MHTCET 11 Apr 2026 Shift 1 Paper

Concept:A galvanometer is converted into an ammeter by connecting a small shunt resistance in parallel to divert most of the current.Explanation:Given full-scale galvanometer current, $I_g = 10\ \mathrm{mA} = 0.01\ \mathrm{A}$.Given full-scale voltage, $V_g = 0.75\ \mathrm{V}$.So, galvanometer resistance is $G = \frac{V_g}{I_g} = \frac{0.75}{0.01} = 75\ \Omega$.Target ammeter range, $I = 10\ \mathrm{A}$.Current through the shunt is $I_s = I - I_g = 10 - 0.01 = 9.99\ \mathrm{A}$.Since the shunt is in parallel with the galvanometer, the voltage across both is equal: $I_g G = I_s S$.Therefore, $S = \frac{I_g G}{I_s} = \frac{0.01 \times 75}{9.99} = \frac{0.75}{9.99}$.Multiplying numerator and denominator by $100$, we get $S = \frac{75}{999}\ \Omega$.Answer:The shunt resistance is $\frac{75}{999}\ \Omega$.Correct option: D.