Physics - Current Electricity Question with Solution | TestHub

Let the voltage of the battery be $V$.

When the switch S is open, the resistors $R$ and the galvanometer $G$ are in series. The current through the galvanometer is:
$I_G=\frac{V}{R+G}$

In this state, since $R$ and $G$ are in the same branch, the current through $R$ is equal to the current through $G$, so $I_R=I_G$.

When the switch S is closed, the potential at node D is determined by the voltage divider formed by the parallel combinations of $P,R$ and $Q,G$. The current through the galvanometer when S is closed is:
$I_G^{\prime }=\frac{V_D}{G}=\frac{V}{G}\left(\frac{\frac{QG}{Q+G}}{\frac{PR}{P+R}+\frac{QG}{Q+G}}\right)$

Given that the reading of the galvanometer remains the same, $I_G=I_G^{\prime }$:
$\frac{V}{R+G}=\frac{V\cdot Q}{(Q+G)\left(\frac{PR}{P+R}+\frac{QG}{Q+G}\right)}$

$\frac{PR}{P+R}+\frac{QG}{Q+G}=\frac{Q(R+G)}{Q+G}$

$\frac{PR}{P+R}=\frac{QR+QG-QG}{Q+G}=\frac{QR}{Q+G}$

$P(Q+G)=Q(P+R)$

$PQ+PG=PQ+QR\Rightarrow PG=QR$

This is the condition for a balanced Wheatstone bridge. Therefore, the potential at node B is equal to the potential at node D ($V_B=V_D$).

Because $V_B=V_D$, no current flows through the branch BD even when the switch S is closed.

Applying Kirchhoff's Current Law at node D, the current entering from $R$ must equal the current leaving through $G$. Thus, $I_R=I_G$ holds true whether the switch S is open or closed.

Answer: ${\mathrm{I}}_{\mathrm{R}}={\mathrm{I}}_{\mathrm{G}}$