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

Answer: 6

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

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