Q. Two cells of equal $e.m.f.$ and of internal resistance $r_{1}$ and $r_{2}\left( r_{1} > r_{2} \right)$ are connected in series. On connecting this combination to an external resistance $R,$ it is observed that the potential difference across the first cell becomes zero. The value of $R$ will be
A) $r_{1} + r_{2}$
B) $r_{1} - r_{2}$
C) $\frac{r_{1} + r_{2}}{2}$
D) $\frac{r_{1} - r_{2}}{2}$
Explanation
(b) Let the voltage across any one cell is $V,$ then
$V = E - ir = E - r_{1}\left( \frac{2E}{r_{1} + r_{2} + R} \right)$
But $V = 0$
$\Rightarrow E - \frac{2Er_{1}}{r_{1} + r_{2} + R} = 0$
$\Rightarrow r_{1} + r_{2} + R = 2r_{1}$
$\Rightarrow R = r_{1} - r_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.