Question

A moving coil galvanometer, having a resistance $G$, produces full scale deflection when a current $I_g$ flows through it. This galvanometer can be converted into (i) an ammeter of range $0$ to $I_0$ $(I_0>I_g$) by connecting a shunt resistance $R_A$ to it and (ii) into a voltmeter of range $0$ to $V(V=G{I}_0$) by connecting a series resistance $R_V$ to it. Then,

• A. $R_A{R}_v=G^2\left(\frac{I_g}{I_0-I_g}\right)aand\frac{R_A}{R_V}=(\frac{I_0-I_g}{I_g}{)}^2$
• B. $R_A{R}_V=G^{2}and\frac{R_A}{R_V}=(\frac{I_g}{I_0-I_g}{)}^2$
• C. $R_A{R}_V=G^{2}and\frac{R_A}{R_V}=\frac{I_g}{I_0-I_g}$
• D. $R_A{R}_V=G^2\left(\frac{I_0-I_g}{I_g}\right)and\frac{R_A}{R_v}=(\frac{I_g}{I_0-I_g}{)}^2$

Answer 1

Answer: (B)

When galvanometer is used as an ammeter
shunt is used in parallel with galvanometer.

$\therefore I_{g}G=(I_o-I_g|)R_A$
$\therefore R_A=\left(\frac{I_g}{I_0-I_g}\right)G$
When galvanometer is used as a voltmeter,
resistance is used in series with galvanometer

$I_g(G+R_V)=V=G{I}_0(givenV=G{I}_0)$
$\therefore R_v=\frac{(I_0-I_g)G}{I_g}$
$\therefore R_A{R}_V=G^2$ & $\frac{R_A}{R_V}=(\frac{I_g}{I_0-I_g}{)}^2$