JEE Main 2016 — Current Electricity Question with Solution

Let full scale deflection of current = I

In case 1, when R = 2400 $$\Omega$$ and deflection of 40 divisions present.

$$\therefore$$   $$\frac{40}{50}\mathrm{I}=\frac{V}{G+R}$$

$$\Rightarrow$$   $$\frac{4}{5}\mathrm{I}$$ = $$\frac{2}{G+2400}$$          . . .(1)

Incase 2, when R = 4900 $$\Omega$$ and deflection of 20 divisions present

$$\therefore$$   $$\frac{20}{50}\mathrm{I}=\frac{V}{G+R}$$

$$\Rightarrow$$   $$\frac{2}{5}\mathrm{I}$$ = $$\frac{2}{G+4900}$$          . . .(2)

From (1) and (2) we get,

$$\frac{4}{2}=\frac{G+4900}{G+2400}$$

$$\Rightarrow$$   2G + 4800 $$=$$ G + 4900

$$\Rightarrow$$   G $$=$$ 100 $$\Omega$$

Putting value of G in equation (1), we get,

$$\frac{4}{5}\mathrm{I}=\frac{2}{100+2400}$$

$$\Rightarrow$$   $$\mathrm{I}=1$$ mA

Current sensitivity $$=$$ $$\frac{\mathrm{I}}{numberofdivisions}$$

$$=$$ $$\frac{1}{50}$$

$$=$$ 0.02 mA / division

$$=$$ 20 $$\mu$$A / division

Resistance required for deflection of 10 divisions

$$\frac{10}{50}\mathrm{I}=\frac{V}{G+R}$$

$$\Rightarrow$$   $$\frac{1}{5}\times 1\times 10^{-3}=\frac{2}{100+R}$$

$$\Rightarrow$$   R $$=$$ 9900 $$\Omega$$