A battery of emf 3 V and internal resistance r is connected in series with resistance of 55 Ω through an ammeter of resistance 1 Ω. The ammeter reads the current 50 mA. The internal resistance r of the battery is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4 Ω
Given:
EMF, $\mathcal{E} = 3 \ \text{V}$
External resistance, $R = 55 \ \Omega$
Ammeter resistance, $R_A = 1 \ \Omega$
Current, $I = 50 \ \text{mA} = 0.05 \ \text{A}$
Total resistance in series: $R_\text{total} = R + R_A + r = 55 + 1 + r = 56 + r$
Using Ohm's law: $I = \frac{\mathcal{E}}{R_\text{total}}$
$0.05 = \frac{3}{56 + r}$
$56 + r = \frac{3}{0.05} = 60 \text{ implies } r = 4 \ \Omega$
Internal resistance of the battery: $r = 4 \ \Omega$