Question

Derive the relation between the resistances of the arms of a Wheatstone bridge in its balance condition. Write its two applications.

Answer 1

Look at the diagram above This is a wheatstone bridge.

You can notice four resistances P,Q,R and S connected on the bridge.

You can also notice a galvanometer connected between points B and D.

The four resistances are adjusted in such a way that the current through the galvanometer becomes zero.

This shows that B and D are at the same electric potential.

$\therefore V_B=V_D$

The potential difference across B will be $V_A-V_B$

Similarly across Q will be $V_A-V_D$

Similarly across R will be $V_B-V_C$

Similarly across S will be $V_D-V_C$

By Ohm's law we know that $V=IR$ where $I$ is the current.

$\therefore V_A-V_B=IP$ $,V_A-V_D=IQ$ $,V_B-V_C=IR$ and $V_D-V_C=IS$

$\because V_B=V_D$ we get

$IP=IQ$ and $IR=IS$

By dividing the 1st by the 2nd, we get

$\frac{IP}{IR}=\frac{IQ}{IS}$

$\therefore \frac{P}{Q}=\frac{R}{S}$ this equation is known as the balancing condition in a wheatstone bridge