Physics - Current Electricity Question with Solution | TestHub

$i=\frac{V}{R_{\mathrm{e}\mathrm{f}\mathrm{f}}}=\frac{V}{1970+30}=\frac{2}{2000}=1\times {10}^{-3}\mathrm{A}=\mathrm{1 }\mathrm{m}\mathrm{A}$
This current provides full scale deflection (i.e. 20 division) In order to limit the deflection 10 divisions,
the resistances needed to connect  can be obtained as
$\theta =\frac{niAB}{k}$$\therefore \frac{{\theta }_1}{{\theta }_2}=\frac{i_1}{i_2}$
$\therefore { \theta }_1 :{\theta }_2=2$
$\therefore i_1:i_2=\mathrm{2:1}$
$\therefore i_2=\frac{i_1}{2}=\frac{1\times {10}^{-3}}{2}=5\times {10}^{-4}\mathrm{m}\mathrm{A}$
$\therefore i=\frac{V}{R_{\mathrm{e}\mathrm{f}\mathrm{f}}+Rs}\Rightarrow Rs=\frac{V}{i}-R_{\mathrm{e}\mathrm{f}\mathrm{f}}$
$=\frac{2}{5\times {10}^{-4}}-2000$
$=\frac{2}{5}\times {10}^4-2000$
$=4\times {10}^3-2000$
$=\mathrm{2000 }\mathrm{\Omega }$
$\therefore$The resistance of$1970 \mathrm{\Omega }$is to be replaced by$1970+2000=\mathrm{3970 }\mathrm{\Omega }$