A battery of emf 3 V and internal resistance r is connected in series with a resistor of 55 Ω. An ammeter of resistance 2 Ω connected in circuit reads 50 mA. The value of r is

3 Ω

The correct answer is Option (2) → 3 Ω

Total current: $I = 50 \, \text{mA} = 0.05 \, \text{A}$

Total series resistance: $R_\text{total} = R + R_A + r = 55 + 2 + r = 57 + r \, \Omega$

Battery EMF: $E = I R_\text{total} \;\Rightarrow\; 3 = 0.05 (57 + r)$

$57 + r = \frac{3}{0.05} = 60 \;\Rightarrow\; r = 3 \, \Omega$

Internal resistance r = 3 Ω