MHTCET 15 Apr 2026 Shift 1 Paper

Concept:A shunt is a parallel resistance that diverts most of the current, while the voltage across the galvanometer and the shunt remains equal.Explanation:Let the galvanometer resistance be $G$.The shunt resistance is given as $S = \frac{G}{n-1}$.Since the shunt is connected in parallel with the galvanometer, the potential difference across both is the same.If $I_g$ is the current through the galvanometer and $I_s$ is the current through the shunt, then:$$I_g G = I_s S$$Therefore,$$\frac{I_s}{I_g} = \frac{G}{S}$$Substituting $S = \frac{G}{n-1}$ gives:$$\frac{I_s}{I_g} = \frac{G}{G/(n-1)} = n-1$$So, $I_s = (n-1) I_g$.The total current is:$$I = I_g + I_s = I_g + (n-1)I_g = n I_g$$Thus, the fraction of the total current passing through the galvanometer is:$$\frac{I_g}{I} = \frac{I_g}{n I_g} = \frac{1}{n}$$This matches option B.Answer:$$\frac{1}{n}$$Option B.