RL Circuits: Difference between revisions

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#How is it connected to your major?
#How is it connected to your major?


In biochemistry, microscopes are used a lot to view cells and other extremely small objects. Some hardware additions to microscope use RL circuits to help images appear more clear to the viewer.
In biology, microscopes are used a lot to view cells and other extremely small objects. Some hardware additions to microscope use RL circuits to help images appear more clear to the viewer.


#Is there an interesting industrial application?
#Is there an interesting industrial application?

Revision as of 21:55, 13 April 2016

RL Circuits CLAIMED BY VJ SERVERA

The Main Idea

An RL circuit is one that contains resistors and inductors. The simplest type of these circuits involves one resistor and one inductor, and is called a first order RL circuit. The order of the circuit is equal to the number of inductors involved in the circuit. In this section we will study the following circuit:

A Mathematical Model

First we must remember the relationship between current going through an inductor and the voltage drop across it. The relationship is simply

[math]\displaystyle{ {L\frac{d\vec{i}}{dt}} = \vec{v(t)} }[/math]

meaning that the derivative of the current going through the inductor is proportional to the voltage drop across it. The proportionality constant is the inductance [math]\displaystyle{ L }[/math].

Before the switch closes, we know there is no current going through the inductor , and therefore no voltage. The initial conditions are:

[math]\displaystyle{ v(0) = 0 }[/math]

[math]\displaystyle{ i(0) = 0 }[/math]


Now applying Kirchoff's Voltage Law (loop-rule) we obtain the first order differential equation:

[math]\displaystyle{ V = iR + L\frac{di}{dt} }[/math]


Now we solve the equation. Rearranging:

[math]\displaystyle{ \frac{di}{dt} = \frac{V-iR}{L} }[/math]

[math]\displaystyle{ \frac{di}{V-iR} = \frac{dt}{L} }[/math]

Integrating both sides:

[math]\displaystyle{ -\frac{\ln(V-iR)}{R} = \frac{t}{L} + K }[/math]


Since [math]\displaystyle{ i = 0 }[/math] when [math]\displaystyle{ t = 0 }[/math], plugging these values in gives us

[math]\displaystyle{ K = -\frac{\ln(V)}{R} }[/math]

Substituting K back into our equation, getting rid of the denominator, and collecting the logarithms, we have

[math]\displaystyle{ \ln(\frac{V-Ri}{V}) = -\frac{R}{L}t }[/math]

Exponentiating both sides and solving for [math]\displaystyle{ i }[/math]

[math]\displaystyle{ i = \frac{V}{R}(1 - \exp(-(\frac{R}{L}t))) }[/math]

// finish up the proof

A Computational Model

Inductor simulation

This simple online applet shows how the current through the inductor behaves as time passes. Eventually, the inductor acts as a straight wire and the curren through it is constant.

Examples

Be sure to show all steps in your solution and include diagrams whenever possible

Simple

Middling

Difficult

Connectedness

  1. How is this topic connected to something that you are interested in?

Im interested in how brain waves are read through EEGs. EEGs involve RL circuits to read brain signals properly and give that information to the viewer.

  1. How is it connected to your major?

In biology, microscopes are used a lot to view cells and other extremely small objects. Some hardware additions to microscope use RL circuits to help images appear more clear to the viewer.

  1. Is there an interesting industrial application?

RL circuits are used universally; anywhere signals from the outside, such as noise or light, are used for data or analysis, RL circuits allow for hardware to recieve these signals and filter out the ones that are important.

History

In the mid-19th century, Joseph Henry discovered the principle of self inductance. By wrapping coil around a magnet and sending a current through the coil, he realized that a magnetic field was created and energy could be stored in it through the "induced" voltage that is generated across the inductor.

See also

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Further reading

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External links

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References

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