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 $(I o>I g)$ 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 $Rv$ to it. Then,choose the correct option

• A. $R_A R_V=G^2 \frac{I_0-I_g}{I_g}$ and $\frac{R_A}{R_V}={\frac{I g}{I_0-I_g}}^2$
• B. $R_A R_V=G^2$ and $\frac{R_A}{R_V}=\frac{I g}{I_0-I_g}$
• C. $R_A R_v=G^2$ and $\frac{R_A}{R_v}={\frac{I g}{I_0-I_g}}^2$
• D. $R_A R_V=G^2 \frac{I_g}{I_0-I g}$ and $\frac{R_A}{R_V}=\frac{I_0-I_g{ }^2}{I g}$

Answer 1

Answer: (C)

$\left(I_0-I_g\right) R_4=I_g G$ and $V _0=I_0 G + R _v$
$I _0=\frac{I_0 G + R _{ A }}{ R _{ A }} \quad V _0=I_0 G + R _{ V }$
$V _0= I _0 G ($ given $) \Rightarrow I_0 G + R _{ V }= I _g$
$\frac{ G ^{+ R _{ A } G }}{ R _{ A }}$
$R_A G+R_V R_A=G^2+R_A G$
$R_V R_A=G^2$
$R_A=\frac{I_g G}{I_0-I_g} ; R_V=\frac{I_0-I_g G}{I_g}$
$\frac{R_A}{R_V}=\frac{I_g^2}{I_0-I_g^2}$