A moving coil galvanometer, having a resistance G, produces full …

MCQ

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
${R_A}{R_V} = {G^2}$ and $\,\frac{{{R_A}}}{{{R_V}}} = \frac{{{I_g}}}{{\left( {{I_0} - {I_g}} \right)}}$
✓
${R_A}{R_V} = {G^2}$ and $\,\,\frac{{{R_A}}}{{{R_V}}} = {\left( {\frac{{{I_g}}}{{{I_0} - {I_g}}}} \right)^2}$
C
${R_A}{R_V} = {G^2}\left( {\frac{{{I_g}}}{{{I_0} - {I_g}}}} \right)\,$ and $\,\,\frac{{{R_A}}}{{{R_V}}} = {\left( {\frac{{{I_0} - {I_g}}}{{{I_g}}}} \right)^2}$
D
${R_A} - {R_V} = {G^2}\left( {\frac{{{I_0} - {I_g}}}{{{I_g}}}} \right)\,$ and $\,\,\frac{{{R_A}}}{{{R_V}}} = {\left( {\frac{{{I_g}}}{{{I_0} - {I_g}}}} \right)^2}$

✓

Answer

Correct option: B.

${R_A}{R_V} = {G^2}$ and $\,\,\frac{{{R_A}}}{{{R_V}}} = {\left( {\frac{{{I_g}}}{{{I_0} - {I_g}}}} \right)^2}$

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

$\therefore \quad \mathrm{I}_{g} \mathrm{G}=\left(\mathrm{I}_{0}-\mathrm{I}_{g}\right) \mathrm{R}_{\mathrm{A}}$

$\therefore \quad R_{A}=\left(\frac{I_{g}}{I_{0}-I_{g}}\right) G$

When galvanometer is used as a voltmeter, resistance is used in series with galvanometer.

$\mathrm{I}_{g}\left(\mathrm{G}+\mathrm{R}_{\mathrm{v}}\right)=\mathrm{V}=\mathrm{GI}_{0}\left(\text { given } \mathrm{V}=\mathrm{GI}_{0}\right)$

$\therefore R_{v}=\frac{\left(I_{0}-I_{g}\right) G}{I_{g}}$

$\therefore \quad {{\text{R}}_{\text{A}}}{{\text{R}}_{\text{v}}} = {{\text{G}}^2}\& \frac{{{{\text{R}}_{\text{A}}}}}{{{{\text{R}}_{\text{v}}}}} = {\left( {\frac{{{{\text{I}}_g}}}{{{{\text{I}}_0} - {{\text{I}}_g}}}} \right)^2}$