A moving coil galvanometer of resistance 100 | vedclass

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(A) Given:Resistance of galvanometer, $G = 100 \Omega$Shunt resistance, $S = 0.1 \Omega$Maximum deflection current in galvanometer, $I_g = 100 \mu A =

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$100.1$

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(A) Given:Resistance of galvanometer, $G = 100 \Omega$Shunt resistance, $S = 0.1 \Omega$Maximum deflection current in galvanometer, $I_g = 100 \mu A = 100 	imes 10^{-6} A = 10^{-4} A$For an ammeter, the galvanometer and shunt resistance are connected in parallel. The potential difference across both is the same:$V_{AB} = I_g G = (I - I_g) S$Substituting the values:$(100 	imes 10^{-6}) 	imes 100 = (I - 100 	imes 10^{-6}) 	imes 0.1$$10^{-2} = (I - 10^{-4}) 	imes 10^{-1}$$0.1 = I - 0.0001$$I = 0.1 + 0.0001 = 0.1001 A$Converting to milliamperes:$I = 0.1001 	imes 1000 mA = 100.1 mA$Thus, the total current in the circuit is $100.1 mA$.

$A$ moving coil galvanometer of resistance $100 \Omega$ is used as an ammeter using a shunt resistance $0.1 \Omega$. The maximum deflection current in the galvanometer is $100 \mu A$. Find the total current in the circuit so that the ammeter shows maximum deflection. (in $mA$)

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