Question

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

• A. 1/n
• B. 1/n+1
• C. n/n+1
• D. n+1/n

Answer 1

The correct option is C n/n+1

$i=\frac{E}{r+nr}=\frac{E}{(n+1)r}$

Terminal difference of cell$=V=E-ir=E-\frac{Er}{nr+r}$

$⟹V=\frac{Enr}{nr+r}=\frac{En}{n+1}$

Ratio of V and E is $\frac{n}{n+1}$

Answer-(C)