Question

In a Wheatstone's network, $P=2 \, \Omega$ , $Q=2 \, \Omega$ , $R=2 \, \Omega$ and $S=3 \, \Omega$ . Find the resistance (in $\Omega$ ) with which $S$ is to be shunted, in order that the bridge gets balanced.

Answer 1

Let a resistance $r \, \Omega$ be shunted with resistance $S$ , so that the bridge is balanced.
Let $S'$ be the resultant resistance of $S$ and $r$ , then
In balanced position

$\frac{P}{Q}=\frac{R}{S^{′}}$
$\frac{2}{2}=\frac{2}{S^{′}}$
$\therefore \, S^{′}=2 \, \Omega$
Now,
$\frac{1}{S^{′}}=\frac{1}{S}+\frac{1}{r}$
$\frac{1}{r}=\frac{1}{S^{′}}-\frac{1}{S}=\frac{1}{2}-\frac{1}{3}=\frac{3 - 2}{6}$
$\frac{1}{r}=\frac{1}{6}$
$r=6 \, \Omega$