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Question: The length of potentiometer wire is 600 cm and a current of 40 mA is flowing in it. When a cell of e...

The length of potentiometer wire is 600 cm and a current of 40 mA is flowing in it. When a cell of emf 2V and internal resistance 10 W is balanced on this potentiometer the balance length is found to be 500 cm. The resistance of potentiometer wire will be-

A

20 W

B

40 W

C

60 W

D

80 W

Answer

60 W

Explanation

Solution

P.D of ext circuit = Potential gradient × Balance length.

2 = IRL×l0\frac{IR}{L} \times \mathcal{l}_{0}

R = 2LIl0=2×60040×10−3×500=12200×10−3=12×103200\frac{2L}{I\mathcal{l}_{0}} = \frac{2 \times 600}{40 \times 10^{- 3} \times 500} = \frac{12}{200 \times 10^{- 3}} = \frac{12 \times 10^{3}}{200}