In the circuit shown $P \ne R$, the reading of the galvanometer is same with switch $S$ open or closed. Then
IIT 1999, Easy
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(a) Reading of galvanometer remains same whether switch S is open or closed, hence no current will flow through the switch i.e. $R$ and $G$ will be in series and same current will flow through them. ${I_R} = {I_G}$.
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