Question

A moving coil galvanometer, having a resistance G. produces full scales deflection when a current Ig flows through it. It can be converted into (i) an ammeter of range 0 to Io $(Io>Ig)$ by connecting a shunt resistance $R_A$ to it and (ii) into a voltmeter of range 0 to $VV=G{I}_0$ by connecting a series resistance $Rv$ to it. Then,choose the correct option

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

Answer 1

Answer: (C)

$\left(I_0-I_g\right)R_4=I_{g}G$ and $V_0=I_{0}G+R_v$
$I_0=\frac{I_{0}G+R_A}{R_A}{V}_0=I_{0}G+R_V$
$V_0=I_{0}G($ given $)\Rightarrow I_{0}G+R_V=I_g$
$\frac{G^{+R_{A}G}}{R_A}$
$R_{A}G+R_V{R}_A=G^2+R_{A}G$
$R_V{R}_A=G^2$
$R_A=\frac{I_{g}G}{I_0-I_g};R_V=\frac{I_0-I_{g}G}{I_g}$
$\frac{R_A}{R_V}=\frac{I_g^2}{I_0-I_g^2}$