Question

A potentiometer wire gives a null point at $3\text{m}$ for a wire of length $3.5\text{m}$. The primary circuit contains only a battery of emf $10\text{V}$. The emf of secondary cell is $8\text{V}$. What is the internal resistance of the battery in the primary circuit (in ohms) if the resistance of the potentiometer wire is $14\Omega$?

Answer 1

Applying KVL for primary circuit,
$V-i\times 3.5x=0$
$\Rightarrow V=3.5x..\left(i\right)$
where $x$ is the resistance per unit length of potentiometer wire.

Similarly for secondary circuit,
$8-3x=0\Rightarrow 8=3x..\left(ii\right)$

Dividing (i) and (ii),
$\frac{8}{V}=\frac{3}{3.5}$
$\Rightarrow V=\frac{28}{3}=10-i\times r$
$\Rightarrow \frac{28}{3}=10-\frac{10r}{14+r}=\frac{140}{14+r}$
$\Rightarrow 14+r=15\Rightarrow r=1\Omega$