Question

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

• A. $1\Omega$
• B. $2\Omega$
• C. $4\Omega$
• D. $6\Omega$

Answer 1

Answer: (D)

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
$\frac{P}{Q}=\frac{R}{S}$
$\frac{P}{O}=\frac{R}{\frac{S\times x}{S+x}}$
$\frac{2}{2}=\frac{2}{\frac{3\times x}{3+x}}$
$1=\frac{(3+x)2}{3x}$
$3x=6+2x$
$3x-2x=6$
$x=6\Omega$