Question

How much resistance should be connected to 15$\Omega$ resistor shown in the circuit in adjoining figure below so that the points M and N are at the same potential.

Answer 1

By balance condition of Wheatstone's bridge
$\frac{P}{Q}=\frac{R}{S}$
$\frac{3}{X}=\frac{30}{60}$
$X=\frac{3\times 60}{30}=6\Omega$
$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$
$\frac{1}{6}=\frac{1}{15}+\frac{1}{R}$
$\frac{1}{R}=\frac{1}{6}-\frac{1}{15}$
$\frac{1}{R}=\frac{5-2}{30}=\frac{3}{30}=\frac{1}{10}$
$R=10\Omega$ in parallel