# 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

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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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.