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 ______ $\Omega$.

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Current Electricity

A null point is found at 200 cm in potentiometer when cell in secondary circuit is shunted by

5Ω

. When a resistance of

15Ω

is used for shunting null point moves to 300 cm. The internal resistance of the cell is ______

Ω

Answer: 5

The length balanced on a potentiometer is proportional to the terminal voltage of the cell,

V = \frac{εR}{R + r}

, for shunt resistance

R

. For the two shunts,

\frac{L _{1}}{L _{2}} = \frac{R _{1} ( R _{2} + r )}{R _{2} ( R _{1} + r )}

. Putting

L_{1} = 200

L_{2} = 300

R_{1} = 5 Ω

and

R_{2} = 15 Ω

\frac{200}{300} = \frac{5 ( 15 + r )}{15 ( 5 + r )}

, which reduces to

2 (5 + r) = 15 + r

, so

10 + 2 r = 15 + r

and

r = 5 Ω

. The internal resistance of the cell is

5 Ω

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Integer answer