A null point is found at $$200$$ cm in potentiometer when cell in secondary circuit is shunted by $$5$$ $$\Omega$$. When a resistance of $$15$$ $$\Omega$$ is used for shunting null point moves to $$300$$ cm. The internal resistance of the cell is ______ $$\Omega$$.

Solution :

For a potentiometer,

$$\frac{E}{V} = \frac{l_E}{l_V}$$

Also,

$$V = \frac{ER}{R+r}$$

where $$r$$ is the internal resistance of the cell.

Therefore,

$$\frac{E}{V} = \frac{R+r}{R}$$

Hence,

$$\frac{R+r}{R} \propto \frac{1}{l_V}$$

For resistance,

$$R_1 = 5\ \Omega$$

balancing length,

$$l_1 = 200\text{ cm}$$

For resistance,

$$R_2 = 15\ \Omega$$

balancing length,

$$l_2 = 300\text{ cm}$$

Thus,

$$\frac{\frac{5+r}{5}}{\frac{15+r}{15}} = \frac{300}{200}$$

$$\frac{5+r}{5} \times \frac{15}{15+r} = \frac{3}{2}$$

$$\frac{3(5+r)}{15+r} = \frac{3}{2}$$

$$\frac{5+r}{15+r} = \frac{1}{2}$$

$$2(5+r)=15+r$$

$$10+2r=15+r$$

$$r=5\ \Omega$$

Final Answer :

$$5\ \Omega$$