Why Does a Wheatstone Bridge Offer Higher Accuracy in Resistance Measurements?

[uchicago2012]

Homework Statement

1. Why is the Wheatstone bridge measurement capable of so much greater accuracy than the measurement or resistance with voltmeters and ammeters, using R = V/I?
and
2. R in this experiment is a resistance box and so is accurate to the nearest ohm. Is this uncertainty in R more or less important than the uncertainties in L1 and L2?

Homework Equations

Attached doc for procedure

The Attempt at a Solution

I'm not sure about these two questions for my post-lab. For 1, I thought that the Wheatstone bridge would be more accurate because it's independent of the supply voltages, other than from thermal inaccuracies. But I don't know why beyond that- maybe that's all the question is getting at? Not sure.
And for 2, I really don't know. Both are in the equation? So why would one be 'more important' than the other, anyway?

Thanks for any insight

 

[darkys]

. :) §§ COMUpdate 1.For question 1: The Wheatstone bridge has greater accuracy because it can measure the resistance of a device more accurately than V/I, since the Wheatstone bridge cancels out any effects from the supply voltage due to the balance in the bridge. This is known as galvanic isolation, and it allows the bridge to measure the resistance of the device more accurately than with a voltmeter and ammeter. For question 2: The uncertainty in R is more important, since it is directly used in the equation to calculate the resistance of the device. The uncertainties in L1 and L2 are not used in the equation, so they do not affect the accuracy of the measurement as much as the uncertainty in R.