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A resistor of unknown resistance can be found by constructing a series circuit, containing a power supply, ammeter and voltmeter across the unknown resistor, then Resistance=Voltage/Current. In a high school exam question, it says by using a potentiometer in the circuit, a more accurate value of the unknown resistance can be obtained by calculating the average value or plotting a graph. But a more detailed explanation was not given. Thanks.

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For each time you vary the current, you can note that $V = I R$. You can do this for several different settings of $I$ and record the voltage drop across the resistor $V$. You can then find an average value for $R$. And for each pair of $V$ and $I$ you can plot a graph of $V$ versus $I$ and the gradient will be $R$.

In both cases you will get a more accurate value for resistance $R$ as oppose to one measurement.

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Every measurement is subject to errors coming from a whole lot of sources: the intrinsic error of the measuring apparatus, mechanical dependencies of the measurement, outside temperature and pressure and so on. A single measurement of a given quantity has an error associated to it and it cannot be taken for granted. For example you could measure the resistence of a resistor using your apparatus and doing it in two different time of the day and probably get a different result, which one is the right one?

The answer is that there's no intrinsically right one, but there's surely a better one. One way of getting a better, more consistent and reliable result is to exploit the linear relation between the current passing through the resistor and the voltage applied to it since $$V = RI$$ If we take measurements of the current at different voltages, for example at $V = 1, 2, 3, 5, 10, 13, 15, 17, 20, 23, 25, 27, 30$ volts (the bigger the dataset, better the result), we would measure different currents. Now let me do a simple simulation of the measurement: I supposed that the current measuring device has a $3\%$ error and the error on the voltage is negligible whereas the resistor has an, unknown, resistence of $300\Omega$

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This is what you would get if you plotted the measured current vs the applied voltage and then by doing a linear regression.

Since we're fitting a line $I(V) = V/R$ the slope of this line is going to give the value of $1/R$. The result of this linear regression gave

$$ \frac{1}{R} = (3.31\pm 0.15 )\times 10^{-3} S\implies R = (301\pm 13)\Omega$$

If this does not seem accurate, just have e look at the simulated data: for every point $R$ came out to be $R=166, 385, 535, 274, 347, 306, 251, 292, 346, 321, 280, 318, 294$.

Obviously this is an extreme case, but I hope it gives you an understanding of why one should exploit the possibly known behaviour of some quantity to aid the experimental results.

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Its not guaranteed that the setups others describe will make a more accurate measurement than merely measuring the resistance in the simple circuit multiple times. All we've really done is add one more source of error -- the potentiometer and its associated wiring. Indeed, its very plausible that it could make things worse! It would be better to simply average multiple results without the potentiometer!

However, there are circuits which use a potentiometer to make more precise readings. Circuits like Wheatstone Bridges and Kelvin Bridges are designed for precisely measuring resistances of unknown resistors.

There's some devils in the details, of course. These potentiometers must be calibrated. This is a sort of chicken-and-the-egg problem, as we would need a bridge to calibrate the potenitometer used in the bridge. But it's okay in practice. Often the issue we are exploring here is not one of maximum precision, but cost-effective precision. One can afford to have a single very expensive calibrated potentiometer, and then use it in a bridge to measure the unknown resistances of many cheaper resistors. The problem, as given, doesn't specify how expensive the potentiometer is, and whether its calibration certificates are up to date, but we'll assume they are!

These bridges are designed to solve problems that arise in practical settings. If you have a reasonable sized resistor, like a 1kohm resistor, the simple test apparatus is fine. However, if you have a 1Mohm unknown resistor, the current flowing through your voltmeter starts to matter. Voltmeters are not ideal -- they do let some current flow through them. Its a small current, but it's there. And in parallel with a 1Mohm resistor, that could actually skew the results. The Whetstone bridge is designed specifically such that no current flows through the meter when you're measuring. Your measurement is taken when the voltage across the meter is 0V.

Likewise, if you're measuring something like a 1ohm unknown resistor, the simple circuit is fraught with challenges. When you have a large amount of current flowing through the system, it gets harder to make good measurements, due to practical issues like wire resistances. In these situations, bridges like the Kelvin Bridge are very effective at minimizing current flow through critical paths to keep the measured quantities in a range that don't cause errors.

Thus, by using not only a calibrated potentiometer, but a few calibrated resistors as well, one can measure these "difficult" unknown resistors with accuracy. Which bridge should you use? Well, it depends on the resistor, but it depends in a very coarse-grained way. You don't need to know that a resistor is 1.24532ohm to decide that the kelvin bridge is the right one. You might measure it with a simple setup and say "whoa... that's below 10ohm. This simple setup probably isn't sufficient, I should swap to a kelvin bridge."

Another advantage which isn't included in the problem at all: with these bridges you can meaningfully control the current through the resistor without saturating your measurement device. You just have to swap out the other resistors to adjust the current through the test device. When you get into more realistic precise measurement of resistors, you are going to start caring about non-ideal behaviors. At high current, resistors heat up, and change resistance. You can test these non-ideal behaviors better in circuits that you have more control over!

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An old fashion voltmeter (as opposed to a digital voltmeter) draws current. That means the reading on the ammeter is not accurate for the resistor. A good potentiometer balances the unknown voltage against an known voltage without drawing current from the unknown.

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