A galvanometer has a resistance of 100 Ω. A potential difference of 100 mV between its terminals gives a full-scale deflection. The shunt resistance required to convert it into an ammeter reading up to 5 A is

$$\begin{gathered} \mathrm{The} \mathrm{current} \mathrm{required} \mathrm{for} \mathrm{full} \mathrm{scale} \mathrm{deflection} \mathrm{is} \\ {\mathrm{i}}_{\mathrm{g}}=\frac{\mathrm{PD} \mathrm{across} \mathrm{galvanometer}}{\mathrm{resistance} \mathrm{of} \mathrm{galvanometer}} \\ =\frac{100\times {10}^{-3}}{100}={10}^{-3} \mathrm{A} \\ \mathrm{The} \mathrm{shunt} \mathrm{resistance} \mathrm{required} \mathrm{is} \\ \mathrm{S}=\left(\frac{{\mathrm{i}}_{\mathrm{g}}}{\mathrm{i}-{\mathrm{i}}_{\mathrm{g}}}\right) \mathrm{G}=\left(\frac{{10}^{-3}}{5-{10}^{-3}}\right)\times 100 \\ =0.02 \mathrm{\Omega } \end{gathered}$$