I don't understand the negative part of the graph. It shows that the resistance is decreasing as the voltage goes from negative towards 0.
What does a negative voltage mean and why does the graph have the shape that it does in that region?
Thanks
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$\begingroup$ What happens to the temperature of the filament as you go from low to high voltage? $\endgroup$– CuriousOneCommented May 11, 2015 at 18:06
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$\begingroup$ @CuriousOne: It increases, so the resistance should increase too. But the increasing gradient in the "negative voltage to 0 voltage" part of the graph means that the resistance is decreasing. $\endgroup$– user45220Commented May 11, 2015 at 18:09
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$\begingroup$ The power heating the filament is the product of voltage and current, both are negative on the left side of the graph, so the power is positive (one can never get power out of a filament, so it has to be positive). The filament is therefor hot at both the right side of the graph and at the left side. Another way of saying this is that a filament operates independently of the polarity of the applied voltage. Is that your question? $\endgroup$– CuriousOneCommented May 11, 2015 at 18:16
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$\begingroup$ @CuriousOne: Sorry, just saw your comment for some reason. Yes, that is my question but I'm still finding difficulty explaining the negative quadrant with reference to the graph. For example, for the positive quadrant I would explain it like this: The gradient is decreasing with voltage, and since the gradient is I/V=1/R, it follows that 1/R is decreasing, meaning that R is increasing. So the resistance is increasing in the positive quadrant of the graph, which is what we expect. But when I try to.. $\endgroup$– user45220Commented May 13, 2015 at 17:40
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1$\begingroup$ Flip the graph by 90 degrees at the diagonal, so that the voltage becomes a function of the current $U=U(I)$. Then you can see that the first derivative $dU/dI$ increases strongly for large currents. It's also always positive. That's the differential resistance of the filament. $\endgroup$– CuriousOneCommented May 13, 2015 at 17:46
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1 Answer
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A negative voltage means that you have hooked your power supply across your device backwards. Purely resistive devices, like resistors and lamp filaments, don't care which way current flows through them, only how much current flows. Such devices will always have current-voltage curves which are "odd" functions, with $I(-V) = -I(V)$, as in your graph. It is safe for you to cover up all but the upper-right quadrant of your graph.
The current-voltage characteristic is curved for large $|V|$ because your filament has a larger resistance when hot (that is, when the power $P=IV$ is large).
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$\begingroup$ That definitely makes sense, but I still don't know why this particular way of reasoning about it fails: "In the negative quadrant the gradient of the graph is increasing with increasing voltage. This means that gradient=I/V=1/R is increasing, implying that R is decreasing. So the resistance is decreasing with voltage." -- which is obviously wrong. How do we correct this reasoning by applying what you said in your answer? Thanks! $\endgroup$ Commented May 13, 2015 at 19:10
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$\begingroup$ @user45220 The voltage is "increasing" towards zero. $\endgroup$ Commented May 13, 2015 at 19:41
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$\begingroup$ @Random832: Yes that's what I said in my argument: As the voltage increases from negative to 0, the gradient of the graph increases. Therefore I/V=1/R increases with increasing voltage, i.e. R decreases with increasing voltage. This is clearly wrong (R has to increase with increasing voltage), but I don't see what's wrong with the argument! $\endgroup$ Commented May 13, 2015 at 19:44
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1$\begingroup$ @user45220 Why is it clearly wrong? In your graph, resistance is at its lowest around zero, and increases as the magnitude of voltage increases in any direction. It's only a trick of math that a negative voltage having its magnitude reduced is "increasing", so I don't see why "R has to increase with increasing voltage". $\endgroup$ Commented May 13, 2015 at 19:45
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$\begingroup$ @Random832: Ah, so it's the magnitude of the current and voltage that matters! I completely missed this point before, maybe I'm just dumb. Thanks! $\endgroup$ Commented May 13, 2015 at 19:47