If $$n$$ represents the actual number of deflections in a converted galvanometer of resistance $$G$$ and shunt resistance $$S$$. Then the total current $$I$$ when its figure of merit is $$K$$ will be

A galvanometer of resistance $$G$$ is converted to an ammeter by connecting a shunt resistance $$S$$ in parallel. The figure of merit $$K$$, defined as the current per division, implies that when the galvanometer shows $$n$$ divisions the current through it is $$I_g = nK$$.

Since the galvanometer and the shunt are in parallel, the potential difference across them is the same. Therefore $$I_g \cdot G = I_s \cdot S$$, where $$I_s$$ is the current through the shunt. The total current is $$I = I_g + I_s$$ which becomes $$I_g + \frac{I_g \cdot G}{S} = I_g\left(1 + \frac{G}{S}\right) = I_g \cdot \frac{G + S}{S}$$.

Substituting $$I_g = nK$$ into this expression gives

$$I = \frac{nK(G + S)}{S}$$

Hence the required option is D.