* http://www.ux1.eiu.edu/~cfadd/1360/28DC/Loop.html

Claimed by Bmock7
[[Category: Simple Circuits]]

Loop Rule

2016-04-16T15:55:12Z

Ldarian3: /* A Mathematical Model */

'''CLAIMED BY LAYLA DARIAN'''

The Loop Rule is a fundamental principle of electric circuits that states when you travel around an entire circuit loop, you will return to the starting voltage.
==Loop Rule==

The loop rule simply states that in any round trip path in a circuit,
[[Electric Potential]] equals zero. Keep in mind that this applies through ANY round trip path; there can be
multiple round trip paths through more complex circuits. This principle deals with the conservation of energy within a circuit. Loop Rule and [[Node Rule]] are the two
fundamental principle of electric circuits and are used to determine the behaviors of electric circuits.
This principle is often used to solve for resistance of the light bulbs or other types of resistors or the
current passing through these resistors.

===A Mathematical Model===
A mathematical representation is:
<math> \Sigma \Delta {V}_{i} = 0 </math>
AND
<math> \Delta {V}_{1} + \Delta {V}_{2} + \space.... = 0 </math>

This can also be represented in a circuit as <math> emf = \Delta {V}_{1} + \Delta {V}_{2} + \space..... </math>

===A Visual Model===

[[File:LoopRule.jpg]]

The total voltage in the circuit is equal to all the individual voltages in the circuit added together.

Another way to think about it is <math> \Delta {V} - (\Delta {V}_{1} + \Delta {V}_{2} + \Delta {V}_{3})= 0 </math>

==Examples==

===Simple===
[[File:SimpleLoopRule.jpg]]

The circuit shown above consists of a single battery and a single resistor. The resistance of the wires is negligible for this problem.

Problem:

If the <math> emf </math> is 5 V and the resistance of the resistor is 10 ohms, what is the current passing through the resistor.

Solution:

Although we can solve this using the V = IR equation for the whole loop, lets examine this problem using the loop rule equation.

The loop rule equation would be <math> {V}_{battery} - {V}_{resistor} = 0 </math>

Since we know the <math> emf </math> of the battery we just need to find the potential difference through the resistor. For this we can use the equation of V = IR.

Thus we now have an loop rule equation of <math> emf - IR = 0 </math>
From here it is a relatively simple process to find the current. We can rewrite the loop rule equation as <math> emf = IR </math> and then plug in 5 for the emf and
10 for the resistance, leaving us with I = .5 amperes.

===Middling===

[[File:MediumExampleLoopRule.jpg]]

Problem:

The circuit shown above consists of a single battery, whose <math> emf </math> is 1.3 V, and three wires made of the same material, but having different cross-sectional areas.
Let the length of the thin wires be <math> {L}_{thick} </math> and the length of the thin wire be <math> {L}_{thin} </math>
Find a loop rule equation that starts at the negative end of the battery and goes counterclockwise through the circuit.

Solution:

When beginning this problem, you must notice that the difference in cross-sectional areas affects the electric field in each wire. Because of this
we will denote the electric field at D. as <math> {E}_{D} </math> and the electric field everywhere else as <math> {E}_{A} </math>.
To begin we will go around the circuit clockwise and add up each component. First, we know that the <math> emf </math> of the battery is 1.3 V. Then, we will add up the potential
voltage of each of the wires.

Remember that the electric potential of a wire is equal to the electric field * length of the wire. From his we can now find the potential difference
of each section of the wires. The electric potential of location A - C is <math> {E}_{A} * {L}_{thick} </math>. This is the same for the electric potential of location E - G of the wire.
For the thin section of the wire, the electric potential is <math> {E}_{D} * {L}_{thin} </math>. From here we just go around the circuit counterclockwise and add each potential difference to the loop rule equation.

Thus we can find that a loop rule equation is: <math> emf - 2 ({E}_{A} * {L}_{thick}) - {E}_{D} * {L}_{thin} = 0 </math>

This can also be rewritten as: <math> emf = 2 ({E}_{A} * {L}_{thick}) + {E}_{D} * {L}_{thin} = 0 </math>

===Difficult===

[[File:HardLoopRule.jpg]]

Problem:

For the circuit above, imagine a situation where the switch has been closed for a long time.
Calculate the current at a,b,c,d,e and charge Q of the capacitor. Answer these using <math> emf, {R}_{1}, {R}_{2},
and \space C </math>

Solution:

First, write loop rule equations for each of the possible loops in the circuit. There are 3 loop equations that are possible.

<math> emf - {I}_{1}{R}_{1} - {I}_{2}{R}_{2} = 0 </math>

<math> emf - {I}_{1}{R}_{1} - Q/C = 0 </math>

<math> {I}_{2}{R}_{2} - Q/C = 0 </math>

From here, we can then solve for the current passing through a,b,d and e. We also know that the current passing through
these points must be the same so <math> {I}_{1} = {I}_{2} </math>

<math> emf - {I}_{1}{R}_{1} - {I}_{1}{R}_{2} = 0 </math>

<math> emf = {I}_{1}({R}_{1} + {R}_{2}) = 0 </math>

<math> emf/({R}_{1} + {R}_{2}) = {I}_{1} </math>

So the current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

You must also know that once a capacitor is charging for a long time, current no longer flows through the capacitor. We can then
easily solve for c because since current is no longer flowing through the capacitor, the current at c = 0.

Lastly, we will use the loop rule equation of <math> {I}_{2}{R}_{2} - Q/C = 0 </math> to solve for Q.

<math> {I}_{2}{R}_{2} = Q/C </math>

<math> C*({I}_{2}{R}_{2}) = Q </math> Since <math> {I}_{1} = {I}_{2} </math>

<math> C*({I}_{1}{R}_{2}) = Q </math> Lastly, we will plug in what we found {I}_{1} equals from before.

<math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}) = Q </math>

Lastly, Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math> and we have now solved the problem.

Answers:

Current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

Current at c = 0

Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math>

==History==

The Loop Rule is formally known as the Kirchhoff Circuit Law, named after Gustav Kirchhoff discovered and defined this fundamental concept of electric circuits. He discovered this during his time as a student at Albertus University of Königsberg in 1845. Kirchoff went on to explore the topics of spectroscopy and black body radiation after his graduation from Albertus. Nowadays, it is used very often in electrical engineering. For more in depth information about Gustav Kirchhoff,
visit the full wiki page on him: [[Gustav Kirchhoff]]

[[File:KirchoffLoopRule.jpg|thumb|right||Gustav Kirchhoff]]

==Connectedness==
Because I am a Computer Science majors, while circuits are not directly connected to programming, computers and other electronics are made of different
circuits. Loop Rule is used in the electrical engineering that is required to create the circuits for the computers I use to program.

One interesting note about Loop Rule is that is does not apply universally to all circuits. In particular, AC (alternating current) circuits, have a fluctuating
electric charge that changes direction. This causes the electric potential of a round trip path around the circuit to no longer be zero. However, DC (direct current)
circuits still follow the loop rule.

== See also ==
Other Circuit Concepts you can check out :

[[Node Rule]]

[[Components]]

[[Resistors and Conductivity]]

[[Current]]

===More Information and External links===

[https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws Kirchoff's Circuit Laws - Wikipedia]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/circuits-and-direct-currents-20/kirchhoff-s-rules-152/the-loop-rule-540-5636/ Loop Rule - Boundless.com Textbook]

[https://www.youtube.com/watch?v=IlyUtYRqMLs Loop Rule - Doc Physics Video Lecture]

==References==

* https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff.27s_voltage_law
* Chabay, Ruth W. Matter and Interactions: Electric and Magnetic Interactions. John Wiley, 2015. Print.
* http://www.ux1.eiu.edu/~cfadd/1360/28DC/Loop.html

Claimed by Bmock7
[[Category: Simple Circuits]]

Loop Rule

2016-04-16T15:54:35Z

Ldarian3: /* A Mathematical Model */

'''CLAIMED BY LAYLA DARIAN'''

The Loop Rule is a fundamental principle of electric circuits that states when you travel around an entire circuit loop, you will return to the starting voltage.
==Loop Rule==

The loop rule simply states that in any round trip path in a circuit,
[[Electric Potential]] equals zero. Keep in mind that this applies through ANY round trip path; there can be
multiple round trip paths through more complex circuits. This principle deals with the conservation of energy within a circuit. Loop Rule and [[Node Rule]] are the two
fundamental principle of electric circuits and are used to determine the behaviors of electric circuits.
This principle is often used to solve for resistance of the light bulbs or other types of resistors or the
current passing through these resistors.

===A Mathematical Model===
A mathematical representation is:
<math> \Sigma \Delta {V}_{i} = 0 </math>
AND
</math> \Delta {V}_{1} + \Delta {V}_{2} + \space.... = 0 </math>

This can also be represented in a circuit as <math> emf = \Delta {V}_{1} + \Delta {V}_{2} + \space..... </math>

===A Visual Model===

[[File:LoopRule.jpg]]

The total voltage in the circuit is equal to all the individual voltages in the circuit added together.

Another way to think about it is <math> \Delta {V} - (\Delta {V}_{1} + \Delta {V}_{2} + \Delta {V}_{3})= 0 </math>

==Examples==

===Simple===
[[File:SimpleLoopRule.jpg]]

The circuit shown above consists of a single battery and a single resistor. The resistance of the wires is negligible for this problem.

Problem:

If the <math> emf </math> is 5 V and the resistance of the resistor is 10 ohms, what is the current passing through the resistor.

Solution:

Although we can solve this using the V = IR equation for the whole loop, lets examine this problem using the loop rule equation.

The loop rule equation would be <math> {V}_{battery} - {V}_{resistor} = 0 </math>

Since we know the <math> emf </math> of the battery we just need to find the potential difference through the resistor. For this we can use the equation of V = IR.

Thus we now have an loop rule equation of <math> emf - IR = 0 </math>
From here it is a relatively simple process to find the current. We can rewrite the loop rule equation as <math> emf = IR </math> and then plug in 5 for the emf and
10 for the resistance, leaving us with I = .5 amperes.

===Middling===

[[File:MediumExampleLoopRule.jpg]]

Problem:

The circuit shown above consists of a single battery, whose <math> emf </math> is 1.3 V, and three wires made of the same material, but having different cross-sectional areas.
Let the length of the thin wires be <math> {L}_{thick} </math> and the length of the thin wire be <math> {L}_{thin} </math>
Find a loop rule equation that starts at the negative end of the battery and goes counterclockwise through the circuit.

Solution:

When beginning this problem, you must notice that the difference in cross-sectional areas affects the electric field in each wire. Because of this
we will denote the electric field at D. as <math> {E}_{D} </math> and the electric field everywhere else as <math> {E}_{A} </math>.
To begin we will go around the circuit clockwise and add up each component. First, we know that the <math> emf </math> of the battery is 1.3 V. Then, we will add up the potential
voltage of each of the wires.

Remember that the electric potential of a wire is equal to the electric field * length of the wire. From his we can now find the potential difference
of each section of the wires. The electric potential of location A - C is <math> {E}_{A} * {L}_{thick} </math>. This is the same for the electric potential of location E - G of the wire.
For the thin section of the wire, the electric potential is <math> {E}_{D} * {L}_{thin} </math>. From here we just go around the circuit counterclockwise and add each potential difference to the loop rule equation.

Thus we can find that a loop rule equation is: <math> emf - 2 ({E}_{A} * {L}_{thick}) - {E}_{D} * {L}_{thin} = 0 </math>

This can also be rewritten as: <math> emf = 2 ({E}_{A} * {L}_{thick}) + {E}_{D} * {L}_{thin} = 0 </math>

===Difficult===

[[File:HardLoopRule.jpg]]

Problem:

For the circuit above, imagine a situation where the switch has been closed for a long time.
Calculate the current at a,b,c,d,e and charge Q of the capacitor. Answer these using <math> emf, {R}_{1}, {R}_{2},
and \space C </math>

Solution:

First, write loop rule equations for each of the possible loops in the circuit. There are 3 loop equations that are possible.

<math> emf - {I}_{1}{R}_{1} - {I}_{2}{R}_{2} = 0 </math>

<math> emf - {I}_{1}{R}_{1} - Q/C = 0 </math>

<math> {I}_{2}{R}_{2} - Q/C = 0 </math>

From here, we can then solve for the current passing through a,b,d and e. We also know that the current passing through
these points must be the same so <math> {I}_{1} = {I}_{2} </math>

<math> emf - {I}_{1}{R}_{1} - {I}_{1}{R}_{2} = 0 </math>

<math> emf = {I}_{1}({R}_{1} + {R}_{2}) = 0 </math>

<math> emf/({R}_{1} + {R}_{2}) = {I}_{1} </math>

So the current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

You must also know that once a capacitor is charging for a long time, current no longer flows through the capacitor. We can then
easily solve for c because since current is no longer flowing through the capacitor, the current at c = 0.

Lastly, we will use the loop rule equation of <math> {I}_{2}{R}_{2} - Q/C = 0 </math> to solve for Q.

<math> {I}_{2}{R}_{2} = Q/C </math>

<math> C*({I}_{2}{R}_{2}) = Q </math> Since <math> {I}_{1} = {I}_{2} </math>

<math> C*({I}_{1}{R}_{2}) = Q </math> Lastly, we will plug in what we found {I}_{1} equals from before.

<math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}) = Q </math>

Lastly, Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math> and we have now solved the problem.

Answers:

Current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

Current at c = 0

Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math>

==History==

The Loop Rule is formally known as the Kirchhoff Circuit Law, named after Gustav Kirchhoff discovered and defined this fundamental concept of electric circuits. He discovered this during his time as a student at Albertus University of Königsberg in 1845. Kirchoff went on to explore the topics of spectroscopy and black body radiation after his graduation from Albertus. Nowadays, it is used very often in electrical engineering. For more in depth information about Gustav Kirchhoff,
visit the full wiki page on him: [[Gustav Kirchhoff]]

[[File:KirchoffLoopRule.jpg|thumb|right||Gustav Kirchhoff]]

==Connectedness==
Because I am a Computer Science majors, while circuits are not directly connected to programming, computers and other electronics are made of different
circuits. Loop Rule is used in the electrical engineering that is required to create the circuits for the computers I use to program.

One interesting note about Loop Rule is that is does not apply universally to all circuits. In particular, AC (alternating current) circuits, have a fluctuating
electric charge that changes direction. This causes the electric potential of a round trip path around the circuit to no longer be zero. However, DC (direct current)
circuits still follow the loop rule.

== See also ==
Other Circuit Concepts you can check out :

[[Node Rule]]

[[Components]]

[[Resistors and Conductivity]]

[[Current]]

===More Information and External links===

[https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws Kirchoff's Circuit Laws - Wikipedia]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/circuits-and-direct-currents-20/kirchhoff-s-rules-152/the-loop-rule-540-5636/ Loop Rule - Boundless.com Textbook]

[https://www.youtube.com/watch?v=IlyUtYRqMLs Loop Rule - Doc Physics Video Lecture]

==References==

* https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff.27s_voltage_law
* Chabay, Ruth W. Matter and Interactions: Electric and Magnetic Interactions. John Wiley, 2015. Print.
* http://www.ux1.eiu.edu/~cfadd/1360/28DC/Loop.html

Claimed by Bmock7
[[Category: Simple Circuits]]

Loop Rule

2016-04-16T15:54:02Z

Ldarian3: /* A Mathematical Model */

'''CLAIMED BY LAYLA DARIAN'''

The Loop Rule is a fundamental principle of electric circuits that states when you travel around an entire circuit loop, you will return to the starting voltage.
==Loop Rule==

The loop rule simply states that in any round trip path in a circuit,
[[Electric Potential]] equals zero. Keep in mind that this applies through ANY round trip path; there can be
multiple round trip paths through more complex circuits. This principle deals with the conservation of energy within a circuit. Loop Rule and [[Node Rule]] are the two
fundamental principle of electric circuits and are used to determine the behaviors of electric circuits.
This principle is often used to solve for resistance of the light bulbs or other types of resistors or the
current passing through these resistors.

===A Mathematical Model===
A mathematical representation is:
<math> \Sigma \Delta {V}_{i} = 0

\Delta {V}_{1} + \Delta {V}_{2} + \space.... = 0 </math>

This can also be represented in a circuit as <math> emf = \Delta {V}_{1} + \Delta {V}_{2} + \space..... </math>

===A Visual Model===

[[File:LoopRule.jpg]]

The total voltage in the circuit is equal to all the individual voltages in the circuit added together.

Another way to think about it is <math> \Delta {V} - (\Delta {V}_{1} + \Delta {V}_{2} + \Delta {V}_{3})= 0 </math>

==Examples==

===Simple===
[[File:SimpleLoopRule.jpg]]

The circuit shown above consists of a single battery and a single resistor. The resistance of the wires is negligible for this problem.

Problem:

If the <math> emf </math> is 5 V and the resistance of the resistor is 10 ohms, what is the current passing through the resistor.

Solution:

Although we can solve this using the V = IR equation for the whole loop, lets examine this problem using the loop rule equation.

The loop rule equation would be <math> {V}_{battery} - {V}_{resistor} = 0 </math>

Since we know the <math> emf </math> of the battery we just need to find the potential difference through the resistor. For this we can use the equation of V = IR.

Thus we now have an loop rule equation of <math> emf - IR = 0 </math>
From here it is a relatively simple process to find the current. We can rewrite the loop rule equation as <math> emf = IR </math> and then plug in 5 for the emf and
10 for the resistance, leaving us with I = .5 amperes.

===Middling===

[[File:MediumExampleLoopRule.jpg]]

Problem:

The circuit shown above consists of a single battery, whose <math> emf </math> is 1.3 V, and three wires made of the same material, but having different cross-sectional areas.
Let the length of the thin wires be <math> {L}_{thick} </math> and the length of the thin wire be <math> {L}_{thin} </math>
Find a loop rule equation that starts at the negative end of the battery and goes counterclockwise through the circuit.

Solution:

When beginning this problem, you must notice that the difference in cross-sectional areas affects the electric field in each wire. Because of this
we will denote the electric field at D. as <math> {E}_{D} </math> and the electric field everywhere else as <math> {E}_{A} </math>.
To begin we will go around the circuit clockwise and add up each component. First, we know that the <math> emf </math> of the battery is 1.3 V. Then, we will add up the potential
voltage of each of the wires.

Remember that the electric potential of a wire is equal to the electric field * length of the wire. From his we can now find the potential difference
of each section of the wires. The electric potential of location A - C is <math> {E}_{A} * {L}_{thick} </math>. This is the same for the electric potential of location E - G of the wire.
For the thin section of the wire, the electric potential is <math> {E}_{D} * {L}_{thin} </math>. From here we just go around the circuit counterclockwise and add each potential difference to the loop rule equation.

Thus we can find that a loop rule equation is: <math> emf - 2 ({E}_{A} * {L}_{thick}) - {E}_{D} * {L}_{thin} = 0 </math>

This can also be rewritten as: <math> emf = 2 ({E}_{A} * {L}_{thick}) + {E}_{D} * {L}_{thin} = 0 </math>

===Difficult===

[[File:HardLoopRule.jpg]]

Problem:

For the circuit above, imagine a situation where the switch has been closed for a long time.
Calculate the current at a,b,c,d,e and charge Q of the capacitor. Answer these using <math> emf, {R}_{1}, {R}_{2},
and \space C </math>

Solution:

First, write loop rule equations for each of the possible loops in the circuit. There are 3 loop equations that are possible.

<math> emf - {I}_{1}{R}_{1} - {I}_{2}{R}_{2} = 0 </math>

<math> emf - {I}_{1}{R}_{1} - Q/C = 0 </math>

<math> {I}_{2}{R}_{2} - Q/C = 0 </math>

From here, we can then solve for the current passing through a,b,d and e. We also know that the current passing through
these points must be the same so <math> {I}_{1} = {I}_{2} </math>

<math> emf - {I}_{1}{R}_{1} - {I}_{1}{R}_{2} = 0 </math>

<math> emf = {I}_{1}({R}_{1} + {R}_{2}) = 0 </math>

<math> emf/({R}_{1} + {R}_{2}) = {I}_{1} </math>

So the current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

You must also know that once a capacitor is charging for a long time, current no longer flows through the capacitor. We can then
easily solve for c because since current is no longer flowing through the capacitor, the current at c = 0.

Lastly, we will use the loop rule equation of <math> {I}_{2}{R}_{2} - Q/C = 0 </math> to solve for Q.

<math> {I}_{2}{R}_{2} = Q/C </math>

<math> C*({I}_{2}{R}_{2}) = Q </math> Since <math> {I}_{1} = {I}_{2} </math>

<math> C*({I}_{1}{R}_{2}) = Q </math> Lastly, we will plug in what we found {I}_{1} equals from before.

<math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}) = Q </math>

Lastly, Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math> and we have now solved the problem.

Answers:

Current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

Current at c = 0

Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math>

==History==

The Loop Rule is formally known as the Kirchhoff Circuit Law, named after Gustav Kirchhoff discovered and defined this fundamental concept of electric circuits. He discovered this during his time as a student at Albertus University of Königsberg in 1845. Kirchoff went on to explore the topics of spectroscopy and black body radiation after his graduation from Albertus. Nowadays, it is used very often in electrical engineering. For more in depth information about Gustav Kirchhoff,
visit the full wiki page on him: [[Gustav Kirchhoff]]

[[File:KirchoffLoopRule.jpg|thumb|right||Gustav Kirchhoff]]

==Connectedness==
Because I am a Computer Science majors, while circuits are not directly connected to programming, computers and other electronics are made of different
circuits. Loop Rule is used in the electrical engineering that is required to create the circuits for the computers I use to program.

One interesting note about Loop Rule is that is does not apply universally to all circuits. In particular, AC (alternating current) circuits, have a fluctuating
electric charge that changes direction. This causes the electric potential of a round trip path around the circuit to no longer be zero. However, DC (direct current)
circuits still follow the loop rule.

== See also ==
Other Circuit Concepts you can check out :

[[Node Rule]]

[[Components]]

[[Resistors and Conductivity]]

[[Current]]

===More Information and External links===

[https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws Kirchoff's Circuit Laws - Wikipedia]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/circuits-and-direct-currents-20/kirchhoff-s-rules-152/the-loop-rule-540-5636/ Loop Rule - Boundless.com Textbook]

[https://www.youtube.com/watch?v=IlyUtYRqMLs Loop Rule - Doc Physics Video Lecture]

==References==

* https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff.27s_voltage_law
* Chabay, Ruth W. Matter and Interactions: Electric and Magnetic Interactions. John Wiley, 2015. Print.
* http://www.ux1.eiu.edu/~cfadd/1360/28DC/Loop.html

Claimed by Bmock7
[[Category: Simple Circuits]]

Loop Rule

2016-04-16T15:53:46Z

Ldarian3: /* A Mathematical Model */

'''CLAIMED BY LAYLA DARIAN'''

The Loop Rule is a fundamental principle of electric circuits that states when you travel around an entire circuit loop, you will return to the starting voltage.
==Loop Rule==

The loop rule simply states that in any round trip path in a circuit,
[[Electric Potential]] equals zero. Keep in mind that this applies through ANY round trip path; there can be
multiple round trip paths through more complex circuits. This principle deals with the conservation of energy within a circuit. Loop Rule and [[Node Rule]] are the two
fundamental principle of electric circuits and are used to determine the behaviors of electric circuits.
This principle is often used to solve for resistance of the light bulbs or other types of resistors or the
current passing through these resistors.

===A Mathematical Model===
A mathematical representation is:
<math> \Sigma \Delta {V}_{i} = 0
\Delta {V}_{1} + \Delta {V}_{2} + \space.... = 0 </math>

This can also be represented in a circuit as <math> emf = \Delta {V}_{1} + \Delta {V}_{2} + \space..... </math>

===A Visual Model===

[[File:LoopRule.jpg]]

The total voltage in the circuit is equal to all the individual voltages in the circuit added together.

Another way to think about it is <math> \Delta {V} - (\Delta {V}_{1} + \Delta {V}_{2} + \Delta {V}_{3})= 0 </math>

==Examples==

===Simple===
[[File:SimpleLoopRule.jpg]]

The circuit shown above consists of a single battery and a single resistor. The resistance of the wires is negligible for this problem.

Problem:

If the <math> emf </math> is 5 V and the resistance of the resistor is 10 ohms, what is the current passing through the resistor.

Solution:

Although we can solve this using the V = IR equation for the whole loop, lets examine this problem using the loop rule equation.

The loop rule equation would be <math> {V}_{battery} - {V}_{resistor} = 0 </math>

Since we know the <math> emf </math> of the battery we just need to find the potential difference through the resistor. For this we can use the equation of V = IR.

Thus we now have an loop rule equation of <math> emf - IR = 0 </math>
From here it is a relatively simple process to find the current. We can rewrite the loop rule equation as <math> emf = IR </math> and then plug in 5 for the emf and
10 for the resistance, leaving us with I = .5 amperes.

===Middling===

[[File:MediumExampleLoopRule.jpg]]

Problem:

The circuit shown above consists of a single battery, whose <math> emf </math> is 1.3 V, and three wires made of the same material, but having different cross-sectional areas.
Let the length of the thin wires be <math> {L}_{thick} </math> and the length of the thin wire be <math> {L}_{thin} </math>
Find a loop rule equation that starts at the negative end of the battery and goes counterclockwise through the circuit.

Solution:

When beginning this problem, you must notice that the difference in cross-sectional areas affects the electric field in each wire. Because of this
we will denote the electric field at D. as <math> {E}_{D} </math> and the electric field everywhere else as <math> {E}_{A} </math>.
To begin we will go around the circuit clockwise and add up each component. First, we know that the <math> emf </math> of the battery is 1.3 V. Then, we will add up the potential
voltage of each of the wires.

Remember that the electric potential of a wire is equal to the electric field * length of the wire. From his we can now find the potential difference
of each section of the wires. The electric potential of location A - C is <math> {E}_{A} * {L}_{thick} </math>. This is the same for the electric potential of location E - G of the wire.
For the thin section of the wire, the electric potential is <math> {E}_{D} * {L}_{thin} </math>. From here we just go around the circuit counterclockwise and add each potential difference to the loop rule equation.

Thus we can find that a loop rule equation is: <math> emf - 2 ({E}_{A} * {L}_{thick}) - {E}_{D} * {L}_{thin} = 0 </math>

This can also be rewritten as: <math> emf = 2 ({E}_{A} * {L}_{thick}) + {E}_{D} * {L}_{thin} = 0 </math>

===Difficult===

[[File:HardLoopRule.jpg]]

Problem:

For the circuit above, imagine a situation where the switch has been closed for a long time.
Calculate the current at a,b,c,d,e and charge Q of the capacitor. Answer these using <math> emf, {R}_{1}, {R}_{2},
and \space C </math>

Solution:

First, write loop rule equations for each of the possible loops in the circuit. There are 3 loop equations that are possible.

<math> emf - {I}_{1}{R}_{1} - {I}_{2}{R}_{2} = 0 </math>

<math> emf - {I}_{1}{R}_{1} - Q/C = 0 </math>

<math> {I}_{2}{R}_{2} - Q/C = 0 </math>

From here, we can then solve for the current passing through a,b,d and e. We also know that the current passing through
these points must be the same so <math> {I}_{1} = {I}_{2} </math>

<math> emf - {I}_{1}{R}_{1} - {I}_{1}{R}_{2} = 0 </math>

<math> emf = {I}_{1}({R}_{1} + {R}_{2}) = 0 </math>

<math> emf/({R}_{1} + {R}_{2}) = {I}_{1} </math>

So the current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

You must also know that once a capacitor is charging for a long time, current no longer flows through the capacitor. We can then
easily solve for c because since current is no longer flowing through the capacitor, the current at c = 0.

Lastly, we will use the loop rule equation of <math> {I}_{2}{R}_{2} - Q/C = 0 </math> to solve for Q.

<math> {I}_{2}{R}_{2} = Q/C </math>

<math> C*({I}_{2}{R}_{2}) = Q </math> Since <math> {I}_{1} = {I}_{2} </math>

<math> C*({I}_{1}{R}_{2}) = Q </math> Lastly, we will plug in what we found {I}_{1} equals from before.

<math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}) = Q </math>

Lastly, Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math> and we have now solved the problem.

Answers:

Current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

Current at c = 0

Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math>

==History==

The Loop Rule is formally known as the Kirchhoff Circuit Law, named after Gustav Kirchhoff discovered and defined this fundamental concept of electric circuits. He discovered this during his time as a student at Albertus University of Königsberg in 1845. Kirchoff went on to explore the topics of spectroscopy and black body radiation after his graduation from Albertus. Nowadays, it is used very often in electrical engineering. For more in depth information about Gustav Kirchhoff,
visit the full wiki page on him: [[Gustav Kirchhoff]]

[[File:KirchoffLoopRule.jpg|thumb|right||Gustav Kirchhoff]]

==Connectedness==
Because I am a Computer Science majors, while circuits are not directly connected to programming, computers and other electronics are made of different
circuits. Loop Rule is used in the electrical engineering that is required to create the circuits for the computers I use to program.

One interesting note about Loop Rule is that is does not apply universally to all circuits. In particular, AC (alternating current) circuits, have a fluctuating
electric charge that changes direction. This causes the electric potential of a round trip path around the circuit to no longer be zero. However, DC (direct current)
circuits still follow the loop rule.

== See also ==
Other Circuit Concepts you can check out :

[[Node Rule]]

[[Components]]

[[Resistors and Conductivity]]

[[Current]]

===More Information and External links===

[https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws Kirchoff's Circuit Laws - Wikipedia]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/circuits-and-direct-currents-20/kirchhoff-s-rules-152/the-loop-rule-540-5636/ Loop Rule - Boundless.com Textbook]

[https://www.youtube.com/watch?v=IlyUtYRqMLs Loop Rule - Doc Physics Video Lecture]

==References==

* https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff.27s_voltage_law
* Chabay, Ruth W. Matter and Interactions: Electric and Magnetic Interactions. John Wiley, 2015. Print.
* http://www.ux1.eiu.edu/~cfadd/1360/28DC/Loop.html

Claimed by Bmock7
[[Category: Simple Circuits]]

Loop Rule

2016-04-16T15:52:56Z

Ldarian3:

'''CLAIMED BY LAYLA DARIAN'''

The Loop Rule is a fundamental principle of electric circuits that states when you travel around an entire circuit loop, you will return to the starting voltage.
==Loop Rule==

The loop rule simply states that in any round trip path in a circuit,
[[Electric Potential]] equals zero. Keep in mind that this applies through ANY round trip path; there can be
multiple round trip paths through more complex circuits. This principle deals with the conservation of energy within a circuit. Loop Rule and [[Node Rule]] are the two
fundamental principle of electric circuits and are used to determine the behaviors of electric circuits.
This principle is often used to solve for resistance of the light bulbs or other types of resistors or the
current passing through these resistors.

===A Mathematical Model===
A mathematical representation is:
<math> \Sigma \Delta {V}_{i} = 0
<math> \Delta {V}_{1} + \Delta {V}_{2} + \space.... = 0 </math>

This can also be represented in a circuit as <math> emf = \Delta {V}_{1} + \Delta {V}_{2} + \space..... </math>

===A Visual Model===

[[File:LoopRule.jpg]]

The total voltage in the circuit is equal to all the individual voltages in the circuit added together.

Another way to think about it is <math> \Delta {V} - (\Delta {V}_{1} + \Delta {V}_{2} + \Delta {V}_{3})= 0 </math>

==Examples==

===Simple===
[[File:SimpleLoopRule.jpg]]

The circuit shown above consists of a single battery and a single resistor. The resistance of the wires is negligible for this problem.

Problem:

If the <math> emf </math> is 5 V and the resistance of the resistor is 10 ohms, what is the current passing through the resistor.

Solution:

Although we can solve this using the V = IR equation for the whole loop, lets examine this problem using the loop rule equation.

The loop rule equation would be <math> {V}_{battery} - {V}_{resistor} = 0 </math>

Since we know the <math> emf </math> of the battery we just need to find the potential difference through the resistor. For this we can use the equation of V = IR.

Thus we now have an loop rule equation of <math> emf - IR = 0 </math>
From here it is a relatively simple process to find the current. We can rewrite the loop rule equation as <math> emf = IR </math> and then plug in 5 for the emf and
10 for the resistance, leaving us with I = .5 amperes.

===Middling===

[[File:MediumExampleLoopRule.jpg]]

Problem:

The circuit shown above consists of a single battery, whose <math> emf </math> is 1.3 V, and three wires made of the same material, but having different cross-sectional areas.
Let the length of the thin wires be <math> {L}_{thick} </math> and the length of the thin wire be <math> {L}_{thin} </math>
Find a loop rule equation that starts at the negative end of the battery and goes counterclockwise through the circuit.

Solution:

When beginning this problem, you must notice that the difference in cross-sectional areas affects the electric field in each wire. Because of this
we will denote the electric field at D. as <math> {E}_{D} </math> and the electric field everywhere else as <math> {E}_{A} </math>.
To begin we will go around the circuit clockwise and add up each component. First, we know that the <math> emf </math> of the battery is 1.3 V. Then, we will add up the potential
voltage of each of the wires.

Remember that the electric potential of a wire is equal to the electric field * length of the wire. From his we can now find the potential difference
of each section of the wires. The electric potential of location A - C is <math> {E}_{A} * {L}_{thick} </math>. This is the same for the electric potential of location E - G of the wire.
For the thin section of the wire, the electric potential is <math> {E}_{D} * {L}_{thin} </math>. From here we just go around the circuit counterclockwise and add each potential difference to the loop rule equation.

Thus we can find that a loop rule equation is: <math> emf - 2 ({E}_{A} * {L}_{thick}) - {E}_{D} * {L}_{thin} = 0 </math>

This can also be rewritten as: <math> emf = 2 ({E}_{A} * {L}_{thick}) + {E}_{D} * {L}_{thin} = 0 </math>

===Difficult===

[[File:HardLoopRule.jpg]]

Problem:

For the circuit above, imagine a situation where the switch has been closed for a long time.
Calculate the current at a,b,c,d,e and charge Q of the capacitor. Answer these using <math> emf, {R}_{1}, {R}_{2},
and \space C </math>

Solution:

First, write loop rule equations for each of the possible loops in the circuit. There are 3 loop equations that are possible.

<math> emf - {I}_{1}{R}_{1} - {I}_{2}{R}_{2} = 0 </math>

<math> emf - {I}_{1}{R}_{1} - Q/C = 0 </math>

<math> {I}_{2}{R}_{2} - Q/C = 0 </math>

From here, we can then solve for the current passing through a,b,d and e. We also know that the current passing through
these points must be the same so <math> {I}_{1} = {I}_{2} </math>

<math> emf - {I}_{1}{R}_{1} - {I}_{1}{R}_{2} = 0 </math>

<math> emf = {I}_{1}({R}_{1} + {R}_{2}) = 0 </math>

<math> emf/({R}_{1} + {R}_{2}) = {I}_{1} </math>

So the current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

You must also know that once a capacitor is charging for a long time, current no longer flows through the capacitor. We can then
easily solve for c because since current is no longer flowing through the capacitor, the current at c = 0.

Lastly, we will use the loop rule equation of <math> {I}_{2}{R}_{2} - Q/C = 0 </math> to solve for Q.

<math> {I}_{2}{R}_{2} = Q/C </math>

<math> C*({I}_{2}{R}_{2}) = Q </math> Since <math> {I}_{1} = {I}_{2} </math>

<math> C*({I}_{1}{R}_{2}) = Q </math> Lastly, we will plug in what we found {I}_{1} equals from before.

<math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}) = Q </math>

Lastly, Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math> and we have now solved the problem.

Answers:

Current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

Current at c = 0

Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math>

==History==

The Loop Rule is formally known as the Kirchhoff Circuit Law, named after Gustav Kirchhoff discovered and defined this fundamental concept of electric circuits. He discovered this during his time as a student at Albertus University of Königsberg in 1845. Kirchoff went on to explore the topics of spectroscopy and black body radiation after his graduation from Albertus. Nowadays, it is used very often in electrical engineering. For more in depth information about Gustav Kirchhoff,
visit the full wiki page on him: [[Gustav Kirchhoff]]

[[File:KirchoffLoopRule.jpg|thumb|right||Gustav Kirchhoff]]

==Connectedness==
Because I am a Computer Science majors, while circuits are not directly connected to programming, computers and other electronics are made of different
circuits. Loop Rule is used in the electrical engineering that is required to create the circuits for the computers I use to program.

One interesting note about Loop Rule is that is does not apply universally to all circuits. In particular, AC (alternating current) circuits, have a fluctuating
electric charge that changes direction. This causes the electric potential of a round trip path around the circuit to no longer be zero. However, DC (direct current)
circuits still follow the loop rule.

== See also ==
Other Circuit Concepts you can check out :

[[Node Rule]]

[[Components]]

[[Resistors and Conductivity]]

[[Current]]

===More Information and External links===

[https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws Kirchoff's Circuit Laws - Wikipedia]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/circuits-and-direct-currents-20/kirchhoff-s-rules-152/the-loop-rule-540-5636/ Loop Rule - Boundless.com Textbook]

[https://www.youtube.com/watch?v=IlyUtYRqMLs Loop Rule - Doc Physics Video Lecture]

==References==

* https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff.27s_voltage_law
* Chabay, Ruth W. Matter and Interactions: Electric and Magnetic Interactions. John Wiley, 2015. Print.
* http://www.ux1.eiu.edu/~cfadd/1360/28DC/Loop.html

Claimed by Bmock7
[[Category: Simple Circuits]]

Loop Rule

2016-03-28T17:44:39Z

Ldarian3:

'''CLAIMED BY LAYLA DARIAN'''

The Loop Rule is a fundamental principle of electric circuits that claims that in any round trip path in a circuit, [[Electric Potential]] equals zero.
==Loop Rule==

The loop rule simply states that in any round trip path in a circuit,
[[Electric Potential]] equals zero. Keep in mind that this applies through ANY round trip path; there can be
multiple round trip paths through more complex circuits. This principle deals with the conservation of energy within a circuit. Loop Rule and [[Node Rule]] are the two
fundamental principle of electric circuits and are used to determine the behaviors of electric circuits.
This principle is often used to solve for resistance of the light bulbs or other types of resistors or the
current passing through these resistors.

===A Mathematical Model===
A mathematical representation is: <math> \Delta {V}_{1} + \Delta {V}_{2} + \space.... = 0 </math>

This can also be represented in a circuit as <math> emf = \Delta {V}_{1} + \Delta {V}_{2} + \space..... </math>

===A Visual Model===

[[File:LoopRule.jpg]]

The total voltage in the circuit is equal to all the individual voltages in the circuit added together.

Another way to think about it is <math> \Delta {V} - (\Delta {V}_{1} + \Delta {V}_{2} + \Delta {V}_{3})= 0 </math>

==Examples==

===Simple===
[[File:SimpleLoopRule.jpg]]

The circuit shown above consists of a single battery and a single resistor. The resistance of the wires is negligible for this problem.

Problem:

If the <math> emf </math> is 5 V and the resistance of the resistor is 10 ohms, what is the current passing through the resistor.

Solution:

Although we can solve this using the V = IR equation for the whole loop, lets examine this problem using the loop rule equation.

The loop rule equation would be <math> {V}_{battery} - {V}_{resistor} = 0 </math>

Since we know the <math> emf </math> of the battery we just need to find the potential difference through the resistor. For this we can use the equation of V = IR.

Thus we now have an loop rule equation of <math> emf - IR = 0 </math>
From here it is a relatively simple process to find the current. We can rewrite the loop rule equation as <math> emf = IR </math> and then plug in 5 for the emf and
10 for the resistance, leaving us with I = .5 amperes.

===Middling===

[[File:MediumExampleLoopRule.jpg]]

Problem:

The circuit shown above consists of a single battery, whose <math> emf </math> is 1.3 V, and three wires made of the same material, but having different cross-sectional areas.
Let the length of the thin wires be <math> {L}_{thick} </math> and the length of the thin wire be <math> {L}_{thin} </math>
Find a loop rule equation that starts at the negative end of the battery and goes counterclockwise through the circuit.

Solution:

When beginning this problem, you must notice that the difference in cross-sectional areas affects the electric field in each wire. Because of this
we will denote the electric field at D. as <math> {E}_{D} </math> and the electric field everywhere else as <math> {E}_{A} </math>.
To begin we will go around the circuit clockwise and add up each component. First, we know that the <math> emf </math> of the battery is 1.3 V. Then, we will add up the potential
voltage of each of the wires.

Remember that the electric potential of a wire is equal to the electric field * length of the wire. From his we can now find the potential difference
of each section of the wires. The electric potential of location A - C is <math> {E}_{A} * {L}_{thick} </math>. This is the same for the electric potential of location E - G of the wire.
For the thin section of the wire, the electric potential is <math> {E}_{D} * {L}_{thin} </math>. From here we just go around the circuit counterclockwise and add each potential difference to the loop rule equation.

Thus we can find that a loop rule equation is: <math> emf - 2 ({E}_{A} * {L}_{thick}) - {E}_{D} * {L}_{thin} = 0 </math>

This can also be rewritten as: <math> emf = 2 ({E}_{A} * {L}_{thick}) + {E}_{D} * {L}_{thin} = 0 </math>

===Difficult===

[[File:HardLoopRule.jpg]]

Problem:

For the circuit above, imagine a situation where the switch has been closed for a long time.
Calculate the current at a,b,c,d,e and charge Q of the capacitor. Answer these using <math> emf, {R}_{1}, {R}_{2},
and \space C </math>

Solution:

First, write loop rule equations for each of the possible loops in the circuit. There are 3 loop equations that are possible.

<math> emf - {I}_{1}{R}_{1} - {I}_{2}{R}_{2} = 0 </math>

<math> emf - {I}_{1}{R}_{1} - Q/C = 0 </math>

<math> {I}_{2}{R}_{2} - Q/C = 0 </math>

From here, we can then solve for the current passing through a,b,d and e. We also know that the current passing through
these points must be the same so <math> {I}_{1} = {I}_{2} </math>

<math> emf - {I}_{1}{R}_{1} - {I}_{1}{R}_{2} = 0 </math>

<math> emf = {I}_{1}({R}_{1} + {R}_{2}) = 0 </math>

<math> emf/({R}_{1} + {R}_{2}) = {I}_{1} </math>

So the current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

You must also know that once a capacitor is charging for a long time, current no longer flows through the capacitor. We can then
easily solve for c because since current is no longer flowing through the capacitor, the current at c = 0.

Lastly, we will use the loop rule equation of <math> {I}_{2}{R}_{2} - Q/C = 0 </math> to solve for Q.

<math> {I}_{2}{R}_{2} = Q/C </math>

<math> C*({I}_{2}{R}_{2}) = Q </math> Since <math> {I}_{1} = {I}_{2} </math>

<math> C*({I}_{1}{R}_{2}) = Q </math> Lastly, we will plug in what we found {I}_{1} equals from before.

<math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}) = Q </math>

Lastly, Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math> and we have now solved the problem.

Answers:

Current at a,b,d,e = <math> emf/({R}_{1} + {R}_{2}) </math>

Current at c = 0

Q = <math> C* (emf/({R}_{1} + {R}_{2})){R}_{2}</math>

==History==

The Loop Rule is formally known as the Kirchhoff Circuit Law, named after Gustav Kirchhoff discovered and defined this fundamental concept of electric circuits. He discovered this during his time as a student at Albertus University of Königsberg in 1845. Kirchoff went on to explore the topics of spectroscopy and black body radiation after his graduation from Albertus. Nowadays, it is used very often in electrical engineering. For more in depth information about Gustav Kirchhoff,
visit the full wiki page on him: [[Gustav Kirchhoff]]

[[File:KirchoffLoopRule.jpg|thumb|right||Gustav Kirchhoff]]

==Connectedness==
Because I am a Computer Science majors, while circuits are not directly connected to programming, computers and other electronics are made of different
circuits. Loop Rule is used in the electrical engineering that is required to create the circuits for the computers I use to program.

One interesting note about Loop Rule is that is does not apply universally to all circuits. In particular, AC (alternating current) circuits, have a fluctuating
electric charge that changes direction. This causes the electric potential of a round trip path around the circuit to no longer be zero. However, DC (direct current)
circuits still follow the loop rule.

== See also ==
Other Circuit Concepts you can check out :

[[Node Rule]]

[[Components]]

[[Resistors and Conductivity]]

[[Current]]

===More Information and External links===

[https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws Kirchoff's Circuit Laws - Wikipedia]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/circuits-and-direct-currents-20/kirchhoff-s-rules-152/the-loop-rule-540-5636/ Loop Rule - Boundless.com Textbook]

[https://www.youtube.com/watch?v=IlyUtYRqMLs Loop Rule - Doc Physics Video Lecture]

==References==

* https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff.27s_voltage_law
* Chabay, Ruth W. Matter and Interactions: Electric and Magnetic Interactions. John Wiley, 2015. Print.
* http://www.ux1.eiu.edu/~cfadd/1360/28DC/Loop.html

Claimed by Bmock7
[[Category: Simple Circuits]]

Right Hand Rule

2015-11-26T19:25:09Z

Ldarian3: /* References */

This topic covers Right Hand Rule.
Claimed by: Layla Darian

==The Main Idea==
The magnitude of the translational angular momentum doesn't tell us everything about the motion of an object. For example, to describe the motion of Earth's orbit if you are standing on the positive z axis looking toward the origin, the Earth is moving counterclockwise in the xy plane. However, this statement cannot be used mathematically. You can't add "counterclockwise in the xy plane" to "clockwise in the yz plane." Instead, we can use the right hand rule to describe the direction of angular momentum as a vector. The direction can be specified like this:
* The plane in which both the position vector from the two objects and the momentum vector for the object of interest lie can be indicated by a unit vector perpendicular to that plane.
* The direction of motion within the plane (clockwise or counterclockwise) can be indicated by establishing a right hand rule (RHR) for this unit vector.
* Right Hand Rule: Using your right arm, point your arm to represent the "r" vector. Now turn your palm in the direction of the momentum vector. Curl your fingers in that direction of the momentum, and extend your thumb outward. The unit vector representing the direction of the angular momentum is defined to point in the direction of your thumb.
*Hint: If the rotational motion is counterclockwise, your right thumb, therefore the unit vector, will point out of the plane. If the rotational motion is clockwise, the unit vector will point into the plane.

===A Mathematical Model===
The direction of the angular momentum can also be solved through calculating the cross product of the '''r''' and '''p''' vectors.
The cross product of two vectors can be solved as so:

[[File:crossproduct2.jpg]]

[[File:Cross product.jpg]]

The first row is the standard basis vectors and must appear in the order given here. The second row is the components of '''r''' and the third row is the components of '''p'''. First, the terms alternate in sign and notice that the 2x2 is missing the column below the standard basis vector that multiplies it as well as the row of standard basis vectors. To solve for the "i" component, you use the expression: bf - ce. Do the same for the "j" and "k" components, and this will give you your vector. Your vector should be <bf-ce, af-cd, ae-bd>.

For more on the explanation of how to calculate cross product, visit this website: https://www.mathsisfun.com/algebra/vectors-cross-product.html

Here is an example problem. Solve first using the right hand rule, and then solve mathematically with the cross product.

[[File:problem1.jpg]]

Answer:
The cross product should be <-12, 0, 0>
So the vector is in the -x direction, which the right hand rule also tells us.

==Examples==

[[File:example1.jpg]]

Ans: 4

[[File:ex2.jpg]]

Ans: 4

[[File:ex3.jpg]]

Ans: 3

[[File:EX4.jpg]]

Ans: 3

[[File:EX5.jpg]]

Ans:
L(A)=L(B)=L(H) = <0, 0, -30>
L(G)=L(C) = <0, 0, 0>
L(D)=L(E)=L(F) = <0, 0, +50>

==Connectedness==
Right Hand Rule is essential for solving all angular momentum problems.
Angular Momentum has been my favorite topic so far this semester. I think the problems and situations are very interesting, and they explain the reason behind why a lot of stuff in real life happen that we don't generally think about.

Here are some interesting videos about the conservation of angular momentum:

https://www.youtube.com/watch?v=Aw5i994n2bw&feature=youtu.be

https://www.youtube.com/watch?v=OKbawIq3w7U

==History==

I wasn't able to find who was the first to apply the Right Hand Rule to angular momentum and torque. However, it seems that the right hand rule is applied to other aspects of physics
as well. For example, André-Marie Ampère, a French physicist and mathematician, created a right hand rule for circuits and electric currents. This is used when a vector must be defined
to represent the rotation of a body, a magnetic field, or a fluid. It reveals a connection between the current and the magnetic field lines in the magnetic field that the current created.
This right hand rule works exactly the same way as the one I have described above.

== See also ==

Once you have learned how to use the Right Hand Rule to determine the direction of angular momentum, you will be able to solve angular momentum and torque problems.

==References==

All pictures and videos were taken from Professor Flavio Fenton's Physics 2211 lecture notes.
https://en.wikipedia.org/wiki/Right-hand_rule

Right Hand Rule

2015-11-26T19:24:58Z

Ldarian3: /* References */

Right Hand Rule

2015-11-26T19:24:13Z

Ldarian3: /* A Mathematical Model */

Right Hand Rule

2015-11-26T19:19:40Z

Ldarian3: /* Connectedness */

This topic covers Right Hand Rule.
Claimed by: Layla Darian

==The Main Idea==
The magnitude of the translational angular momentum doesn't tell us everything about the motion of an object. For example, to describe the motion of Earth's orbit if you are standing on the positive z axis looking toward the origin, the Earth is moving counterclockwise in the xy plane. However, this statement cannot be used mathematically. You can't add "counterclockwise in the xy plane" to "clockwise in the yz plane." Instead, we can use the right hand rule to describe the direction of angular momentum as a vector. The direction can be specified like this:
* The plane in which both the position vector from the two objects and the momentum vector for the object of interest lie can be indicated by a unit vector perpendicular to that plane.
* The direction of motion within the plane (clockwise or counterclockwise) can be indicated by establishing a right hand rule (RHR) for this unit vector.
* Right Hand Rule: Using your right arm, point your arm to represent the "r" vector. Now turn your palm in the direction of the momentum vector. Curl your fingers in that direction of the momentum, and extend your thumb outward. The unit vector representing the direction of the angular momentum is defined to point in the direction of your thumb.
*Hint: If the rotational motion is counterclockwise, your right thumb, therefore the unit vector, will point out of the plane. If the rotational motion is clockwise, the unit vector will point into the plane.

===A Mathematical Model===
The direction of the angular momentum can also be solved through calculating the cross product of the '''r''' and '''p''' vectors.
The cross product of two vectors can be solved as so:

[[File:crossproduct2.jpg]]

[[File:Cross product.jpg]]

The first row is the standard basis vectors and must appear in the order given here. The second row is the components of '''r''' and the third row is the components of '''p'''. First, the terms alternate in sign and notice that the 2x2 is missing the column below the standard basis vector that multiplies it as well as the row of standard basis vectors. To solve for the "i" component, you use the expression: bf - ce. Do the same for the "j" and "k" components, and this will give you your vector. Your vector should be <bf-ce, af-cd, ae-bd>.

For more on the explanation of how to calculate cross product, visit this website: [https://www.mathsisfun.com/algebra/vectors-cross-product.html]

Here is an example problem. Solve first using the right hand rule, and then solve mathematically with the cross product.

[[File:problem1.jpg]]

Answer:
The cross product should be <-12, 0, 0>
So the vector is in the -x direction, which the right hand rule also tells us.

==Examples==

[[File:example1.jpg]]

Ans: 4

[[File:ex2.jpg]]

Ans: 4

[[File:ex3.jpg]]

Ans: 3

[[File:EX4.jpg]]

Ans: 3

[[File:EX5.jpg]]

Ans:
L(A)=L(B)=L(H) = <0, 0, -30>
L(G)=L(C) = <0, 0, 0>
L(D)=L(E)=L(F) = <0, 0, +50>

==Connectedness==
Right Hand Rule is essential for solving all angular momentum problems.
Angular Momentum has been my favorite topic so far this semester. I think the problems and situations are very interesting, and they explain the reason behind why a lot of stuff in real life happen that we don't generally think about.

Here are some interesting videos about the conservation of angular momentum:

https://www.youtube.com/watch?v=Aw5i994n2bw&feature=youtu.be

https://www.youtube.com/watch?v=OKbawIq3w7U

==History==

I wasn't able to find who was the first to apply the Right Hand Rule to angular momentum and torque. However, it seems that the right hand rule is applied to other aspects of physics
as well. For example, André-Marie Ampère, a French physicist and mathematician, created a right hand rule for circuits and electric currents. This is used when a vector must be defined
to represent the rotation of a body, a magnetic field, or a fluid. It reveals a connection between the current and the magnetic field lines in the magnetic field that the current created.
This right hand rule works exactly the same way as the one I have described above.

== See also ==

Once you have learned how to use the Right Hand Rule to determine the direction of angular momentum, you will be able to solve angular momentum and torque problems.

==References==

All pictures and videos were taken from Professor Flavio Fenton's Physics 2211 lecture notes.

[[Category:Which Category did you place this in?]]

Right Hand Rule

2015-11-26T19:17:16Z

Ldarian3: /* References */

This topic covers Right Hand Rule.
Claimed by: Layla Darian

==The Main Idea==
The magnitude of the translational angular momentum doesn't tell us everything about the motion of an object. For example, to describe the motion of Earth's orbit if you are standing on the positive z axis looking toward the origin, the Earth is moving counterclockwise in the xy plane. However, this statement cannot be used mathematically. You can't add "counterclockwise in the xy plane" to "clockwise in the yz plane." Instead, we can use the right hand rule to describe the direction of angular momentum as a vector. The direction can be specified like this:
* The plane in which both the position vector from the two objects and the momentum vector for the object of interest lie can be indicated by a unit vector perpendicular to that plane.
* The direction of motion within the plane (clockwise or counterclockwise) can be indicated by establishing a right hand rule (RHR) for this unit vector.
* Right Hand Rule: Using your right arm, point your arm to represent the "r" vector. Now turn your palm in the direction of the momentum vector. Curl your fingers in that direction of the momentum, and extend your thumb outward. The unit vector representing the direction of the angular momentum is defined to point in the direction of your thumb.
*Hint: If the rotational motion is counterclockwise, your right thumb, therefore the unit vector, will point out of the plane. If the rotational motion is clockwise, the unit vector will point into the plane.

===A Mathematical Model===
The direction of the angular momentum can also be solved through calculating the cross product of the '''r''' and '''p''' vectors.
The cross product of two vectors can be solved as so:

[[File:crossproduct2.jpg]]

[[File:Cross product.jpg]]

The first row is the standard basis vectors and must appear in the order given here. The second row is the components of '''r''' and the third row is the components of '''p'''. First, the terms alternate in sign and notice that the 2x2 is missing the column below the standard basis vector that multiplies it as well as the row of standard basis vectors. To solve for the "i" component, you use the expression: bf - ce. Do the same for the "j" and "k" components, and this will give you your vector. Your vector should be <bf-ce, af-cd, ae-bd>.

For more on the explanation of how to calculate cross product, visit this website: [https://www.mathsisfun.com/algebra/vectors-cross-product.html]

Here is an example problem. Solve first using the right hand rule, and then solve mathematically with the cross product.

[[File:problem1.jpg]]

Answer:
The cross product should be <-12, 0, 0>
So the vector is in the -x direction, which the right hand rule also tells us.

==Examples==

[[File:example1.jpg]]

Ans: 4

[[File:ex2.jpg]]

Ans: 4

[[File:ex3.jpg]]

Ans: 3

[[File:EX4.jpg]]

Ans: 3

[[File:EX5.jpg]]

Ans:
L(A)=L(B)=L(H) = <0, 0, -30>
L(G)=L(C) = <0, 0, 0>
L(D)=L(E)=L(F) = <0, 0, +50>

==Connectedness==
Right Hand Rule is essential for solving all angular momentum problems.
Angular Momentum has been my favorite topic so far this semester. I think the problems and situations are very interesting, and they explain the reason behind why a lot of stuff in real life happen that we don't generally think about.

Here are some interesting videos about the conservation of angular momentum:

[[https://www.youtube.com/watch?v=Aw5i994n2bw&feature=youtu.be]]

[[https://www.youtube.com/watch?v=OKbawIq3w7U]]

==History==

I wasn't able to find who was the first to apply the Right Hand Rule to angular momentum and torque. However, it seems that the right hand rule is applied to other aspects of physics
as well. For example, André-Marie Ampère, a French physicist and mathematician, created a right hand rule for circuits and electric currents. This is used when a vector must be defined
to represent the rotation of a body, a magnetic field, or a fluid. It reveals a connection between the current and the magnetic field lines in the magnetic field that the current created.
This right hand rule works exactly the same way as the one I have described above.

== See also ==

Once you have learned how to use the Right Hand Rule to determine the direction of angular momentum, you will be able to solve angular momentum and torque problems.

==References==

All pictures and videos were taken from Professor Flavio Fenton's Physics 2211 lecture notes.

[[Category:Which Category did you place this in?]]

Right Hand Rule

2015-11-26T19:14:52Z

Ldarian3: /* History */

This topic covers Right Hand Rule.
Claimed by: Layla Darian

==The Main Idea==
The magnitude of the translational angular momentum doesn't tell us everything about the motion of an object. For example, to describe the motion of Earth's orbit if you are standing on the positive z axis looking toward the origin, the Earth is moving counterclockwise in the xy plane. However, this statement cannot be used mathematically. You can't add "counterclockwise in the xy plane" to "clockwise in the yz plane." Instead, we can use the right hand rule to describe the direction of angular momentum as a vector. The direction can be specified like this:
* The plane in which both the position vector from the two objects and the momentum vector for the object of interest lie can be indicated by a unit vector perpendicular to that plane.
* The direction of motion within the plane (clockwise or counterclockwise) can be indicated by establishing a right hand rule (RHR) for this unit vector.
* Right Hand Rule: Using your right arm, point your arm to represent the "r" vector. Now turn your palm in the direction of the momentum vector. Curl your fingers in that direction of the momentum, and extend your thumb outward. The unit vector representing the direction of the angular momentum is defined to point in the direction of your thumb.
*Hint: If the rotational motion is counterclockwise, your right thumb, therefore the unit vector, will point out of the plane. If the rotational motion is clockwise, the unit vector will point into the plane.

===A Mathematical Model===
The direction of the angular momentum can also be solved through calculating the cross product of the '''r''' and '''p''' vectors.
The cross product of two vectors can be solved as so:

[[File:crossproduct2.jpg]]

[[File:Cross product.jpg]]

The first row is the standard basis vectors and must appear in the order given here. The second row is the components of '''r''' and the third row is the components of '''p'''. First, the terms alternate in sign and notice that the 2x2 is missing the column below the standard basis vector that multiplies it as well as the row of standard basis vectors. To solve for the "i" component, you use the expression: bf - ce. Do the same for the "j" and "k" components, and this will give you your vector. Your vector should be <bf-ce, af-cd, ae-bd>.

For more on the explanation of how to calculate cross product, visit this website: [https://www.mathsisfun.com/algebra/vectors-cross-product.html]

Here is an example problem. Solve first using the right hand rule, and then solve mathematically with the cross product.

[[File:problem1.jpg]]

Answer:
The cross product should be <-12, 0, 0>
So the vector is in the -x direction, which the right hand rule also tells us.

==Examples==

[[File:example1.jpg]]

Ans: 4

[[File:ex2.jpg]]

Ans: 4

[[File:ex3.jpg]]

Ans: 3

[[File:EX4.jpg]]

Ans: 3

[[File:EX5.jpg]]

Ans:
L(A)=L(B)=L(H) = <0, 0, -30>
L(G)=L(C) = <0, 0, 0>
L(D)=L(E)=L(F) = <0, 0, +50>

==Connectedness==
Right Hand Rule is essential for solving all angular momentum problems.
Angular Momentum has been my favorite topic so far this semester. I think the problems and situations are very interesting, and they explain the reason behind why a lot of stuff in real life happen that we don't generally think about.

Here are some interesting videos about the conservation of angular momentum:

[[https://www.youtube.com/watch?v=Aw5i994n2bw&feature=youtu.be]]

[[https://www.youtube.com/watch?v=OKbawIq3w7U]]

==History==

I wasn't able to find who was the first to apply the Right Hand Rule to angular momentum and torque. However, it seems that the right hand rule is applied to other aspects of physics
as well. For example, André-Marie Ampère, a French physicist and mathematician, created a right hand rule for circuits and electric currents. This is used when a vector must be defined
to represent the rotation of a body, a magnetic field, or a fluid. It reveals a connection between the current and the magnetic field lines in the magnetic field that the current created.
This right hand rule works exactly the same way as the one I have described above.

== See also ==

Once you have learned how to use the Right Hand Rule to determine the direction of angular momentum, you will be able to solve angular momentum and torque problems.

==References==

All pictures were taken from Professor Flavio Fenton's Physics 2211 lecture notes.

[[Category:Which Category did you place this in?]]

Right Hand Rule

2015-11-26T19:03:03Z

Ldarian3: /* Examples */

This topic covers Right Hand Rule.
Claimed by: Layla Darian

==The Main Idea==
The magnitude of the translational angular momentum doesn't tell us everything about the motion of an object. For example, to describe the motion of Earth's orbit if you are standing on the positive z axis looking toward the origin, the Earth is moving counterclockwise in the xy plane. However, this statement cannot be used mathematically. You can't add "counterclockwise in the xy plane" to "clockwise in the yz plane." Instead, we can use the right hand rule to describe the direction of angular momentum as a vector. The direction can be specified like this:
* The plane in which both the position vector from the two objects and the momentum vector for the object of interest lie can be indicated by a unit vector perpendicular to that plane.
* The direction of motion within the plane (clockwise or counterclockwise) can be indicated by establishing a right hand rule (RHR) for this unit vector.
* Right Hand Rule: Using your right arm, point your arm to represent the "r" vector. Now turn your palm in the direction of the momentum vector. Curl your fingers in that direction of the momentum, and extend your thumb outward. The unit vector representing the direction of the angular momentum is defined to point in the direction of your thumb.
*Hint: If the rotational motion is counterclockwise, your right thumb, therefore the unit vector, will point out of the plane. If the rotational motion is clockwise, the unit vector will point into the plane.

===A Mathematical Model===
The direction of the angular momentum can also be solved through calculating the cross product of the '''r''' and '''p''' vectors.
The cross product of two vectors can be solved as so:

[[File:crossproduct2.jpg]]

[[File:Cross product.jpg]]

The first row is the standard basis vectors and must appear in the order given here. The second row is the components of '''r''' and the third row is the components of '''p'''. First, the terms alternate in sign and notice that the 2x2 is missing the column below the standard basis vector that multiplies it as well as the row of standard basis vectors. To solve for the "i" component, you use the expression: bf - ce. Do the same for the "j" and "k" components, and this will give you your vector. Your vector should be <bf-ce, af-cd, ae-bd>.

For more on the explanation of how to calculate cross product, visit this website: [https://www.mathsisfun.com/algebra/vectors-cross-product.html]

Here is an example problem. Solve first using the right hand rule, and then solve mathematically with the cross product.

[[File:problem1.jpg]]

Answer:
The cross product should be <-12, 0, 0>
So the vector is in the -x direction, which the right hand rule also tells us.

==Examples==

[[File:example1.jpg]]

Ans: 4

[[File:ex2.jpg]]

Ans: 4

[[File:ex3.jpg]]

Ans: 3

[[File:EX4.jpg]]

Ans: 3

[[File:EX5.jpg]]

Ans:
L(A)=L(B)=L(H) = <0, 0, -30>
L(G)=L(C) = <0, 0, 0>
L(D)=L(E)=L(F) = <0, 0, +50>

==Connectedness==
Right Hand Rule is essential for solving all angular momentum problems.
Angular Momentum has been my favorite topic so far this semester. I think the problems and situations are very interesting, and they explain the reason behind why a lot of stuff in real life happen that we don't generally think about.

Here are some interesting videos about the conservation of angular momentum:

[[https://www.youtube.com/watch?v=Aw5i994n2bw&feature=youtu.be]]

[[https://www.youtube.com/watch?v=OKbawIq3w7U]]

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Once you have learned how to use the Right Hand Rule to determine the direction of angular momentum, you will be able to solve angular momentum and torque problems.

==References==

All pictures were taken from Professor Flavio Fenton's Physics 2211 lecture notes.

[[Category:Which Category did you place this in?]]

File:EX5.jpg

2015-11-26T18:59:02Z

Ldarian3:

File:EX4.jpg

2015-11-26T18:58:33Z

Ldarian3:

Right Hand Rule

2015-11-26T18:53:55Z

Ldarian3: /* Examples */

This topic covers Right Hand Rule.
Claimed by: Layla Darian

==The Main Idea==
The magnitude of the translational angular momentum doesn't tell us everything about the motion of an object. For example, to describe the motion of Earth's orbit if you are standing on the positive z axis looking toward the origin, the Earth is moving counterclockwise in the xy plane. However, this statement cannot be used mathematically. You can't add "counterclockwise in the xy plane" to "clockwise in the yz plane." Instead, we can use the right hand rule to describe the direction of angular momentum as a vector. The direction can be specified like this:
* The plane in which both the position vector from the two objects and the momentum vector for the object of interest lie can be indicated by a unit vector perpendicular to that plane.
* The direction of motion within the plane (clockwise or counterclockwise) can be indicated by establishing a right hand rule (RHR) for this unit vector.
* Right Hand Rule: Using your right arm, point your arm to represent the "r" vector. Now turn your palm in the direction of the momentum vector. Curl your fingers in that direction of the momentum, and extend your thumb outward. The unit vector representing the direction of the angular momentum is defined to point in the direction of your thumb.
*Hint: If the rotational motion is counterclockwise, your right thumb, therefore the unit vector, will point out of the plane. If the rotational motion is clockwise, the unit vector will point into the plane.

===A Mathematical Model===
The direction of the angular momentum can also be solved through calculating the cross product of the '''r''' and '''p''' vectors.
The cross product of two vectors can be solved as so:

[[File:crossproduct2.jpg]]

[[File:Cross product.jpg]]

The first row is the standard basis vectors and must appear in the order given here. The second row is the components of '''r''' and the third row is the components of '''p'''. First, the terms alternate in sign and notice that the 2x2 is missing the column below the standard basis vector that multiplies it as well as the row of standard basis vectors. To solve for the "i" component, you use the expression: bf - ce. Do the same for the "j" and "k" components, and this will give you your vector. Your vector should be <bf-ce, af-cd, ae-bd>.

For more on the explanation of how to calculate cross product, visit this website: [https://www.mathsisfun.com/algebra/vectors-cross-product.html]

Here is an example problem. Solve first using the right hand rule, and then solve mathematically with the cross product.

[[File:problem1.jpg]]

Answer:
The cross product should be <-12, 0, 0>
So the vector is in the -x direction, which the right hand rule also tells us.

==Examples==

[[File:example1.jpg]]

Ans: 4

[[File:ex2.jpg]]

Ans: 4

[[File:ex3.jpg]]

Ans: 3

[[File:EX4.jpg]]

Ans:

[[File:EX5.jpg]]

==Connectedness==
Right Hand Rule is essential for solving all angular momentum problems.
Angular Momentum has been my favorite topic so far this semester. I think the problems and situations are very interesting, and they explain the reason behind why a lot of stuff in real life happen that we don't generally think about.

Here are some interesting videos about the conservation of angular momentum:

[[https://www.youtube.com/watch?v=Aw5i994n2bw&feature=youtu.be]]

[[https://www.youtube.com/watch?v=OKbawIq3w7U]]

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Once you have learned how to use the Right Hand Rule to determine the direction of angular momentum, you will be able to solve angular momentum and torque problems.

==References==

All pictures were taken from Professor Flavio Fenton's Physics 2211 lecture notes.

[[Category:Which Category did you place this in?]]

Right Hand Rule

2015-11-26T18:47:25Z

Ldarian3: /* A Mathematical Model */

This topic covers Right Hand Rule.
Claimed by: Layla Darian

==The Main Idea==
The magnitude of the translational angular momentum doesn't tell us everything about the motion of an object. For example, to describe the motion of Earth's orbit if you are standing on the positive z axis looking toward the origin, the Earth is moving counterclockwise in the xy plane. However, this statement cannot be used mathematically. You can't add "counterclockwise in the xy plane" to "clockwise in the yz plane." Instead, we can use the right hand rule to describe the direction of angular momentum as a vector. The direction can be specified like this:
* The plane in which both the position vector from the two objects and the momentum vector for the object of interest lie can be indicated by a unit vector perpendicular to that plane.
* The direction of motion within the plane (clockwise or counterclockwise) can be indicated by establishing a right hand rule (RHR) for this unit vector.
* Right Hand Rule: Using your right arm, point your arm to represent the "r" vector. Now turn your palm in the direction of the momentum vector. Curl your fingers in that direction of the momentum, and extend your thumb outward. The unit vector representing the direction of the angular momentum is defined to point in the direction of your thumb.
*Hint: If the rotational motion is counterclockwise, your right thumb, therefore the unit vector, will point out of the plane. If the rotational motion is clockwise, the unit vector will point into the plane.

===A Mathematical Model===
The direction of the angular momentum can also be solved through calculating the cross product of the '''r''' and '''p''' vectors.
The cross product of two vectors can be solved as so:

[[File:crossproduct2.jpg]]

[[File:Cross product.jpg]]

The first row is the standard basis vectors and must appear in the order given here. The second row is the components of '''r''' and the third row is the components of '''p'''. First, the terms alternate in sign and notice that the 2x2 is missing the column below the standard basis vector that multiplies it as well as the row of standard basis vectors. To solve for the "i" component, you use the expression: bf - ce. Do the same for the "j" and "k" components, and this will give you your vector. Your vector should be <bf-ce, af-cd, ae-bd>.

For more on the explanation of how to calculate cross product, visit this website: [https://www.mathsisfun.com/algebra/vectors-cross-product.html]

Here is an example problem. Solve first using the right hand rule, and then solve mathematically with the cross product.

[[File:problem1.jpg]]

Answer:
The cross product should be <-12, 0, 0>
So the vector is in the -x direction, which the right hand rule also tells us.

==Examples==

[[File:example1.jpg]]

Ans: 4

[[File:ex2.jpg]]

Ans: 4

[[File:ex3.jpg]]

Ans: 3

==Connectedness==
Right Hand Rule is essential for solving all angular momentum problems.
Angular Momentum has been my favorite topic so far this semester. I think the problems and situations are very interesting, and they explain the reason behind why a lot of stuff in real life happen that we don't generally think about.

Here are some interesting videos about the conservation of angular momentum:

[[https://www.youtube.com/watch?v=Aw5i994n2bw&feature=youtu.be]]

[[https://www.youtube.com/watch?v=OKbawIq3w7U]]

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Once you have learned how to use the Right Hand Rule to determine the direction of angular momentum, you will be able to solve angular momentum and torque problems.

==References==

All pictures were taken from Professor Flavio Fenton's Physics 2211 lecture notes.

[[Category:Which Category did you place this in?]]

Right Hand Rule

2015-11-26T18:47:10Z

Ldarian3: /* A Mathematical Model */

This topic covers Right Hand Rule.
Claimed by: Layla Darian

==The Main Idea==
The magnitude of the translational angular momentum doesn't tell us everything about the motion of an object. For example, to describe the motion of Earth's orbit if you are standing on the positive z axis looking toward the origin, the Earth is moving counterclockwise in the xy plane. However, this statement cannot be used mathematically. You can't add "counterclockwise in the xy plane" to "clockwise in the yz plane." Instead, we can use the right hand rule to describe the direction of angular momentum as a vector. The direction can be specified like this:
* The plane in which both the position vector from the two objects and the momentum vector for the object of interest lie can be indicated by a unit vector perpendicular to that plane.
* The direction of motion within the plane (clockwise or counterclockwise) can be indicated by establishing a right hand rule (RHR) for this unit vector.
* Right Hand Rule: Using your right arm, point your arm to represent the "r" vector. Now turn your palm in the direction of the momentum vector. Curl your fingers in that direction of the momentum, and extend your thumb outward. The unit vector representing the direction of the angular momentum is defined to point in the direction of your thumb.
*Hint: If the rotational motion is counterclockwise, your right thumb, therefore the unit vector, will point out of the plane. If the rotational motion is clockwise, the unit vector will point into the plane.

===A Mathematical Model===
The direction of the angular momentum can also be solved through calculating the cross product of the '''r''' and '''p''' vectors.
The cross product of two vectors can be solved as so:

[[File:crossproduct2.jpg]]

[[File:Cross product.jpg]]

The first row is the standard basis vectors and must appear in the order given here. The second row is the components of '''r''' and the third row is the components of '''p'''. First, the terms alternate in sign and notice that the 2x2 is missing the column below the standard basis vector that multiplies it as well as the row of standard basis vectors. To solve for the "i" component, you use the expression: bf - ce. Do the same for the "j" and "k" components, and this will give you your vector. Your vector should be <bf-ce, af-cd, ae-bd>.

For more on the explanation of how to calculate cross product, visit this website: [https://www.mathsisfun.com/algebra/vectors-cross-product.html]

Here is an example problem. Solve first using the right hand rule, and then solve mathematically with the cross product.

[[File:problem1.jpg]]
Answer
The cross product should be <-12, 0, 0>
So the vector is in the -x direction, which the right hand rule also tells us.

==Examples==

[[File:example1.jpg]]

Ans: 4

[[File:ex2.jpg]]

Ans: 4

[[File:ex3.jpg]]

Ans: 3

==Connectedness==
Right Hand Rule is essential for solving all angular momentum problems.
Angular Momentum has been my favorite topic so far this semester. I think the problems and situations are very interesting, and they explain the reason behind why a lot of stuff in real life happen that we don't generally think about.

Here are some interesting videos about the conservation of angular momentum:

[[https://www.youtube.com/watch?v=Aw5i994n2bw&feature=youtu.be]]

[[https://www.youtube.com/watch?v=OKbawIq3w7U]]

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Once you have learned how to use the Right Hand Rule to determine the direction of angular momentum, you will be able to solve angular momentum and torque problems.

==References==

All pictures were taken from Professor Flavio Fenton's Physics 2211 lecture notes.

[[Category:Which Category did you place this in?]]

Right Hand Rule

2015-11-26T18:46:16Z

Ldarian3: /* A Mathematical Model */

This topic covers Right Hand Rule.
Claimed by: Layla Darian

==The Main Idea==
The magnitude of the translational angular momentum doesn't tell us everything about the motion of an object. For example, to describe the motion of Earth's orbit if you are standing on the positive z axis looking toward the origin, the Earth is moving counterclockwise in the xy plane. However, this statement cannot be used mathematically. You can't add "counterclockwise in the xy plane" to "clockwise in the yz plane." Instead, we can use the right hand rule to describe the direction of angular momentum as a vector. The direction can be specified like this:
* The plane in which both the position vector from the two objects and the momentum vector for the object of interest lie can be indicated by a unit vector perpendicular to that plane.
* The direction of motion within the plane (clockwise or counterclockwise) can be indicated by establishing a right hand rule (RHR) for this unit vector.
* Right Hand Rule: Using your right arm, point your arm to represent the "r" vector. Now turn your palm in the direction of the momentum vector. Curl your fingers in that direction of the momentum, and extend your thumb outward. The unit vector representing the direction of the angular momentum is defined to point in the direction of your thumb.
*Hint: If the rotational motion is counterclockwise, your right thumb, therefore the unit vector, will point out of the plane. If the rotational motion is clockwise, the unit vector will point into the plane.

===A Mathematical Model===

The direction of the angular momentum can also be solved through calculating the cross product of the '''r''' and '''p''' vectors.

The cross product of two vectors can be solved as so:

[[File:crossproduct2.jpg]]

[[File:Cross product.jpg]]

The first row is the standard basis vectors and must appear in the order given here. The second row is the components of '''r''' and the third row is the components of '''p'''. First, the terms alternate in sign and notice that the 2x2 is missing the column below the standard basis vector that multiplies it as well as the row of standard basis vectors. To solve for the "i" component, you use the expression: bf - ce. Do the same for the "j" and "k" components, and this will give you your vector. Your vector should be <bf-ce, af-cd, ae-bd>.

For more on the explanation of how to calculate cross product, visit this website: [https://www.mathsisfun.com/algebra/vectors-cross-product.html]

Here is an example problem. Solve first using the right hand rule, and then solve mathematically with the cross product.

[[File:problem1.jpg]]

==Examples==

[[File:example1.jpg]]

Ans: 4

[[File:ex2.jpg]]

Ans: 4

[[File:ex3.jpg]]

Ans: 3

==Connectedness==
Right Hand Rule is essential for solving all angular momentum problems.
Angular Momentum has been my favorite topic so far this semester. I think the problems and situations are very interesting, and they explain the reason behind why a lot of stuff in real life happen that we don't generally think about.

Here are some interesting videos about the conservation of angular momentum:

[[https://www.youtube.com/watch?v=Aw5i994n2bw&feature=youtu.be]]

[[https://www.youtube.com/watch?v=OKbawIq3w7U]]

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Once you have learned how to use the Right Hand Rule to determine the direction of angular momentum, you will be able to solve angular momentum and torque problems.

==References==

All pictures were taken from Professor Flavio Fenton's Physics 2211 lecture notes.

[[Category:Which Category did you place this in?]]

Right Hand Rule

2015-11-26T18:42:58Z

Ldarian3: /* A Mathematical Model */

This topic covers Right Hand Rule.
Claimed by: Layla Darian

==The Main Idea==
The magnitude of the translational angular momentum doesn't tell us everything about the motion of an object. For example, to describe the motion of Earth's orbit if you are standing on the positive z axis looking toward the origin, the Earth is moving counterclockwise in the xy plane. However, this statement cannot be used mathematically. You can't add "counterclockwise in the xy plane" to "clockwise in the yz plane." Instead, we can use the right hand rule to describe the direction of angular momentum as a vector. The direction can be specified like this:
* The plane in which both the position vector from the two objects and the momentum vector for the object of interest lie can be indicated by a unit vector perpendicular to that plane.
* The direction of motion within the plane (clockwise or counterclockwise) can be indicated by establishing a right hand rule (RHR) for this unit vector.
* Right Hand Rule: Using your right arm, point your arm to represent the "r" vector. Now turn your palm in the direction of the momentum vector. Curl your fingers in that direction of the momentum, and extend your thumb outward. The unit vector representing the direction of the angular momentum is defined to point in the direction of your thumb.
*Hint: If the rotational motion is counterclockwise, your right thumb, therefore the unit vector, will point out of the plane. If the rotational motion is clockwise, the unit vector will point into the plane.

===A Mathematical Model===

The direction of the angular momentum can also be solved through calculating the cross product of the '''r''' and '''p''' vectors.

The cross product of two vectors can be solved as so:

[[File:crossproduct2.jpg]]

[[File:Cross product.jpg]]

The first row is the standard basis vectors and must appear in the order given here. The second row is the components of '''r''' and the third row is the components of '''p'''. First, the terms alternate in sign and notice that the 2x2 is missing the column below the standard basis vector that multiplies it as well as the row of standard basis vectors. To solve for the "i" component, you use the expression: bf - ce. Do the same for the "j" and "k" components, and this will give you your vector. Your vector should be <bf-ce, af-cd, ae-bd>.

For more on the explanation of how to calculate cross product, visit this website: [https://www.mathsisfun.com/algebra/vectors-cross-product.html]

Here are some example problems. Try solving the cross product:

[[File:problem1.jpg]]

==Examples==

[[File:example1.jpg]]

Ans: 4

[[File:ex2.jpg]]

Ans: 4

[[File:ex3.jpg]]

Ans: 3

==Connectedness==
Right Hand Rule is essential for solving all angular momentum problems.
Angular Momentum has been my favorite topic so far this semester. I think the problems and situations are very interesting, and they explain the reason behind why a lot of stuff in real life happen that we don't generally think about.

Here are some interesting videos about the conservation of angular momentum:

[[https://www.youtube.com/watch?v=Aw5i994n2bw&feature=youtu.be]]

[[https://www.youtube.com/watch?v=OKbawIq3w7U]]

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Once you have learned how to use the Right Hand Rule to determine the direction of angular momentum, you will be able to solve angular momentum and torque problems.

==References==

All pictures were taken from Professor Flavio Fenton's Physics 2211 lecture notes.

[[Category:Which Category did you place this in?]]

Right Hand Rule

2015-11-25T21:34:13Z

Ldarian3: /* References */

This topic covers Right Hand Rule.
Claimed by: Layla Darian

==The Main Idea==
The magnitude of the translational angular momentum doesn't tell us everything about the motion of an object. For example, to describe the motion of Earth's orbit if you are standing on the positive z axis looking toward the origin, the Earth is moving counterclockwise in the xy plane. However, this statement cannot be used mathematically. You can't add "counterclockwise in the xy plane" to "clockwise in the yz plane." Instead, we can use the right hand rule to describe the direction of angular momentum as a vector. The direction can be specified like this:
* The plane in which both the position vector from the two objects and the momentum vector for the object of interest lie can be indicated by a unit vector perpendicular to that plane.
* The direction of motion within the plane (clockwise or counterclockwise) can be indicated by establishing a right hand rule (RHR) for this unit vector.
* Right Hand Rule: Using your right arm, point your arm to represent the "r" vector. Now turn your palm in the direction of the momentum vector. Curl your fingers in that direction of the momentum, and extend your thumb outward. The unit vector representing the direction of the angular momentum is defined to point in the direction of your thumb.
*Hint: If the rotational motion is counterclockwise, your right thumb, therefore the unit vector, will point out of the plane. If the rotational motion is clockwise, the unit vector will point into the plane.

===A Mathematical Model===

The direction of the angular momentum can also be solved through calculating the cross product of the '''r''' and '''p''' vectors.

The cross product of two vectors can be solved as so:

[[File:crossproduct2.jpg]]

[[File:Cross product.jpg]]

The first row is the standard basis vectors and must appear in the order given here. The second row is the components of '''r''' and the third row is the components of '''p'''. First, the terms alternate in sign and notice that the 2x2 is missing the column below the standard basis vector that multiplies it as well as the row of standard basis vectors. To solve for the "i" component, you use the expression: bf - ce. Do the same for the "j" and "k" components, and this will give you your vector. Your vector should be <bf-ce, af-cd, ae-bd>.

For more on the explanation of how to calculate cross product, visit this website: [https://www.mathsisfun.com/algebra/vectors-cross-product.html]

Here are some example problems. Try solving the cross product:

[[File:problem1.jpg]]

Click [[File:problem1ans.jpg]] for the answer

==Examples==

[[File:example1.jpg]]

Ans: 4

[[File:ex2.jpg]]

Ans: 4

[[File:ex3.jpg]]

Ans: 3

==Connectedness==
Right Hand Rule is essential for solving all angular momentum problems.
Angular Momentum has been my favorite topic so far this semester. I think the problems and situations are very interesting, and they explain the reason behind why a lot of stuff in real life happen that we don't generally think about.

Here are some interesting videos about the conservation of angular momentum:

[[https://www.youtube.com/watch?v=Aw5i994n2bw&feature=youtu.be]]

[[https://www.youtube.com/watch?v=OKbawIq3w7U]]

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Once you have learned how to use the Right Hand Rule to determine the direction of angular momentum, you will be able to solve angular momentum and torque problems.

==References==

All pictures were taken from Professor Flavio Fenton's Physics 2211 lecture notes.

[[Category:Which Category did you place this in?]]

Right Hand Rule

2015-11-25T21:33:39Z

Ldarian3: /* See also */

This topic covers Right Hand Rule.
Claimed by: Layla Darian

==The Main Idea==
The magnitude of the translational angular momentum doesn't tell us everything about the motion of an object. For example, to describe the motion of Earth's orbit if you are standing on the positive z axis looking toward the origin, the Earth is moving counterclockwise in the xy plane. However, this statement cannot be used mathematically. You can't add "counterclockwise in the xy plane" to "clockwise in the yz plane." Instead, we can use the right hand rule to describe the direction of angular momentum as a vector. The direction can be specified like this:
* The plane in which both the position vector from the two objects and the momentum vector for the object of interest lie can be indicated by a unit vector perpendicular to that plane.
* The direction of motion within the plane (clockwise or counterclockwise) can be indicated by establishing a right hand rule (RHR) for this unit vector.
* Right Hand Rule: Using your right arm, point your arm to represent the "r" vector. Now turn your palm in the direction of the momentum vector. Curl your fingers in that direction of the momentum, and extend your thumb outward. The unit vector representing the direction of the angular momentum is defined to point in the direction of your thumb.
*Hint: If the rotational motion is counterclockwise, your right thumb, therefore the unit vector, will point out of the plane. If the rotational motion is clockwise, the unit vector will point into the plane.

===A Mathematical Model===

The direction of the angular momentum can also be solved through calculating the cross product of the '''r''' and '''p''' vectors.

The cross product of two vectors can be solved as so:

[[File:crossproduct2.jpg]]

[[File:Cross product.jpg]]

The first row is the standard basis vectors and must appear in the order given here. The second row is the components of '''r''' and the third row is the components of '''p'''. First, the terms alternate in sign and notice that the 2x2 is missing the column below the standard basis vector that multiplies it as well as the row of standard basis vectors. To solve for the "i" component, you use the expression: bf - ce. Do the same for the "j" and "k" components, and this will give you your vector. Your vector should be <bf-ce, af-cd, ae-bd>.

For more on the explanation of how to calculate cross product, visit this website: [https://www.mathsisfun.com/algebra/vectors-cross-product.html]

Here are some example problems. Try solving the cross product:

[[File:problem1.jpg]]

Click [[File:problem1ans.jpg]] for the answer

==Examples==

[[File:example1.jpg]]

Ans: 4

[[File:ex2.jpg]]

Ans: 4

[[File:ex3.jpg]]

Ans: 3

==Connectedness==
Right Hand Rule is essential for solving all angular momentum problems.
Angular Momentum has been my favorite topic so far this semester. I think the problems and situations are very interesting, and they explain the reason behind why a lot of stuff in real life happen that we don't generally think about.

Here are some interesting videos about the conservation of angular momentum:

[[https://www.youtube.com/watch?v=Aw5i994n2bw&feature=youtu.be]]

[[https://www.youtube.com/watch?v=OKbawIq3w7U]]

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Once you have learned how to use the Right Hand Rule to determine the direction of angular momentum, you will be able to solve angular momentum and torque problems.

==References==

This section contains the the references you used while writing this page

[[Category:Which Category did you place this in?]]

Right Hand Rule

2015-11-25T21:32:17Z

Ldarian3: /* Connectedness */

This topic covers Right Hand Rule.
Claimed by: Layla Darian

==The Main Idea==
The magnitude of the translational angular momentum doesn't tell us everything about the motion of an object. For example, to describe the motion of Earth's orbit if you are standing on the positive z axis looking toward the origin, the Earth is moving counterclockwise in the xy plane. However, this statement cannot be used mathematically. You can't add "counterclockwise in the xy plane" to "clockwise in the yz plane." Instead, we can use the right hand rule to describe the direction of angular momentum as a vector. The direction can be specified like this:
* The plane in which both the position vector from the two objects and the momentum vector for the object of interest lie can be indicated by a unit vector perpendicular to that plane.
* The direction of motion within the plane (clockwise or counterclockwise) can be indicated by establishing a right hand rule (RHR) for this unit vector.
* Right Hand Rule: Using your right arm, point your arm to represent the "r" vector. Now turn your palm in the direction of the momentum vector. Curl your fingers in that direction of the momentum, and extend your thumb outward. The unit vector representing the direction of the angular momentum is defined to point in the direction of your thumb.
*Hint: If the rotational motion is counterclockwise, your right thumb, therefore the unit vector, will point out of the plane. If the rotational motion is clockwise, the unit vector will point into the plane.

===A Mathematical Model===

The direction of the angular momentum can also be solved through calculating the cross product of the '''r''' and '''p''' vectors.

The cross product of two vectors can be solved as so:

[[File:crossproduct2.jpg]]

[[File:Cross product.jpg]]

The first row is the standard basis vectors and must appear in the order given here. The second row is the components of '''r''' and the third row is the components of '''p'''. First, the terms alternate in sign and notice that the 2x2 is missing the column below the standard basis vector that multiplies it as well as the row of standard basis vectors. To solve for the "i" component, you use the expression: bf - ce. Do the same for the "j" and "k" components, and this will give you your vector. Your vector should be <bf-ce, af-cd, ae-bd>.

For more on the explanation of how to calculate cross product, visit this website: [https://www.mathsisfun.com/algebra/vectors-cross-product.html]

Here are some example problems. Try solving the cross product:

[[File:problem1.jpg]]

Click [[File:problem1ans.jpg]] for the answer

==Examples==

[[File:example1.jpg]]

Ans: 4

[[File:ex2.jpg]]

Ans: 4

[[File:ex3.jpg]]

Ans: 3

==Connectedness==
Right Hand Rule is essential for solving all angular momentum problems.
Angular Momentum has been my favorite topic so far this semester. I think the problems and situations are very interesting, and they explain the reason behind why a lot of stuff in real life happen that we don't generally think about.

Here are some interesting videos about the conservation of angular momentum:

[[https://www.youtube.com/watch?v=Aw5i994n2bw&feature=youtu.be]]

[[https://www.youtube.com/watch?v=OKbawIq3w7U]]

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Which Category did you place this in?]]

Right Hand Rule

2015-11-25T21:28:08Z

Ldarian3: /* Connectedness */

This topic covers Right Hand Rule.
Claimed by: Layla Darian

==The Main Idea==
The magnitude of the translational angular momentum doesn't tell us everything about the motion of an object. For example, to describe the motion of Earth's orbit if you are standing on the positive z axis looking toward the origin, the Earth is moving counterclockwise in the xy plane. However, this statement cannot be used mathematically. You can't add "counterclockwise in the xy plane" to "clockwise in the yz plane." Instead, we can use the right hand rule to describe the direction of angular momentum as a vector. The direction can be specified like this:
* The plane in which both the position vector from the two objects and the momentum vector for the object of interest lie can be indicated by a unit vector perpendicular to that plane.
* The direction of motion within the plane (clockwise or counterclockwise) can be indicated by establishing a right hand rule (RHR) for this unit vector.
* Right Hand Rule: Using your right arm, point your arm to represent the "r" vector. Now turn your palm in the direction of the momentum vector. Curl your fingers in that direction of the momentum, and extend your thumb outward. The unit vector representing the direction of the angular momentum is defined to point in the direction of your thumb.
*Hint: If the rotational motion is counterclockwise, your right thumb, therefore the unit vector, will point out of the plane. If the rotational motion is clockwise, the unit vector will point into the plane.

===A Mathematical Model===

The direction of the angular momentum can also be solved through calculating the cross product of the '''r''' and '''p''' vectors.

The cross product of two vectors can be solved as so:

[[File:crossproduct2.jpg]]

[[File:Cross product.jpg]]

The first row is the standard basis vectors and must appear in the order given here. The second row is the components of '''r''' and the third row is the components of '''p'''. First, the terms alternate in sign and notice that the 2x2 is missing the column below the standard basis vector that multiplies it as well as the row of standard basis vectors. To solve for the "i" component, you use the expression: bf - ce. Do the same for the "j" and "k" components, and this will give you your vector. Your vector should be <bf-ce, af-cd, ae-bd>.

For more on the explanation of how to calculate cross product, visit this website: [https://www.mathsisfun.com/algebra/vectors-cross-product.html]

Here are some example problems. Try solving the cross product:

[[File:problem1.jpg]]

Click [[File:problem1ans.jpg]] for the answer

==Examples==

[[File:example1.jpg]]

Ans: 4

[[File:ex2.jpg]]

Ans: 4

[[File:ex3.jpg]]

Ans: 3

==Connectedness==
Right Hand Rule is essential for solving all angular momentum problems.
Angular Momentum has been my favorite topic so far this semester. I think the problems and situations are very interesting, and they explain the reason behind why a lot of stuff in real life happen that we don't generally think about.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Which Category did you place this in?]]

Right Hand Rule

2015-11-25T21:20:14Z

Ldarian3: /* Examples */

This topic covers Right Hand Rule.
Claimed by: Layla Darian

==The Main Idea==
The magnitude of the translational angular momentum doesn't tell us everything about the motion of an object. For example, to describe the motion of Earth's orbit if you are standing on the positive z axis looking toward the origin, the Earth is moving counterclockwise in the xy plane. However, this statement cannot be used mathematically. You can't add "counterclockwise in the xy plane" to "clockwise in the yz plane." Instead, we can use the right hand rule to describe the direction of angular momentum as a vector. The direction can be specified like this:
* The plane in which both the position vector from the two objects and the momentum vector for the object of interest lie can be indicated by a unit vector perpendicular to that plane.
* The direction of motion within the plane (clockwise or counterclockwise) can be indicated by establishing a right hand rule (RHR) for this unit vector.
* Right Hand Rule: Using your right arm, point your arm to represent the "r" vector. Now turn your palm in the direction of the momentum vector. Curl your fingers in that direction of the momentum, and extend your thumb outward. The unit vector representing the direction of the angular momentum is defined to point in the direction of your thumb.
*Hint: If the rotational motion is counterclockwise, your right thumb, therefore the unit vector, will point out of the plane. If the rotational motion is clockwise, the unit vector will point into the plane.

===A Mathematical Model===

The direction of the angular momentum can also be solved through calculating the cross product of the '''r''' and '''p''' vectors.

The cross product of two vectors can be solved as so:

[[File:crossproduct2.jpg]]

[[File:Cross product.jpg]]

The first row is the standard basis vectors and must appear in the order given here. The second row is the components of '''r''' and the third row is the components of '''p'''. First, the terms alternate in sign and notice that the 2x2 is missing the column below the standard basis vector that multiplies it as well as the row of standard basis vectors. To solve for the "i" component, you use the expression: bf - ce. Do the same for the "j" and "k" components, and this will give you your vector. Your vector should be <bf-ce, af-cd, ae-bd>.

For more on the explanation of how to calculate cross product, visit this website: [https://www.mathsisfun.com/algebra/vectors-cross-product.html]

Here are some example problems. Try solving the cross product:

[[File:problem1.jpg]]

Click [[File:problem1ans.jpg]] for the answer

==Examples==

[[File:example1.jpg]]

Ans: 4

[[File:ex2.jpg]]

Ans: 4

[[File:ex3.jpg]]

Ans: 3

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Which Category did you place this in?]]

File:Ex3.jpg

2015-11-25T21:18:58Z

Ldarian3:

File:Example1.jpg

2015-11-25T21:17:59Z

Ldarian3:

Right Hand Rule

2015-11-25T21:17:28Z

Ldarian3: /* Examples */

This topic covers Right Hand Rule.
Claimed by: Layla Darian

==The Main Idea==
The magnitude of the translational angular momentum doesn't tell us everything about the motion of an object. For example, to describe the motion of Earth's orbit if you are standing on the positive z axis looking toward the origin, the Earth is moving counterclockwise in the xy plane. However, this statement cannot be used mathematically. You can't add "counterclockwise in the xy plane" to "clockwise in the yz plane." Instead, we can use the right hand rule to describe the direction of angular momentum as a vector. The direction can be specified like this:
* The plane in which both the position vector from the two objects and the momentum vector for the object of interest lie can be indicated by a unit vector perpendicular to that plane.
* The direction of motion within the plane (clockwise or counterclockwise) can be indicated by establishing a right hand rule (RHR) for this unit vector.
* Right Hand Rule: Using your right arm, point your arm to represent the "r" vector. Now turn your palm in the direction of the momentum vector. Curl your fingers in that direction of the momentum, and extend your thumb outward. The unit vector representing the direction of the angular momentum is defined to point in the direction of your thumb.
*Hint: If the rotational motion is counterclockwise, your right thumb, therefore the unit vector, will point out of the plane. If the rotational motion is clockwise, the unit vector will point into the plane.

===A Mathematical Model===

The direction of the angular momentum can also be solved through calculating the cross product of the '''r''' and '''p''' vectors.

The cross product of two vectors can be solved as so:

[[File:crossproduct2.jpg]]

[[File:Cross product.jpg]]

The first row is the standard basis vectors and must appear in the order given here. The second row is the components of '''r''' and the third row is the components of '''p'''. First, the terms alternate in sign and notice that the 2x2 is missing the column below the standard basis vector that multiplies it as well as the row of standard basis vectors. To solve for the "i" component, you use the expression: bf - ce. Do the same for the "j" and "k" components, and this will give you your vector. Your vector should be <bf-ce, af-cd, ae-bd>.

For more on the explanation of how to calculate cross product, visit this website: [https://www.mathsisfun.com/algebra/vectors-cross-product.html]

Here are some example problems. Try solving the cross product:

[[File:problem1.jpg]]

Click [[File:problem1ans.jpg]] for the answer

==Examples==

[[File:example1.jpg]]

Ans:

[[File:ex2.jpg]]

Ans:

[[File:ex3.jpg]]

Ans:

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Which Category did you place this in?]]

File:Ex2.jpg

2015-11-25T21:16:54Z

Ldarian3:

Right Hand Rule

2015-11-25T21:16:04Z

Ldarian3: /* Examples */

This topic covers Right Hand Rule.
Claimed by: Layla Darian

==The Main Idea==
The magnitude of the translational angular momentum doesn't tell us everything about the motion of an object. For example, to describe the motion of Earth's orbit if you are standing on the positive z axis looking toward the origin, the Earth is moving counterclockwise in the xy plane. However, this statement cannot be used mathematically. You can't add "counterclockwise in the xy plane" to "clockwise in the yz plane." Instead, we can use the right hand rule to describe the direction of angular momentum as a vector. The direction can be specified like this:
* The plane in which both the position vector from the two objects and the momentum vector for the object of interest lie can be indicated by a unit vector perpendicular to that plane.
* The direction of motion within the plane (clockwise or counterclockwise) can be indicated by establishing a right hand rule (RHR) for this unit vector.
* Right Hand Rule: Using your right arm, point your arm to represent the "r" vector. Now turn your palm in the direction of the momentum vector. Curl your fingers in that direction of the momentum, and extend your thumb outward. The unit vector representing the direction of the angular momentum is defined to point in the direction of your thumb.
*Hint: If the rotational motion is counterclockwise, your right thumb, therefore the unit vector, will point out of the plane. If the rotational motion is clockwise, the unit vector will point into the plane.

===A Mathematical Model===

The direction of the angular momentum can also be solved through calculating the cross product of the '''r''' and '''p''' vectors.

The cross product of two vectors can be solved as so:

[[File:crossproduct2.jpg]]

[[File:Cross product.jpg]]

The first row is the standard basis vectors and must appear in the order given here. The second row is the components of '''r''' and the third row is the components of '''p'''. First, the terms alternate in sign and notice that the 2x2 is missing the column below the standard basis vector that multiplies it as well as the row of standard basis vectors. To solve for the "i" component, you use the expression: bf - ce. Do the same for the "j" and "k" components, and this will give you your vector. Your vector should be <bf-ce, af-cd, ae-bd>.

For more on the explanation of how to calculate cross product, visit this website: [https://www.mathsisfun.com/algebra/vectors-cross-product.html]

Here are some example problems. Try solving the cross product:

[[File:problem1.jpg]]

Click [[File:problem1ans.jpg]] for the answer

==Examples==

[[File:ex1.jpg]]

Ans:

[[File:ex2.jpg]]

Ans:

[[File:ex3.jpg]]

Ans:

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Which Category did you place this in?]]

Right Hand Rule

2015-11-25T20:57:56Z

Ldarian3: /* A Computational Model */

This topic covers Right Hand Rule.
Claimed by: Layla Darian

==The Main Idea==
The magnitude of the translational angular momentum doesn't tell us everything about the motion of an object. For example, to describe the motion of Earth's orbit if you are standing on the positive z axis looking toward the origin, the Earth is moving counterclockwise in the xy plane. However, this statement cannot be used mathematically. You can't add "counterclockwise in the xy plane" to "clockwise in the yz plane." Instead, we can use the right hand rule to describe the direction of angular momentum as a vector. The direction can be specified like this:
* The plane in which both the position vector from the two objects and the momentum vector for the object of interest lie can be indicated by a unit vector perpendicular to that plane.
* The direction of motion within the plane (clockwise or counterclockwise) can be indicated by establishing a right hand rule (RHR) for this unit vector.
* Right Hand Rule: Using your right arm, point your arm to represent the "r" vector. Now turn your palm in the direction of the momentum vector. Curl your fingers in that direction of the momentum, and extend your thumb outward. The unit vector representing the direction of the angular momentum is defined to point in the direction of your thumb.
*Hint: If the rotational motion is counterclockwise, your right thumb, therefore the unit vector, will point out of the plane. If the rotational motion is clockwise, the unit vector will point into the plane.

===A Mathematical Model===

The direction of the angular momentum can also be solved through calculating the cross product of the '''r''' and '''p''' vectors.

The cross product of two vectors can be solved as so:

[[File:crossproduct2.jpg]]

[[File:Cross product.jpg]]

The first row is the standard basis vectors and must appear in the order given here. The second row is the components of '''r''' and the third row is the components of '''p'''. First, the terms alternate in sign and notice that the 2x2 is missing the column below the standard basis vector that multiplies it as well as the row of standard basis vectors. To solve for the "i" component, you use the expression: bf - ce. Do the same for the "j" and "k" components, and this will give you your vector. Your vector should be <bf-ce, af-cd, ae-bd>.

For more on the explanation of how to calculate cross product, visit this website: [https://www.mathsisfun.com/algebra/vectors-cross-product.html]

Here are some example problems. Try solving the cross product:

[[File:problem1.jpg]]

Click [[File:problem1ans.jpg]] for the answer

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
===Middling===
===Difficult===

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Which Category did you place this in?]]

Right Hand Rule

2015-11-25T20:54:02Z

Ldarian3: /* A Computational Model */

This topic covers Right Hand Rule.
Claimed by: Layla Darian

==The Main Idea==
The magnitude of the translational angular momentum doesn't tell us everything about the motion of an object. For example, to describe the motion of Earth's orbit if you are standing on the positive z axis looking toward the origin, the Earth is moving counterclockwise in the xy plane. However, this statement cannot be used mathematically. You can't add "counterclockwise in the xy plane" to "clockwise in the yz plane." Instead, we can use the right hand rule to describe the direction of angular momentum as a vector. The direction can be specified like this:
* The plane in which both the position vector from the two objects and the momentum vector for the object of interest lie can be indicated by a unit vector perpendicular to that plane.
* The direction of motion within the plane (clockwise or counterclockwise) can be indicated by establishing a right hand rule (RHR) for this unit vector.
* Right Hand Rule: Using your right arm, point your arm to represent the "r" vector. Now turn your palm in the direction of the momentum vector. Curl your fingers in that direction of the momentum, and extend your thumb outward. The unit vector representing the direction of the angular momentum is defined to point in the direction of your thumb.
*Hint: If the rotational motion is counterclockwise, your right thumb, therefore the unit vector, will point out of the plane. If the rotational motion is clockwise, the unit vector will point into the plane.

===A Mathematical Model===

The direction of the angular momentum can also be solved through calculating the cross product of the '''r''' and '''p''' vectors.

The cross product of two vectors can be solved as so:

[[File:crossproduct2.jpg]]

[[File:Cross product.jpg]]

The first row is the standard basis vectors and must appear in the order given here. The second row is the components of '''r''' and the third row is the components of '''p'''. First, the terms alternate in sign and notice that the 2x2 is missing the column below the standard basis vector that multiplies it as well as the row of standard basis vectors. To solve for the "i" component, you use the expression: bf - ce. Do the same for the "j" and "k" components, and this will give you your vector. Your vector should be <bf-ce, af-cd, ae-bd>.

For more on the explanation of how to calculate cross product, visit this website: [https://www.mathsisfun.com/algebra/vectors-cross-product.html]

Here are some example problems. Try solving the cross product:

[[File:problem1.jpg]]

Click [[File:problem1ans.jpg]] for the answer

===A Computational Model===

How do we visualize or predict using this topic. Consider embedding some vpython code here [https://trinket.io/glowscript/31d0f9ad9e Teach hands-on with GlowScript]
IHIOHO

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
===Middling===
===Difficult===

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Which Category did you place this in?]]

Right Hand Rule

2015-11-25T20:45:20Z

Ldarian3: /* A Mathematical Model */

This topic covers Right Hand Rule.
Claimed by: Layla Darian

==The Main Idea==
The magnitude of the translational angular momentum doesn't tell us everything about the motion of an object. For example, to describe the motion of Earth's orbit if you are standing on the positive z axis looking toward the origin, the Earth is moving counterclockwise in the xy plane. However, this statement cannot be used mathematically. You can't add "counterclockwise in the xy plane" to "clockwise in the yz plane." Instead, we can use the right hand rule to describe the direction of angular momentum as a vector. The direction can be specified like this:
* The plane in which both the position vector from the two objects and the momentum vector for the object of interest lie can be indicated by a unit vector perpendicular to that plane.
* The direction of motion within the plane (clockwise or counterclockwise) can be indicated by establishing a right hand rule (RHR) for this unit vector.
* Right Hand Rule: Using your right arm, point your arm to represent the "r" vector. Now turn your palm in the direction of the momentum vector. Curl your fingers in that direction of the momentum, and extend your thumb outward. The unit vector representing the direction of the angular momentum is defined to point in the direction of your thumb.
*Hint: If the rotational motion is counterclockwise, your right thumb, therefore the unit vector, will point out of the plane. If the rotational motion is clockwise, the unit vector will point into the plane.

===A Mathematical Model===

The direction of the angular momentum can also be solved through calculating the cross product of the '''r''' and '''p''' vectors.

The cross product of two vectors can be solved as so:

[[File:crossproduct2.jpg]]

[[File:Cross product.jpg]]

The first row is the standard basis vectors and must appear in the order given here. The second row is the components of '''r''' and the third row is the components of '''p'''. First, the terms alternate in sign and notice that the 2x2 is missing the column below the standard basis vector that multiplies it as well as the row of standard basis vectors. To solve for the "i" component, you use the expression: bf - ce. Do the same for the "j" and "k" components, and this will give you your vector. Your vector should be <bf-ce, af-cd, ae-bd>.

For more on the explanation of how to calculate cross product, visit this website: [https://www.mathsisfun.com/algebra/vectors-cross-product.html]

Here are some example problems. Try solving the cross product:

[[File:problem1.jpg]]

Click [[File:problem1ans.jpg]] for the answer

===A Computational Model===

How do we visualize or predict using this topic. Consider embedding some vpython code here [https://trinket.io/glowscript/31d0f9ad9e Teach hands-on with GlowScript]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
===Middling===
===Difficult===

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Which Category did you place this in?]]

Right Hand Rule

2015-11-25T20:39:22Z

Ldarian3: /* A Mathematical Model */

This topic covers Right Hand Rule.
Claimed by: Layla Darian

==The Main Idea==
The magnitude of the translational angular momentum doesn't tell us everything about the motion of an object. For example, to describe the motion of Earth's orbit if you are standing on the positive z axis looking toward the origin, the Earth is moving counterclockwise in the xy plane. However, this statement cannot be used mathematically. You can't add "counterclockwise in the xy plane" to "clockwise in the yz plane." Instead, we can use the right hand rule to describe the direction of angular momentum as a vector. The direction can be specified like this:
* The plane in which both the position vector from the two objects and the momentum vector for the object of interest lie can be indicated by a unit vector perpendicular to that plane.
* The direction of motion within the plane (clockwise or counterclockwise) can be indicated by establishing a right hand rule (RHR) for this unit vector.
* Right Hand Rule: Using your right arm, point your arm to represent the "r" vector. Now turn your palm in the direction of the momentum vector. Curl your fingers in that direction of the momentum, and extend your thumb outward. The unit vector representing the direction of the angular momentum is defined to point in the direction of your thumb.
*Hint: If the rotational motion is counterclockwise, your right thumb, therefore the unit vector, will point out of the plane. If the rotational motion is clockwise, the unit vector will point into the plane.

===A Mathematical Model===

The direction of the angular momentum can also be solved through calculating the cross product of the '''r''' and '''p''' vectors.

The cross product of two vectors can be solved as so:

[[File:crossproduct2.jpg]]

[[File:Cross product.jpg]]

The first row is the standard basis vectors and must appear in the order given here. The second row is the components of '''r''' and the third row is the components of '''p'''. First, the terms alternate in sign and notice that the 2x2 is missing the column below the standard basis vector that multiplies it as well as the row of standard basis vectors. To solve for the "i" component, you use the expression: bf - ce. Do the same for the "j" and "k" components, and this will give you your vector. Your vector should be <bf-ce, af-cd, ae-bd>.

For more on the explanation of how to calculate cross product, visit this website: [https://www.mathsisfun.com/algebra/vectors-cross-product.html]

Here are some example problems. Try solving the cross product:

[[File:problem1.jpg]]
Click [[File:problem1ans.jpg]] for the answer

===A Computational Model===

How do we visualize or predict using this topic. Consider embedding some vpython code here [https://trinket.io/glowscript/31d0f9ad9e Teach hands-on with GlowScript]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
===Middling===
===Difficult===

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Which Category did you place this in?]]

File:Problem1ans.jpg

2015-11-18T05:35:02Z

Ldarian3: