# Wheatstone Bridge Experiment.

jvdamdshdt
Hello Forum, I did an experiment on a Wheatstone bridge. We are to calculate the theoretical resistance of a given set of wires, then by using a value closest to the theoretical we are to set the resistance on a decade box. After the resistances are set we try to find the point on the bridge where the current is balanced using a galvanometer. My question is below.

## Homework Statement

Why is it more accurate to start a Wheatstone bridge at the middle?

None

## The Attempt at a Solution

It it because the potential will be different?

## Answers and Replies

Homework Helper
I think you used a potentiometer to set the balance of the Bridge. You have chosen the constant resistance ( set by the decade box) closest to the theoretical value of the unknown resistance. You expect them only slightly different. The balance condition is Rx/Ro=R1/R2, where Ro is the decade resistance and R1 and R2 are the resistances of the two parts of the potentiometer. If Rx=Ro R1=R2 follows, the two parts of the potentiometer are equal, its slide contact is at the middle. So you set it at the middle initially, and then move slightly in both directions checking if the galvanometer current decreases.
It can be derived that the bridge will give most accurate results (i.e. the fractional error in Rx, dRx/Rx, will be the smallest for a small change of the bridge ratio if the bridge balances near its center. That is why you set the constant resistance about equal to the unknown one.

ehild

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