Thank you for reporting, we will resolve it shortly
Q.
In a Wheatstone’s network, $P = 2 \, \Omega , Q = 2\, \Omega , R = 2\, \Omega$ and$ S = 3\,\Omega$. The resistance with which $S$ is to be shunted in order that the bridge may be balanced is
Given, $ P=2 \Omega$
$Q=2 \Omega $
$R=2 \Omega $
$S=3 \Omega$
Let by using of $x$ resistance. $S$ to be shunted.
So, according to the Wheatstones network, we have
$\frac{P}{Q} =\frac{R}{S} $
$\frac{P}{O} =\frac{R}{\frac{S \times x}{S+x}} $
$\frac{2}{2} =\frac{2}{\frac{3 \times x}{3+x}} $
$1 =\frac{(3+x) 2}{3 x} $
$3 x =6+2 x$
$3 x-2 x =6 $
$x =6 \Omega$