To measure the internal resistance of a battery, potentiometer is used. For R=10 Ω, the balance point is observed at l=500 cm and for R=1 Ω the balance point is observed at l=400 cm. The internal resistance of the battery is approximately :

Solution

Physics sir · 4 steps

Balance length tracks terminal voltage, so take the ratio of the two lengths and the emf cancels.

1. What balances. With resistance R across the battery, the potentiometer balances the terminal voltage V = ER/(R+r). The balance length is proportional to this voltage.
2. Take the ratio. Divide the two cases. The emf E and the potential gradient both cancel, so r is the only unknown left.

   l₁/l₂ = (10/(10+r))/(1/(1+r)) = 10(1+r)/(10+r) = 500/400 = 5/4

3. Solve for r. Cross-multiply: 40(1+r) = 5(10+r), so 40 + 40r = 50 + 5r. That gives 35r = 10, so r = 2/7 ≈ 0.29 Ω.
4. So. The internal resistance is about 0.3 Ω, option (b).

The idea to keep. A potentiometer balance length is proportional to the voltage it balances. For internal resistance, write V = ER/(R+r) for each case and take the ratio so that E drops out.

Answer from the NTA final answer key. Solution by PrepWiser.