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

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

Tardigrade Admin · Accepted Answer

I=(E/r+nr) =(E/r(n+1)) V=E-Ir =E-(E/r(n+1))r=(nE/n+1) ∴ (V/E)=(n/n+1)

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

$(1/ n)$

$1/ (n + 1)$

$n / (n + 1)$

$(n + 1) / n$

$I = \frac{E}{r + n r}$
$= \frac{E}{r ( n + 1 )}$
$V = E - I r$
$= E - \frac{E}{r ( n + 1 )} r = \frac{n E}{n + 1}$
$∴$ $\frac{V}{E} = \frac{n}{n + 1}$

Q. 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

(1/n)

1/(n+1)

n/(n+1)

(n+1)/n