Conversion Of Galvanometer To Ammeter

NCERT Class 12, Chapter 4 (Moving Charges and Magnetism) describes converting a galvanometer into an ammeter by connecting a low resistance shunt $S$ in parallel, so most current bypasses the sensitive coil. The shunt satisfies $I_{g}G=(I-I_g)S$, hence $S=\frac{I_{g}G}{I-I_g}$, where $G$ is the galvanometer resistance, $I_g$ the full-scale current, and $I$ the maximum range. Substituting $G=50\Omega$, $I_g=2\times 10^{-3}$ A, $I=1$ A: $S=\frac{(2\times 10^{-3})(50)}{1-2\times 10^{-3}}=\frac{0.1}{0.998}\approx 0.1\Omega$. The value $1.0\Omega$ ignores the current ratio. The value $0.05\Omega$ divides $G$ by the wrong factor. The value $0.5\Omega$ misplaces a decimal. Plausibility check: a shunt must be far smaller than $G$ so that nearly all the large current is diverted around the coil, and $0.1\Omega \ll 50\Omega$ satisfies this expectation. Physically, the ratio $I/I_g=1/0.002=500$ tells us the meter's range is boosted five-hundred-fold, and the shunt shares the current inversely with resistance, so its value being roughly $G/500\approx 0.1\Omega$ is exactly the order of magnitude we would predict, giving confidence in the result.