Question

Use Kirchhoff's rules to obtain condition for the balance condition in a Wheatstone bridge

Answer 1

When the bridge is balanced, no current will flow through galvanometer G.

Now using the Kirchhoff's junction law the currents are distributed in the four arms of the Wheatstone bridge as shown in figure.

Applying Kirchhoff's voltage law

for loop ABDA, $i_{1}P-i_{2}R=0\Rightarrow \frac{i_1}{i_2}=\frac{R}{P}....\left(1\right)$

and for loop BCDB, $i_{1}Q-i_{2}S=0\Rightarrow \frac{i_1}{i_2}=\frac{S}{Q}....\left(2\right)$

From (1) and (2), $\frac{P}{R}=\frac{Q}{S}$, this is the balanced condition for the bridge.