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. $\left( \frac{1}{n} \right)$
• B. $\frac{1}{n + 1}$
• C. $\frac{n}{n + 1}$
• D. $\frac{n + 1}{n}$

Answer 1

Answer: (C)

$\frac{n}{n + 1}$

Answer 2

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

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

$\frac{V_{T}}{E}$ = $\frac{n}{n + 1}$