Question

A null point is obtained at 200 cm on potentiometer wire when cell in secondary circuit is shunted by 52. When a resistance of 152 is used for shunting, null point moves to 300 cm. The internal resistance of the cell is ______.

पर्याय

• 3Ω

• 4Ω

• 5Ω

• 6Ω

Answer 1

A null point is obtained at 200 cm on potentiometer wire 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 5Ω.

Explanation:

For determining the internal resistance of a cell, we have 

r = R\[\left({\frac{l_{1}}{l_{2}}}-1\right)\]

When shunted by 5 Ω, l₂ = 200 cm = 2 m

\[\therefore\] r = 5\[\left({\frac{l_{1}}{2}}-1\right)\]

r = \[\frac {5l_1}{2}\] - 5   ...(i)

When shunted by 15 Ω, l₂ = 300 cm = 3m

\[\therefore\] r = 15 \[\left({\frac{l_{1}}{3}}-1\right)\]

r = 5l₁ - 15   ...(ii)

Equating (i) and (ii),

\[\frac {5l_1}{2}\] - 5 = 5l₁ - 15

\[\therefore\] l₁ = 4 m

Putting this value in (ii), we get r = 5(4) - 15 = 5Ω