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 $V(V = GI_0$) by connecting a series resistance $R_V$ to it. Then,

• A. $R_A R_v = G^2 \, \bigg(\frac{I_g}{I_0 - I_g}\bigg) \, aand \, \, \frac{R_A}{R_V} \, = \, \bigg(\frac{I_0 - I_g}{I_g}\bigg)^2$
• B. $R_A R_V = G^2 \, and \, \frac{R_A}{R_V} = \bigg(\frac{I_g}{I_0 - I_g}\bigg)^2$
• C. $R_A R_V = G^2 \, and \, \frac{R_A}{R_V} = \frac{I_g}{I_0 - I_g}$
• D. $R_A R_V = G^2 \bigg(\frac{I_0 - I_g}{I_g}\bigg) \, and \, \frac{R_A}{R_v} = \bigg(\frac{I_g}{I_0 - I_g}\bigg)^2$

Answer 1

Answer: (B)

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

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

$I_g (G+R_V) = V = GI_0 (given \, V \, = \, GI_0)$
$\therefore \, \, \, R_v = \frac{(I_0 - I_g)G}{I_g}$
$\therefore \, \, R_A R_V = G^2 $ & $\frac{R_A}{R_V} = \bigg(\frac{I_g}{I_0 - I_g}\bigg)^2$