Calculating resistance from a graph

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To calculate resistance from a graph of voltage versus current, Ohm's law (R = V/I) applies to linear components like resistors, where resistance equals the gradient. For non-linear components like lamps, resistance varies with voltage and current, requiring the use of ΔV/ΔI at a specific operating point instead of a simple V/I calculation. This does not contradict Ohm's law, as it can be generalized to describe impedance, which accounts for non-linear behavior. The resistance of non-linear components can change with different operating conditions, but for small signals, they can often be treated as having a constant resistance over a limited range. Understanding these concepts is crucial for analyzing circuits effectively.
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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
 
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JizzaDaMan said:
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

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?
 
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?
 
JizzaDaMan said:
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?

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

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

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