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Step-by-Step Solution
Step 1: Understand the Experimental Setup
A galvanometer (of unknown resistance $G$) is connected in series with a known resistance $R$ and a battery of emf $V_E$. This setup produces a deflection $\theta$ in the galvanometer.
Step 2: Current Through the Galvanometer Initially
When only the galvanometer $G$ and resistor $R$ are connected in series, the total resistance is $(R + G)$. The current through the galvanometer is
$$
I_G = \frac{V_E}{R + G}.
$$
Step 3: Introducing the Shunt Resistor $S$
A shunt of resistance $S$ is connected in parallel with the galvanometer. This reduces the current through the galvanometer to half its original value (i.e., the deflection is halved).
In this new arrangement, the combined parallel section ($G$ in parallel with $S$) is in series with $R$.
The current through the overall circuit is
$$
I = \frac{V_E}{\,R + \frac{GS}{G + S}\,}.
$$
Step 4: Condition for Half Deflection
With the shunt connected, the current through the galvanometer becomes
$$
I_G' = \frac{I_G}{2} = \frac{1}{2}\cdot \frac{V_E}{R + G}.
$$
Since $G$ and $S$ are in parallel, they have the same voltage across them, so the currents are related by:
$$
I_G'\,G = (I - I_G')\,S.
$$
Simplifying, we get
$$
I_G'(G + S) = I\,S.
$$
Substituting $I_G' = \frac{I_G}{2}$ and the expression for $I$, one arrives at the condition for half deflection.
Step 5: Deriving the Relationship Among $G$, $R$, and $S$
By carefully simplifying the expressions for currents with and without the shunt, one obtains the key final step:
$$
RG = S(R + G).
$$
Rearranging gives
$$
S (R + G) = RG.
$$
Step 6: Final Result
Thus, for a galvanometer of resistance $G$ connected in series with $R$ (and a battery of emf $V_E$) to have its deflection halved by adding a shunt $S$, the required relationship is:
$$
\boxed{S \,(R + G) = R\,G.}
$$
