Question

In the circuit shown, the value of resistance $X$ in order that the potential difference between the points $B$ and $D$ is zero, will be

• A. $4\Omega$
• B. $6\Omega$
• C. $8\Omega$
• D. $9\Omega$

Answer 1

Answer: (C)

$8\Omega$

The given circuit is equivalent to the balanced Wheatstone's network.

Let, resistance in arm $AB$ is $P,P=15+6=21\Omega$

Resistance in arm $BC$ is $Q,Q=\left(\frac{8X}{8+X}\right)+\left(3\right)\Omega$

Resistance in arm $DA$ is $R,R=\left(\frac{\left(6\right)\left(6\right)}{6+6}\right)+\left(15\right)=3+15=18\Omega$

Resistance in arm $CD$ is $S,S=\left(\frac{\left(4\right)\left(4\right)}{4+4}\right)+\left(4\right)=2+4=6\Omega$

Using balancing condition for Wheatstone's network.

$\frac{P}{Q}=\frac{R}{S}$
$\frac{21}{\left(\frac{8X}{8+X}\right)+\left(3\right)}=\frac{18}{6}$
$\frac{21}{\left(\frac{8X}{8+X}\right)+\left(3\right)}=3$
$\left(\frac{8X}{8+X}\right)+\left(3\right)=\frac{21}{3}$

$\left(\frac{8X}{8+X}\right)+\left(3\right)=7$

$\frac{8X}{8+X}=4$

$8X=32+4X$

$4X=32$

$X=8\Omega$