# Wheatstone bridge galvanometer error

We had to measure the resistance of $R_x$, we balanced the Wheatstone bridge and did calculations. My question is: we didn't include galvanometer error into calculations. Why is that? I read that it's very precise, but that doesn't seem like a good enough explanation in exact sciences :/

Edit: The precision is not the case as I was told, I need to go into more detail. When I measured, I set a value on the potentiometer, then I adjusted the adjustable resistor, so that the galvanometer would show zero. Is the error somehow compensated?

ɛ - electromotive force. K - switch. R - adjustable resistor. C - potentiometer.

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–  Qmechanic Nov 11 '12 at 18:11

## 2 Answers

What it means by very precise is that the error from the galvanometer is significantly less then the error within your system. The Wheatstone Bridge calculations do not require serious precision; the galvanometer are the least of your concerns. IF you were using this device in some type of research environment that required highly calibrated and precise equipment, this device may need to be considered if critical to the experiment (it probably would not be).

The error of the galvonometer may be on the order of $\pm .0001$ or smaller while the rest of your error may be more worthy of mention around $\pm .1$ or $\pm.01$ or something to that effect. I would not worry about it, but if you are writing a lab report it would be a good idea to write about the possibility of it being considered; if the device tells you the error associated with it (on a sticker on the device or in the manuel), I would write it down.

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I updated my question, the precision is not the case. –  Indrek R Nov 11 '12 at 14:06

The huge advantage of bridge measurements is that you're only using the galvanometer to determine when the current between nodes C and D is 0. For this particular case, it's easy to calibrate the galvanometer exactly (or as good as your eyesight, anyway): before you apply any voltage to the circuit, note the galvanometer reading: that's 0!

Compare this technique to a straightforward V/I measurement, where the current meter could read any value, and will in general have some error.

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