Question

Use Kirchhoff's rule to obtain conditions for the balance condition in a Wheatstone bridge.

Answer 1

We apply Kirchoff's current law in the shown circuit.

At junction B,

$i_1=i_g+i_3$

At junction D,

$i_2+i_g=i_4$

If current through the galvanometer is zero,

$i_g=0$

thus $i_1=i_3$

and $i_2=i_4$

Applying Kirchoff's voltage law for loop ABDA,

$i_{1}P+i_{g}G=i_{2}R$

Applying Kirchoff's voltage law for loop BCDB,

$i_{3}Q=i_{4}S+i_{g}G$

When $i_g=0$,

$i_{1}P=i_{2}R$

and $i_{3}Q=i_{4}S$

But $i_1=i_3$ and $i_2=i_4$,

Therefore $\frac{P}{Q}=\frac{R}{S}$