Resistance from I vs. V graph


by nokia8650
Tags: graph, resistance
nokia8650
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#1
May21-08, 10:33 AM
P: 228
Hi,

I had a query regarding the resistance from and I vs. V graph. Am i correct in thinking that the resistance is the recipricol of the GRADIENT at any given point? Or is the resistance simply the voltage through it at that point divided by the current? The two are obviously the same for a linear graph, but what about for non-linear graphs?

Thanks
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astrorob
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May21-08, 10:54 AM
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Since the gradient's defined as [tex]\frac{\delta{y}}{\delta{x}}[/tex], if your y axis represents voltage and your x axis represents current then the gradient will simply be the resistance. If your axes are reversed, then indeed it will be the reciprocal of the gradient.

In terms of non-linear plots, the resistance will vary, and so taking the derivative of the graph at each point will give you the resistance at that point.
Hootenanny
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May21-08, 10:58 AM
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Quote Quote by nokia8650 View Post
Hi,

I had a query regarding the resistance from and I vs. V graph. Am i correct in thinking that the resistance is the recipricol of the GRADIENT at any given point? Or is the resistance simply the voltage through it at that point divided by the current? The two are obviously the same for a linear graph, but what about for non-linear graphs?

Thanks
From Ohm's law one may write

[tex]I = \frac{1}{R}\cdot V[/tex]

So yes, where the graph of I vs. V is linear then the gradient represents the resistance of the component. Components that exhibit linear I vs. V graphs are called Ohmic conductors, since they obey Ohm's law.

Non-Ohmic conductors have non-linear I vs. V plots and therefore do not obey Ohm's law. Needless to say, one cannot apply Ohm's law to non-ohmic conductors and so it doesn't really make sense to say that a non-ohmic conductor has a resistance R because it's value depends on the current and voltage through the component.

nokia8650
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#4
May21-08, 12:16 PM
P: 228

Resistance from I vs. V graph


Thank you both for the help. So if I was given say a graph of a filament lamp for I vs. V, and asked to find the resistance at V=15V, would I finding the gradient of the graph, and then find the recipricol - would simply dividing the voltage by the current at that point give an incorrect answer?

Thanks
DeanBH
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#5
May21-08, 01:01 PM
P: 83
R = V/I

just follow the graph across at 15 volts and find the Watts. then the volts by the watts.
alphysicist
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#6
May21-08, 01:20 PM
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Hi nokia8650,

Quote Quote by nokia8650 View Post
Thank you both for the help. So if I was given say a graph of a filament lamp for I vs. V, and asked to find the resistance at V=15V, would I finding the gradient of the graph, and then find the recipricol - would simply dividing the voltage by the current at that point give an incorrect answer?

Thanks
Dividing the voltage by the current would give the correct answer (to within the uncertainties in your measurement), because that's the definition of resistance: the ratio of the voltage to the current caused by that voltage being applied. This resistance will normally be different at different voltages unless the object obeys Ohm's law (and filament lamps do not obey Ohm's law).

The slope of your curve (or 1/slope depending on what's on your y-axis) will not give the resistance except if the material obeys Ohms law. There is a quantity (differential resistance?) that is given by the slope of the curve, but again it's only equal to the resistance for Ohmic resistors.

I believe the reason the slope is often used to calculate the resistance for ohmic resistors is that it minimizes the effects of experimental errors from any one measurement.


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