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)

$6\, \Omega$

Answer 2

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