Q. In the adjoining circuit, the battery $E_{1}$ has an $e.m.f.$ of $12volt$ and zero internal resistance while the battery $E$ has an $e.m.f.$ of $2volt.$ If the galvanometer $G$ reads zero, then the value of the resistance $X$ in $ohm$ is
A) 10
B) 100
C) 500
D) 200
Explanation
(b) For no current through galvanometer, we have
$\left( \frac{E_{1}}{500 + X} \right)X = E \Rightarrow \left( \frac{12}{500 + X} \right)X = 2 \Rightarrow X = 100\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.