When a resistance of 5,Omega is shunted with | vedclass

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(C) Let $I_g$ be the full-scale deflection current of the galvanometer and $R_G$ be its resistance.Case $1$: Shunting with $5\,\Omega$ resistance.The

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$50$

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(C) Let $I_g$ be the full-scale deflection current of the galvanometer and $R_G$ be its resistance.Case $1$: Shunting with $5\,\Omega$ resistance.The total current is $I = 250\,mA = 0.25\,A$. The shunt resistance $S = 5\,\Omega$.Using the current division rule: $I_g = I 	imes \frac{S}{S + R_G}$$I_g = 0.25 	imes \frac{5}{5 + R_G} \dots(i)$Case $2$: Connecting $1050\,\Omega$ in series.The total voltage is $V = 25\,V$. The series resistance $R = 1050\,\Omega$.Using Ohm's law: $I_g = \frac{V}{R + R_G}$$I_g = \frac{25}{1050 + R_G} \dots(ii)$Equating $(i)$ and $(ii)$:$0.25 	imes \frac{5}{5 + R_G} = \frac{25}{1050 + R_G}$$\frac{1.25}{5 + R_G} = \frac{25}{1050 + R_G}$$1.25(1050 + R_G) = 25(5 + R_G)$$1312.5 + 1.25 R_G = 125 + 25 R_G$$1312.5 - 125 = 25 R_G - 1.25 R_G$$1187.5 = 23.75 R_G$$R_G = \frac{1187.5}{23.75} = 50\,\Omega$

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