Q. In Wheatstone’s bridge $P = 9\ ohm,\ Q = 11\ ohm,\ R = 4\ ohm$ and $S = 6\ ohm.$ How much resistance must be put in parallel to the resistance $S$ to balance the bridge
A) $24\ ohm$
B) $\frac{44}{9}ohm$
C) $26.4\ ohm$
D) $18.7\ ohm$
Explanation
(c) $\frac{P}{Q} = \frac{R}{S^{'}}$ [For balancing bridge]
$\Rightarrow S^{'} = \frac{4 \times 11}{9} = \frac{44}{9}$
$\Rightarrow \frac{1}{S^{'}} = \frac{1}{r} + \frac{1}{6}$
$\Rightarrow \frac{9}{44} - \frac{1}{6} = \frac{1}{r}$
$\Rightarrow r = \frac{132}{5} = 26.4\ \Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.