WAEC Physics 2024 Paper

Concept:The shunt is connected in parallel with the galvanometer, so the voltage across both is equal.Explanation:Full-scale deflection current is $I_g = 10\,\text{mA} = 0.01\,\text{A}$.Galvanometer resistance is $R_g = 5\,\Omega$.Total current to be measured is $I = 2\,\text{A}$.Current through shunt is $I_s = I - I_g = 2 - 0.01 = 1.99\,\text{A}$.Since shunt and galvanometer are in parallel, $I_s R_s = I_g R_g$.So, $R_s = \frac{I_g R_g}{I_s}$.Substitute values: $R_s = \frac{0.01 \times 5}{1.99} = \frac{0.05}{1.99}$.Thus, $R_s \approx 0.025\,\Omega$.Answer:The required shunt is $\boxed{0.025\,\Omega}$, which is option A.