A cell of internal resistance r is connected across an external resistance nr. Then the ratio of the terminal voltage to the emf of the cell is

Question

A cell of internal resistance r is connected across an external resistance nr. Then the ratio of the terminal voltage to the emf of the cell is (A) (1/n) (B) (1/n+1) (C) (n/n+1) (D) ( n -1/ n )

Tardigrade Admin · Accepted Answer

Internal resistance = r, External resistance = nr. Let terminal voltage = V text then V=E-I r ⇒ V=E-(E r/(n+1) r) V=(n E/n+1) ⇒ (V/E)=(n/n+1)

Q. A cell of internal resistance $r$ is connected across an external resistance $n r$ . Then the ratio of the terminal voltage to the emf of the cell is

$\frac{1}{n}$

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

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

$\frac{n - 1}{n}$

Internal resistance $= r$ ,
External resistance $= n r$ .
Let terminal voltage $= V$
$then V = E - I r \Rightarrow V = E - \frac{E r}{( n + 1 ) r}$
$V = \frac{n E}{n + 1} \Rightarrow \frac{V}{E} = \frac{n}{n + 1}$

Q. A cell of internal resistance r is connected across an external resistance nr. Then the ratio of the terminal voltage to the emf of the cell is

n1​

n+11​

n+1n​

nn−1​