The galvanometer shown in fig. has resistance $50\mathrm{\Omega }$ and current required for full scale deflection is $1 \mathrm{mA}$. Find the resistances R₁, R₂ and R₃ required to convert it into ammeter having ranges as indicated.

$\text{ For the range }0.1\mathrm{A},\mathrm{G}\text{ is in parallel to }\left({\mathrm{R}}_1+{\mathrm{R}}_2+{\mathrm{R}}_3\right)$

$I_{g}G=\left(I-I_g\right)\left(R_1+R_2+R_3\right)$

$\Rightarrow {10}^{-3}\left(50\right)=\left(0.1-{10}^{-3}\right)\left(R_1+R_2+R_3\right)$

$\Rightarrow R_1+R_2+R_3=\frac{100}{198}$

$\text{ For the range }1\mathrm{A},\left(\mathrm{G}+{\mathrm{R}}_3\right)\text{ is in parallel to }\left({\mathrm{R}}_1+{\mathrm{R}}_2\right)$

$I_g\left(G+R_3\right)=\left(I-I_g\right)\left(R_1+R_2\right)$

$\Rightarrow \left({10}^{-3}\right)\left(50+R_3\right)=\left(1-{10}^{-3}\right)\left(R_1+R_2\right)$

$\Rightarrow R_1+R_2=\frac{10}{198}$

$\text{ For the range }10\mathrm{A},\left(\mathrm{G}+{\mathrm{R}}_2+{\mathrm{R}}_3\right)\text{ is in parallel to }{\mathrm{R}}_1$

$I_g\left(G+R_2+R_3\right)=\left(I-I_g\right)R_1$

$\Rightarrow \left({10}^{-3}\right)\left(50+R_2+R_3\right)=\left(10-{10}^{-3}\right)R_1$

$\Rightarrow R_1=\frac{1}{198}\Omega$

Thus, $R_2=\frac{9}{198}=\frac{1}{22}\Omega$

and $R_3=\frac{90}{198}=\frac{15}{33}\Omega$