Q. In a Wheatstone’s bridge, three resistances $P,\ Q\ and\ R$ connected in the three arms and the fourth arm is formed by two resistances $S_{1}\ and\ S_{2}$ connected in parallel. The condition for the bridge to be balanced will be
A) $\frac{P}{Q} = \frac{2R}{S_{1} + S_{2}}$
B) $\frac{P}{Q} = \frac{R(S_{1} + S_{2})}{S_{1}S_{2}}$
C) $\frac{P}{Q} = \frac{R(S_{1} + S_{2})}{{2S}_{1}S_{2}}$
D) $\frac{P}{Q} = \frac{R}{S_{1} + S_{2}}$
Explanation
(b) Here $S$ consist of $S_{1}$and $S_{2}$ arranged in parallel, hence
$S = \frac{S_{1}S_{2}}{S_{1} + S_{2}}$
So, the balance condition will be $\frac{P}{Q} = \frac{R}{S} = \frac{R(S_{1} + S_{2})}{S_{1}S_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.