Q. Two cells of same emf $E$ but of different internal resistances $r_{1}$and $r_{2}$are connected in series with an external resistance $R$. The potential drop across the first cell is found to be zero. The external resistance $R$ is
A) $r_{1} + r_{2}$
B) $r_{1} - r_{2}$
C) $r_{2} - r_{1}$
D) $r_{1}r_{2}$
Explanation
(b) As both cells are in series, the circuit current
$i = \frac{E + E}{r_{1} + r_{2} + R} = \frac{2E}{r_{1} + r_{2} = R}$
As terminal potential drop across 1st cell is zero, hence
$V_{1} = E - ir_{1} = E - \frac{2E}{(r_{1} + r_{2} + R)}r_{1} = 0$
[image:3_image181.emf]
$\Rightarrow \ E = \frac{2Er_{1}}{(r_{1} + r_{2} + R)}or\ r_{1} + r_{2} + R = 2r_{1} = R = (r_{1} - r_{2})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.