A galvanometer having coil resistance 10 Ome | vedclass

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Question

(C) Given:Galvanometer resistance,$G = 10 \ \Omega$Full scale deflection current,$I_g = 3 \ mA = 3 	imes 10^{-3} \ A$Range of ammeter,$I = 8 \ A$To c

Vedclass · Accepted Answer

$3.75 	imes 10^{-3} \ \Omega$

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(C) Given:Galvanometer resistance,$G = 10 \ \Omega$Full scale deflection current,$I_g = 3 \ mA = 3 	imes 10^{-3} \ A$Range of ammeter,$I = 8 \ A$To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel with the galvanometer.The potential difference across the galvanometer and the shunt must be equal:$I_g G = (I - I_g) S$Rearranging for $S$:$S = \frac{I_g G}{I - I_g}$Substituting the values:$S = \frac{(3 	imes 10^{-3} \ A) 	imes 10 \ \Omega}{8 \ A - 3 	imes 10^{-3} \ A}$$S = \frac{0.03}{8 - 0.003} \ \Omega$$S = \frac{0.03}{7.997} \ \Omega \approx 3.75 	imes 10^{-3} \ \Omega$Thus,the required shunt resistance is $3.75 	imes 10^{-3} \ \Omega$.

$A$ galvanometer having coil resistance $10 \ \Omega$ shows a full scale deflection for a current of $3 \ mA$. For it to measure a current of $8 \ A$,the value of the shunt should be:

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