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Why is using a Wheatstone bridge such an accurate way of calculating an unknown resistance? What are the benefits of using it over Ohm's law?

It seems that it has something to do with the wires heating up during ohm's law calculations, and the fact that an ammeter/voltmeter still draw some current and provide some resistance which makes results unreliable.

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A Wheatstone bridge is only a very sensitive way of finding a match between two resistances. You have got to have a nice, well calibrated variable resistance on the opposite side to the unknown resistance. The accuracy with which you can judge the unknown is limited by the calibration accuracy of the variable resistance.

Using Ohm's law to find an unknown resistance is a great way to do it. You just match the voltage you apply to the unknown based on the range of resistance you expect to encounter. Suppose you had a short piece of heavy copper buss bar you wanted to measure the resistance of. I have a nice AC welder that can put out 400 amps for brief periods. I will short the output with the buss bar, put 400 amps to it and while the current is flowing, I will measure the voltage difference between each end of the bar. With such a small resistance in the unknown resistor, it takes a lot of amps to generate a nice measurable voltage.

Conversely, the mega high resistance of thick layers of insulation can be measured by putting a couple of tens of kilovolts to it and reading the milliamperage. I use neon sign transformers and large resistor banks for stepping down and for amp measurements.

I have a series of calibrated shunts that are useful. They run a millivolt per amp. You read .5 volts of potential from one end to the other of the shunt - that is 500 amps. I have a big brass shunt on my car dash that gives me a reading of starting amps. I ran old welding cable to bring the starter amps into the passenger compartment. I have a giant switch on the dash, custom made of 1/4" x 2" copper bussbar in several pieces up to a foot long. No auto thief would ever have the balls to close that switch. I have skulls and red warnings. High voltage, etc.

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Good information, but most of it doesn't really go towards answering the question... –  David Z May 30 '11 at 4:34
    
His premise that the bridge is is an accurate way of calculating an unknown resistance is false. It cannot calculate anything. It can only compare to a standard. If you want to find an unknown resistance and you do not happen to have an exact match laying around that you can use in a bridge configuration then you are forced to use Ohms law. –  Bill Slugg May 30 '11 at 23:28
    
Admittedly it's loose terminology, but comparing to a standard can be referred to as "calculating". In any case, yes, it's fair to point out that using Ohm's law (actually: using a voltmeter and ammeter, since even the bridge method is based on Ohm's law) is more practical, but your answer winds up being mostly about how to apply Ohm's law, which I don't see as particularly relevant. Everything after the second sentence of your second paragraph is unnecessary. (Not that it makes your answer bad or anything like that.) –  David Z May 30 '11 at 23:48

The reasons can be found here: Even in the most simplistic case when all Rs (they are reference resistors with low tolerance error) are equal the sensitivity to any deviation is very large (last eq in the page). See the structure of the eq put some numbers to see that the numerator goes near zero. It is temperature balanced because the effects in the ensemble cancel (the ratio of resistance changes are near 1).
A lot of photos here, interesting. Years ago I saw a very sophisticated Wheatstone bridge used to measure the parameters of a transatlantic submarine communications cable. The maintenance team had to know where (at what distance in the ocean) the cable was faulty.

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Advantages: You can use DC or AC as your source.. bridge rectifier is there.

You are using null instrument method.. highly sensitive and better resolution than deflection type.

You already know rest of the parameters and your instruments are already calibrated (zeroed) based on the same, so calculating a single unknown variable is not a pain.

You can do compensation even for lead wire resistances.

Actually you do not encounter errors of voltage or current measurement, the main emphasis is on zeroing the line current.

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......Absolutely right. –  Curious Nov 11 '12 at 18:57

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