Question

With the help of a neatly drawn and labelled diagram, obtain balancing condition of a Wheatstone bridge.

Answer 1

The potential at B is equal to potential at D. So no, current flows through the galvanometer. This is called balanced condition of Wheatstone's bridge.
In closed circuit ABDA
$i_{1}P-i_{2}R=0$
$i_{1}P=i_{2}R$ .....$\left(1\right)$
In closed mash BCDB.
$i_{1}Q-i_{2}S=0$
$i_{1}Q=i_{2}S$ ......$\left(2\right)$
Divide Equation $\left(1\right)$ by equation $\left(2\right)$
$\frac{i_{1}P}{i_{1}Q}=\frac{i_{2}R}{i_{2}S}$
We get,

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