Q. $A,\ B,\ C$ and $D$ are four resistances of $2\Omega,\ 2\Omega,\ 2\Omega\ and\ 3\Omega$ respectively. They are used to form a Wheatstone bridge. The resistance $D$ is short circuted with a resistances R in order to get the bridge balanced. The value of $R$ will be
A) $4\ \Omega$
B) $6\ \Omega$
C) $8\ \Omega$
D) $3\ \Omega$
Explanation
(b) The bridge will be balanced when the shunted resistance of value $2\Omega\ ie,\ 2 = \frac{3 \times S}{3 + S}$. On solving $S = 6\ \Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.