Q. $n$ identical cells each of $e.m.f.\ E$ and internal resistance $r$ are connected in series. An external resistance $R$ is connected in series to this combination. The current through $R$ is
A) $\frac{nE}{R + nr}$
B) $\frac{nE}{nR + r}$
C) $\frac{E}{R + nr}$
D) $\frac{nE}{R + r}$
Explanation
(a) Total e.m.f. $= nE,$ Total resistance $R + nr\ \Rightarrow i = \frac{nE}{R + nr}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.