Conversion Of Galvanometer To Ammeter

Question

A galvanometer of resistance $50\ \Omega$ shows full-scale deflection at $2$ mA; what shunt resistance is required to convert it into an ammeter reading up to $1$ A?

RankGuru · Accepted Answer

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 $S = \dfrac{I_g G}{I - I_g}$, where $G$ is the galvanometer resistance, $I_g$ the full-scale current, and $I$ the maximum range. Substituting $G = 50\ \Omega$, $I_g = 2 	imes 10^{-3}$ A, $I = 1$ A: $S = \dfrac{(2 	imes 10^{-3})(50)}{1 - 2 	imes 10^{-3}} = \dfrac{0.1}{0.998} \approx 0.1\ \Omega$. 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.

Conversion Of Galvanometer To Ammeter

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Conversion Of Galvanometer To Ammeter

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