Calculating resistance from a graph

JizzaDaMan

Hi, i have a graph with voltage (y axis) plotted against current (x axis) and I need to calculate the resistance of a resistor and a lamp from this.

My physics teacher has told me that by Ohm's law, R = V/I, and he has also told me that on the graph, the resistance is the gradient of the graph.

The graph for the resistor is linear, so V/I is the same as the gradient, I have no problem with this. However, the graph for the lamp is non linear, so V/I isn't the same as the gradient.

What's the correct answer here? Am I missing something blindingly obvious? Thanks for any help

berkeman

Good question. You would need to figure out what the operating point is (what current are you running at -- what point are you at on the graph?), and use ΔV/ΔI at that point. Does that make sense?

JizzaDaMan

It makes perfect sense to me :D but what doesn't is that if I calculate the gradient at that point, it isn't the same as V/I. surely this contradicts Ohm's law?

berkeman

Why do you say that? And don't call me Shirley.

JizzaDaMan

Well if the resistance of the lamp is ΔV/ΔI at any point on the graph, then given that the relationship between the voltage and current for the lamp is non-linear, this will not be V/I which by Ohm's law is the formula for resistance.

berkeman

The original definition of Ohms Law was just the linear equation V=IR. But it can be generalized to describe an Impedance Z which is ∂V/∂I. This is used all the time in EE. We talk about the "Impedance at a point" pretty frequently (because there are many non-linear circuits that you will work with...).

sophiecentaur

Ohm's Law only describes the behaviour of a metal at constant temperature. That's all. The Resistance of a component is defined as V/I but, of course, as V changes, this WILL change for a non-linear component (by definition).
There is no violation of any law here because Ohm's Law is not really a physical 'Law', in any case - it is just an observation of how a certain material tends to behave under given conditions.
For 'small signals', many non-linear components behave as if they have a 'resistance' because ∂V/∂I happens to be more or less constant over a particular range of V. So, under certain circumstances, we can assign a value for R and work with it.
EE does this all the time. We use 'current sources' and 'voltage sources' all the time and it works very well in practical terms and it's the same sort of thing. You must not let it spoil your day - just be glad that the system works so well.

JizzaDaMan

Thank you guys, I'm pretty sure I understand that now :)