A moving coil galvanometer of resistance 100 | vedclass

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Question

(A) The galvanometer resistance $G = 100 \,\Omega$ and shunt resistance $S = 0.1 \,\Omega$.The maximum current through the galvanometer is $I_G = 100

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$100.1$

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(A) The galvanometer resistance $G = 100 \,\Omega$ and shunt resistance $S = 0.1 \,\Omega$.The maximum current through the galvanometer is $I_G = 100 \,\mu A = 100 	imes 10^{-6} \, A = 0.1 \, mA$.For an ammeter,the shunt resistance $S$ is connected in parallel with the galvanometer.The potential difference across the galvanometer and the shunt is the same: $I_G 	imes G = (I - I_G) 	imes S$.Rearranging for the total current $I$: $I = I_G \left( 1 + \frac{G}{S} ight)$.Substituting the values: $I = 0.1 \, mA 	imes \left( 1 + \frac{100}{0.1} ight)$.$I = 0.1 \, mA 	imes (1 + 1000) = 0.1 	imes 1001 \, mA$.$I = 100.1 \, mA$.

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

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