Physics - Current Electricity Question with Solution | TestHub

Key Idea To convert a galvanometer to ammeter a small resistance is connected in parallel to the coil of the galvanometer. $$\text{ Here, }\begin{array}{rl}& {\mathrm{G}}_1=60\Omega ,{\mathrm{I}}_{\mathrm{g}}=1.0\mathrm{A},\mathrm{I}=5\mathrm{A}\\ & {\mathrm{I}}_{\mathrm{g}}{\mathrm{G}}_1=\left(\mathrm{I}-{\mathrm{I}}_{\mathrm{g}}\right)\mathrm{S}\\ & \mathrm{S}=\frac{{\mathrm{I}}_{\mathrm{g}}{\mathrm{G}}_1}{\mathrm{I}-{\mathrm{I}}_{\mathrm{g}}}=\frac{1}{5-1}\times 60=15\Omega \end{array}$$ Putting $15\Omega$ resistance in parallel.