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So I collected current and voltage data from a simple circuit with a power source and a resistor, using a multimeter. I created a graph for this data using excel and got the y-intercept (which is basically the value of the current when the voltage was zero) and found it to be a small negative number:

enter image description here

What I'm wondering about is what is the physical explanation of this value, which by ohm's law should have been zero. Is it simply experimental error in the data collected by the multimeter? Or does it have any deeper physical meaning?

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The answer is almost certainly experimental error, but it will help if you can post the image. –  Nathaniel Mar 1 '13 at 11:55
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It's strange to call it "experimental error". There are systematic errors and statistical error and this is likely to be an ordinary statistical error, an unavoidable one. Experimental measurements simply can't ever be exact and linear regression has no reason to exactly cross the origin, either. If you know that $U=0$ for $I=0$ and vice versa, good for you, you may also insert it as an extra condition for your laws. But if you don't assume it and do linear regression, there's no reason why you should exactly derive this law. You will derive it at most within an error margin. –  Luboš Motl Mar 1 '13 at 12:47
    
To expand on Lubos's comment, if you look at the confidence interval (or p-value or some other reasonable test statistic) you will almost certainly find that the fit is consistent with a line going through the origin with some high degree of confidence. If not then there is probably some systematic error. Can you get a plot which includes the origin? –  Michael Brown Mar 1 '13 at 12:52
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If you are confident on your physical intuition that the line should pass through origin, you can set that condition in Microsoft Excel. If the $R^2$ is still above 0.97, it's good enough. But here's the probable reason for the non-zero y-intercept: Ohm's law assumes that in your apparatus, multimeter has ideally zero resistance when measuring current and ideally infinite resistance when measuring voltage. The deviation you get from zero i.e. 0.015 is explained from this if the least count of your instrument is comparable to this value. Is this the case? –  Cheeku Mar 1 '13 at 14:03
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2 Answers 2

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Assuming your circuit is as simple as sounds, then if the voltage across the resistor is zero the steady state current through it must be zero. The negative intercept is likely to be due to experimental error, especially since you are extrapolating quite a long way down to zero. You really need to measure more point between $V = 0$ and $V = 2$ to get any further.

There are circumstances in which you need a finite voltage before the current increases from zero, for example if your circuit contains a forward biased diode. This would give you a negative $y$ intercept.

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Like John Rennie, I'm assuming the circuit is just a resistor and a voltage supply. In that case, you are correct: there should be no current at zero voltage.

The most obvious (to me) source of this "discrepancy" is the uncertainty in your current measurements. (The uncertainties in your voltage measurements is also relevant, but those are probably small enough that you can ignore them.) Your fitted slope and intercept will have uncertainties. Those uncertainties are functions of your raw data points and the uncertainties in those raw data points. I don't know if Excel gives you the uncertainties in your fit parameters; if it does, I wouldn't know how to get it. But you could work that for yourself. Wolfram gives the form of those uncertainties, but I've also seen it in introductory error analysis books. Normally you would want to have your fitting software do that, or write a program to do it. But if Excel won't give you that information, and with only 6 data points, it might not be too difficult to do it by hand. My guess is that you'll find that the uncertainty in the y-intercept is larger than the fitted value you got for the y-intercept, meaning that your data is consistent with a y-intercept of zero.

If you still have access to the equipment, you might try measuring the current with zero applied voltage to see what your multimeter gives you. You might also fill in some more of the region in your plot between $V=0$ and $V=2$. The first would indicate if there is any systematic bias in your multimeter. The second would change your fit parameters and their uncertainties, which might make that discrepancy go away.

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