MHTCET 16 May 2024 Shift 2 Paper

Concept:A shunt resistance is connected in parallel with a galvanometer to extend its current range.For a larger current range, the required shunt is smaller.Thus, an ammeter needs a smaller shunt than a milliammeter for the same galvanometer.Explanation:Let the galvanometer have full-scale deflection current $I_g$ and internal resistance $R_g$.When converted to measure a maximum current $I$, the shunt $R_s$ satisfies:$$I_g R_g = (I - I_g) R_s$$So,$$R_s = \frac{I_g R_g}{I - I_g}$$For an ammeter, the full-scale current $I_A$ is very large, so $I_A - I_g$ is large.Hence, the ammeter shunt is:$$R_{s,A} = \frac{I_g R_g}{I_A - I_g}$$For a milliammeter, the full-scale current $I_m$ is small and only slightly greater than $I_g$.Hence, the milliammeter shunt is:$$R_{s,m} = \frac{I_g R_g}{I_m - I_g}$$Since $I_A \gg I_m$, the denominator for the ammeter is much larger.Therefore, $R_{s,A} < R_{s,m}$.So, the shunt of the ammeter is less than that of the milliammeter.Answer:Option A: less.