JAMB Physics 2024 Paper

Concept:To use a galvanometer as an ammeter, a low resistance shunt is connected in parallel so that most of the current bypasses the galvanometer.Explanation:The galvanometer gives full-scale deflection at $I_g = 50\ \text{mA} = 0.05\ \text{A}$.The total current to be measured is $I = 2\ \text{A}$.The galvanometer resistance from the circuit is $G = 10\ \Omega$.Current through the shunt is $I_s = I - I_g = 2 - 0.05 = 1.95\ \text{A}$.In parallel, the voltage across the galvanometer equals the voltage across the shunt:$$I_g G = I_s R$$Thus,$$R = \frac{I_g G}{I_s} = \frac{0.05 \times 10}{1.95} = \frac{0.5}{1.95} = 0.2564\ \Omega$$Answer:B. $0.256\ \Omega$