A moving coil galvanometer has 50 turns and e | vedclass

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(C) Given:Number of turns $n = 50$Area $A = 2 	imes 10^{-4} \ m^2$Magnetic field $B = 0.02 \ T$Torsional constant $C = 10^{-4} \ N \ m \ rad^{-1}$Ful

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$5.56$

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(C) Given:Number of turns $n = 50$Area $A = 2 	imes 10^{-4} \ m^2$Magnetic field $B = 0.02 \ T$Torsional constant $C = 10^{-4} \ N \ m \ rad^{-1}$Full-scale deflection angle $	heta = 0.2 \ rad$Galvanometer resistance $G = 50 \ \Omega$The torque balance equation for a moving coil galvanometer is given by:$C 	heta = n i_g A B$Where $i_g$ is the full-scale deflection current.$i_g = \frac{C 	heta}{n A B} = \frac{10^{-4} 	imes 0.2}{50 	imes 2 	imes 10^{-4} 	imes 0.02} = \frac{0.2 	imes 10^{-4}}{200 	imes 10^{-4} 	imes 0.02} = \frac{0.2}{4} = 0.05 \ A$Wait,recalculating: $i_g = \frac{0.2 	imes 10^{-4}}{50 	imes 2 	imes 10^{-4} 	imes 0.02} = \frac{0.2}{50 	imes 2 	imes 0.02} = \frac{0.2}{2} = 0.1 \ A$To convert the galvanometer into an ammeter of range $I = 1.0 \ A$,a shunt resistance $S$ is connected in parallel.The current through the shunt is $I_s = I - i_g = 1.0 - 0.1 = 0.9 \ A$.Since the galvanometer and shunt are in parallel,the voltage across them is equal:$i_g G = I_s S$$0.1 	imes 50 = 0.9 	imes S$$S = \frac{5}{0.9} = \frac{50}{9} \approx 5.56 \ \Omega$

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