Q. Three resistance $P,\ Q,\ R$ each of $2\Omega$ and an unknown resistance $S$ form the four arms of a wheatstone bridge circuit. When a resistance of $6\Omega$ is connected in parallel to $S$ the bridge gets balanced. What is the value of $S$
A) $2\ \Omega$
B) $3\ \Omega$
C) $6\ \Omega$
D) $1\ \Omega$
Explanation
(b) By using $\frac{P}{Q} = \frac{R}{S} \Rightarrow \frac{2}{2} = \frac{2}{\frac{6S}{(6 + S)}} \Rightarrow S = 3\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.