Q. In the Wheatstone’s bridge shown, $P = 2\Omega,\ Q = 3\Omega,\ R = 6\Omega$ and $S = 8\ \Omega.$ In order to obtain balance, shunt resistance across $'S'$ must be
A) $2\ \Omega$
B) $3\ \Omega$
C) $6\ \Omega$
D) $8\ \Omega$
Explanation
(d) Let the value of shunt be $r.$ Hence the equivalent resistance of branch containing $S$ will be $\frac{Sr}{S + r}$
In balance condition, $\frac{P}{Q} = \frac{Sr/(S + r)}{R}.$ This gives $r = 8\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.