(3)

Let N = number of turns in galvanometer, A = Area,

B = magnetic field

$k$ = the restoring torque per unit twist.

Current sensitivity, $I_S=\frac{NBA}{k}$

Voltage sensitivity, $V_S=\frac{NBA}{k{R}_G}$

So, resistance of galvanometer

$R_G=\frac{I_S}{V_S}=\frac{5\times 1}{20\times {10}^{-3}}=\frac{5000}{20}=250$$\mathrm{\Omega }$

(3)

[IMAGE 227]

Here, resistance of the galvanometer $=G$

Current through the galvanometer,

$I_G=0.2\%\text{ of }I=\frac{0.2}{100}I=\frac{1}{500}I$

$\therefore$   Current through the shunt,

        $I_S=I-I_G=I-\frac{1}{500}I=\frac{499}{500}I$

As shunt and galvanometer are in parallel

$\therefore$   $I_{G}G=I_{S}S$

        $\left(\frac{{\displaystyle 1}}{{\displaystyle 500}}I\right)G=\left(\frac{{\displaystyle 499}}{{\displaystyle 500}}\right)S \text{or} S=\frac{G}{499}$

Resistance of the ammeter $R_A$ is

        $\frac{1}{R_A}=\frac{1}{G}+\frac{1}{S}=\frac{1}{G}+\frac{1}{\frac{G}{499}}=\frac{500}{G} ; R_A=\frac{1}{500}G$