Fundamentals of Resistance

claimed by Benjamin Flamm

Resistors are elements that are involved in circuits and oppose the flow of current. This page gives examples of computing resistance as well as the history and applications of resistors.

The Main Idea

Resistors have many forms throughout modern technology and are applied in electronic industries ranging from basic manufacturing (lightbulbs, portable devices, etc.) to advanced biomedical instrumentation such as electrocardiogram devices. (electronicdesign.com) In almost everything, resistance exists but for this page only wires and manufactured resistors are used as examples.

The primary goal of a resistor is to limit the current that flows through a circuit. For example, a lightbulb is a very simple application of Tungsten or another material that has a high resistance. As electrons flow into the lightbulb, they begin to collide with themselves and the high number of charge carriers in the high-resistance filament. The result of these collisions is energy released as light and heat. See the Mathematical Model section for the relationship of these factors and how they determine resistance.

A Mathematical Model

Resistance can be modeled by starting at the fundamental concept [math]\displaystyle{ {I = |q|nA\bar{v}} }[/math] where [math]\displaystyle{ I }[/math] is conventional current, [math]\displaystyle{ |q| }[/math] is the magnitude of the charge being carried, [math]\displaystyle{ n }[/math] is the number of charge carriers, [math]\displaystyle{ A }[/math] is the area of the resistor, and [math]\displaystyle{ \bar{v} }[/math] is the drift speed of the charge.

[math]\displaystyle{ {I = |q|nA\bar{v} = |q|nAuE} }[/math] and [math]\displaystyle{ {J = \frac IA } }[/math] the equation for current density

Grouping the properties of the material together and utilizing the equation for conductivity [math]\displaystyle{ {\sigma = |q|nu} }[/math]:

[math]\displaystyle{ {I = (|q|nu)AE = {\sigma}AE} }[/math]

[math]\displaystyle{ {J = \frac IA = {\sigma}E} }[/math]

Substituting in the equation for electric field we get [math]\displaystyle{ {\frac IA = {\sigma}\frac {{\Delta}V}{L}} }[/math]

Finally, using algebra we attain [math]\displaystyle{ {I = \dfrac{{\Delta}V}{\dfrac{L}{{\sigma}A}} = \frac {{\Delta}V}{R}} }[/math]

Resulting in the definition of resistance being [math]\displaystyle{ {R = \dfrac {L}{{\sigma}{A}}} }[/math]

Although resistance can be easily derived and calculated, the majority of problems that contain resistance involve circuit analysis and Ohm's Law [math]\displaystyle{ {{\Delta}V = IR} }[/math] in which resistance is usually provided beforehand.

A Computational Model

Given a diagram consisting of a thinner wire leading into a thicker wire and then back through a thinner wire, it is possible to visualize how the cross-sectional area of a wire affects its resistance. The two thin sections of wire are identical and labeled with areas A\sub{1} and the thick section of wire has an area of A2.

Charge carriers (electrons in the diagram above) are flowing through the three wires with the same charge but are forced to pass through regions with varying cross-sectional areas. Analyzing the equation (R=l/sigA) it is obvious that resistance is inversely proportional to its cross-sectional area but observing the above diagram gives a more solidifying explanation of why this relationship exists.

So let's say electron flow is from left to right; traveling through the first thin wire, the thick wire, and then finally exiting through the second thin wire. The charge of these electrons are equal throughout each individual wire, as well as the number of charge carriers and the mobility of these charges. Observing the diagram, you can see that the electron flow is more dense in the thinner wires. This is due to the fact that the electrons have less room to move around and are constantly colliding with each other (makes sense, less area:more crowded). In the thick wire, the same amount of electrons are passing through but with much more space which means less collisions and ultimately less resistance.

Examples

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

Simple

Calculate the resistance of a copper wire with length of 10m and a diameter of 1.63mm (conductivity of copper is sigma = 1.7e-8)

Given that the diameter = 1.63mm, find the cross-sectional area of the wire [math]\displaystyle{ {A = \pi{r^2} = \pi{.815^-3}} }[/math]

Now we have the length, conductivity, and area which is sufficient to solve the equation r = yada

Middling

Calculate the internal resistance of a battery that has a terminal voltage of 7.992V, an emf of 8V, and is connected in series with a .7 (ohm) resistor.

Connectedness

Resistors are used all throughout industry specifically power electronics. Although they have a simple purpose (limiting current flow) they have a very important application in power system projects around the world. Engineers return to the fundamentals of voltage and resistance when designing basic grid systems in developing countries.

Electrical and computer engineering students use wire wound resistors every semester in order to build circuits involving push-buttons or any type of active-low/active-high switches.

Resistors specific purpose in industry is particularly boring; however, they play a major role in many electronics such as the laptop you are reading this on and the jets flying overhead.

History

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

See also

Series Circuits

Parallel Circuits

Parallel vs. Series Circuits

Further reading

Introduction to Resistors

Wirewound Resistors

Resistance Fundamentals

Buy Resistors for Experiments

References

Chabay, Ruth W. Matter and Interactions: Electric and Magnetic Interactions. Place of Publication Not Identified: John Wiley, 2015. Print.

Resistivity of Copper