Figure of merit of galvanometer

If the figure of merit is $K$, then current required for one deflection is $K$.

Hence, for $n$ deflections, the current through the galvanometer coil is $I_g = nK$

Converted galvanometer into ammeter using shunt

Let:

• galvanometer resistance $= G$
• shunt resistance $= S$
• total current through the combination $= I$

Since galvanometer and shunt are in parallel, the potential difference across them is same: $I_g G = I_s S$ where $I_s$ is current through shunt.

Therefore, $I_s = \frac{I_g G}{S}$

Total current

Total current is the sum of galvanometer current and shunt current: $I = I_g + I_s$ $I = I_g + \frac{I_g G}{S}$ $I = I_g\left(1 + \frac{G}{S}\right)$ $I = I_g\frac{S+G}{S}$

Substitute $I_g = nK$

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

Match with options

This corresponds to: $\boxed{\text{D: } \frac{nK(G+S)}{S}}$