A cell of e.m.f. E and internal resistance r is connected in series with an external resistance nr then the ratio of the terminal potential difference to E.M.F. is
Answer & explanation
Correct answer: option 3
$I = \frac{E}{nr+r} $
$ \text{Terminal potential } V = E -ir = E - \frac{E}{n+1} = \frac{nE}{n+1}$