Question

A galvanometer of resistance 25 ohm is connevted to a battery of 2V along with a resistance in series. When the value of this resistance is 3000 ohm a full scale deflection of 30 units is obtained in the galvanometer. In order to reduce this deflection to 20 units, the resistance in series will be?

Answer 1

$$\begin{gathered} \mathrm{Current}\mathrm{through}\mathrm{the}\mathrm{galvanometer}=\frac{2}{3000+25}=\frac{2}{3025}\mathrm{A} \\ \mathrm{This}\mathrm{produces}\mathrm{a}\mathrm{deflection}\mathrm{of}30\mathrm{divisions}.\mathrm{To}\mathrm{reduce}\mathrm{the}\mathrm{deflection}\mathrm{from}30\mathrm{divisions}\mathrm{to}20\mathrm{divisions},\mathrm{the}\mathrm{current}\mathrm{should}\mathrm{be}, \\ \frac{20}{30}\times \frac{2}{3025}=\frac{4}{3\times 3025}\mathrm{A} \\ \mathrm{If}\mathrm{R}\mathrm{be}\mathrm{the}\mathrm{series}\mathrm{resistance}\mathrm{now},\mathrm{then}, \\ \frac{4}{3\times 3025}=\frac{2}{\mathrm{R}+25} \\ \mathrm{or}4\mathrm{R}+100=18150 \\ \mathrm{Or},\mathrm{R}=4512.5\mathrm{ohm} \end{gathered}$$