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Electric Charges And Fields

Question

When a resistance of α Ω is connected at the ends of a battery, its potential difference decreases from 40 V to 30V. The internal resistance of the battery is

3Ω

6Ω

1.5Ω

4Ω

Solution

3Ω

Given:- E=40 V, V=30V , r=α Ω

Internal resistance

$r = (\frac{E}{V} - 1) R = (\frac{40}{30} - 1) \times 9 = \frac{1}{3} \times 9 = 3 Ω$