A moving coil galvanometer, when shunted with | vedclass

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(C) Let $G$ be the galvanometer resistance and $I_g$ be the current for full-scale deflection.Case $1$: Shunt $S = 2	ext{ }\Omega$, total current $I

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(C) Let $G$ be the galvanometer resistance and $I_g$ be the current for full-scale deflection.Case $1$: Shunt $S = 2	ext{ }\Omega$, total current $I = 500	ext{ mA} = 0.5	ext{ A}$.Using the shunt formula: $I_g G = (I - I_g)S$$I_g G = (0.5 - I_g) 	imes 2$$I_g(G + 2) = 1 \Rightarrow I_g = \frac{1}{G + 2}$ ... (Equation $1$)Case $2$: Series resistance $R_s = 470	ext{ }\Omega$, voltage $V = 10	ext{ V}$.Using Ohm's law: $I_g = \frac{V}{G + R_s}$$I_g = \frac{10}{G + 470}$ ... (Equation $2$)Equating Equation $1$ and Equation $2$:$\frac{1}{G + 2} = \frac{10}{G + 470}$$G + 470 = 10(G + 2)$$G + 470 = 10G + 20$$9G = 450$$G = 50	ext{ }\Omega$.

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