A moving coil galvanometer converted into an ammeter reads up to 0.03 A by connecting a shunt of resistance 4r across it and ammeter reads up to 0.06 A, when a shunt of resistance r is used. What is the maximum current which can be sent through this galvanometer if no shunt is used?

Let G be the resistance of galvanometer and i_g the  current which on passing through the galvanometer produces full-scale deflection.

Since G and S are in parallel, the potential difference across them will be

$\begin{array}{l}{\mathrm{i}}_{\mathrm{g}}\times \mathrm{G}=\left(\mathrm{i}-{\mathrm{i}}_{\mathrm{g}}\right)\times \mathrm{S}\\ \frac{{\mathrm{i}}_{\mathrm{g}}}{\mathrm{i}}=\frac{\mathrm{S}}{\mathrm{S}+\mathrm{G}}\end{array}$

Hence,   ${\mathrm{i}}_{\mathrm{g}}=\frac{4\mathrm{r}}{4\mathrm{r}+\mathrm{G}}\times 0.03 ....\left(1\right)$

${\mathrm{i}}_{\mathrm{g}}=\frac{\mathrm{r}}{\mathrm{r}+\mathrm{G}}\times 0.06 ....\left(2\right)$

Equating Eqs. (1) and (2), we get

4r + G = 2(r + G)

$\begin{array}{l}\Rightarrow \mathrm{G}=2\mathrm{r}\\ \therefore {\mathrm{i}}_{\mathrm{g}}=\frac{4\mathrm{r}}{4\mathrm{r}+2\mathrm{r}}\times 0.03=0.02\mathrm{A}\left[\mathrm{from}\mathrm{Eq}.\left(1\right)\right]\end{array}$