Resistance from I vs. V graph problem

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Homework Help Overview

The discussion revolves around understanding resistance as derived from an I vs. V graph, particularly focusing on the implications for linear versus non-linear relationships. Participants explore the definitions and calculations of resistance in different contexts.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants question whether resistance is the reciprocal of the gradient or simply the voltage divided by current, especially in non-linear scenarios. There is discussion about the implications of Ohm's law and how it applies to different types of conductors.

Discussion Status

Some participants provide insights into the definitions of resistance and its calculation, noting that for non-linear graphs, resistance can vary and may not be directly derived from the slope. There is an acknowledgment of the differences between Ohmic and non-Ohmic conductors, with some guidance offered on how to approach resistance calculations in these contexts.

Contextual Notes

Participants are considering specific cases, such as a filament lamp, and the potential for different resistance values at varying voltages. The discussion reflects on the definitions and assumptions underlying resistance measurements.

nokia8650
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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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Since the gradient's defined as \frac{\delta{y}}{\delta{x}}, 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.
 
nokia8650 said:
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

I = \frac{1}{R}\cdot V

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.
 
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
 
R = V/I

just follow the graph across at 15 volts and find the Watts. then the volts by the watts.
 
Hi nokia8650,

nokia8650 said:
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 anyone measurement.
 

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