Question

The emf of a cell is $E\text{volt}$ and its internal resistance is $1\Omega$. A resistance of $4\Omega$ is joined to battery in parallel. This is connected in secondary circuit of potentiometer. The balancing length is $160\text{cm}$. If $1\text{V}$ cell balances for $100\text{cm}$ of potentiometer wire, then the emf of the cell is (Give integer value)

Answer 1

For the secondary circuit, emf is,
$E=I(R+r)$

$E=I(4+1)$

$E=5I$

$I=\frac{E}{5}...\left(1\right)$

The terminal voltage across the cell is,

$V=IR=\frac{4E}{5}$

Potential gradient $=\frac{V}{100}=0.01\text{V/cm}$

$\therefore V=0.01\times l_1$

$=0.01\times 160$

$\Rightarrow V=1.6\text{volts}$

$\because V=\frac{4E}{5}$

$E=\frac{1.6\times 5}{4}$

$\Rightarrow E=2\text{V}$

Accepted answer : $2$