Q. In a Wheatstone’s network $\ P = 2\Omega,\ \mathcal{Q} = 2\Omega,\ R = 2\Omega\ and\ S = 3\Omega$. 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$\ \Omega$
Explanation
(d) Let a resistance r ohm be shunted with resistance S, so that the bridge is balanced.
If S’ is the resultant resistance of S and r, then
In balanced position
$\frac{P}{Q} = \frac{R}{S^{'}}$
$\frac{2}{2} = \frac{2}{S^{'}}$
$\therefore\ S^{'} = 2\Omega$
Now,
$\frac{1}{S^{'}} = \frac{1}{S} + \frac{1}{r}$
$or\frac{1}{r} = \frac{1}{S^{'}} - \frac{1}{S} = \frac{1}{2} - \frac{1}{3} = \frac{3 - 2}{6}$
$\frac{1}{r} = \frac{1}{6}$
$r = 6\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.