Question

A cell of emf $E$ and internal resistance $r$ is connected in series with an external resistance $nr$. Then the ratio of the terminal potential difference to emf is

• A. $ (1/n) $
• B. $ 1/(n+1) $
• C. $ n/(n+1) $
• D. $ (n+1)/n $

Answer 1

Answer: (C)

$ I=\frac{E}{r+nr} $
$ =\frac{E}{r(n+1)} $
$ V=E-Ir $
$ =E-\frac{E}{r(n+1)}r=\frac{nE}{n+1} $
$ \therefore $ $ \frac{V}{E}=\frac{n}{n+1} $