A moving coil galvanometer of resistance 100 | vedclass

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(C) Given:Resistance of the galvanometer,$G = 100 \ \Omega$Full-scale deflection current,$I_g = 100 \ \mu A = 100 	imes 10^{-6} \ A = 0.1 	imes 10^{

Vedclass · Accepted Answer

$\frac{100}{9}$

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(C) Given:Resistance of the galvanometer,$G = 100 \ \Omega$Full-scale deflection current,$I_g = 100 \ \mu A = 100 	imes 10^{-6} \ A = 0.1 	imes 10^{-3} \ A$Required total current,$I = 1 \ mA = 1 	imes 10^{-3} \ A$To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel with it.The formula for shunt resistance is given by:$S = \frac{I_g G}{I - I_g}$Substituting the values:$S = \frac{(0.1 	imes 10^{-3}) 	imes 100}{1 	imes 10^{-3} - 0.1 	imes 10^{-3}}$$S = \frac{0.1 	imes 10^{-1}}{0.9 	imes 10^{-3}}$$S = \frac{0.01}{0.0009} = \frac{100}{9} \ \Omega$Thus,the required shunt resistance is $\frac{100}{9} \ \Omega$.

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