In the circuit shown in the figure,P neq R. T | vedclass

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(C) Let the potential at node $B$ be $V_B$ when the switch $S$ is open. When the switch $S$ is closed,the potential at node $B$ remains the same if th

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$I_{R}=I_{G}$

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(C) Let the potential at node $B$ be $V_B$ when the switch $S$ is open. When the switch $S$ is closed,the potential at node $B$ remains the same if the current through the switch is zero. Since the galvanometer reading remains the same,the potential difference across the galvanometer must be independent of the state of the switch $S$. This implies that the potential at node $B$ must be equal to the potential at node $D$ (i.e.,$V_B = V_D$) when the switch is closed,or the current through the switch is zero. Applying Kirchhoff's Current Law at node $B$: When $S$ is open,the current $I_P$ flows through $P$ and $I_Q$ flows through $Q$. When $S$ is closed,if the galvanometer reading is unchanged,it implies that the potential at $B$ does not change. For the current through the switch to be zero,the potential at $B$ must be equal to the potential at $D$. Looking at the circuit,the current $I_R$ flows through $R$ and $I_G$ flows through the galvanometer. By applying the condition for the galvanometer reading to remain unchanged,we find that $I_R = I_G$.

In the circuit shown in the figure,$P \neq R$. The reading of the galvanometer remains the same with switch $S$ open or closed. Then

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