Question

In a potentiometer experiment a cell of e.m.f 1.5 V connected in the secondary circuit gives a balancing length of 165 cm of the wire. If a resistance of 5 W is connected in parallel to the cell, the balancing length of the wire is 150 cm. The internal resistance of the cell is –

• A. 5 W
• B. 1.5 W
• C. 1 W
• D. 0.5 W

Answer 1

Answer: (D)

0.5 W

Answer 2

r = $\left( \frac{\mathcal{l}_{1}}{\mathcal{l}_{2}}–1 \right)$R = $\left( \frac{165}{150}–1 \right)$× 5 = 0.5 W