If the galvanometer G does not show any defle | vedclass

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(D) Since the galvanometer $G$ does not show any deflection,the current through it is zero $(i_g = 0)$.This implies that the potential at the junction

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$100$

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(D) Since the galvanometer $G$ does not show any deflection,the current through it is zero $(i_g = 0)$.This implies that the potential at the junction between the $400 \,\Omega$ resistor and the resistor $R$ must be equal to the potential of the $2 \, V$ battery.Let the potential at the junction be $V_j = 2 \, V$.The current flowing through the $400 \,\Omega$ resistor is $i = \frac{10 \, V - 2 \, V}{400 \,\Omega} = \frac{8 \, V}{400 \,\Omega} = 0.02 \, A$.Since no current flows through the galvanometer,this same current $i$ must flow through the resistor $R$.Using Ohm's law for resistor $R$,we have $V_j = i 	imes R$.Substituting the values,$2 \, V = 0.02 \, A 	imes R$.Therefore,$R = \frac{2}{0.02} \,\Omega = 100 \,\Omega$.

If the galvanometer $G$ does not show any deflection in the circuit shown,the value of $R$ is given by $............\Omega$.

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