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

A moving coil galvanometer, having a resistance

G

produces full scales deflection when a current

{I}_{g}

flows through it. It can be converted into& an ammeter of range

0

{I}_{0}

\left(> {I}_{g}\right)

by connecting a shunt of resistance

{R}_{A}

to it. If it can also be converted into a voltmeter of range

to

V = {I}_{0}G

by connecting a resistance

{R}_{V} to it in series, then which among the following is correct?

Options

A{R}_{A}{R}_{V} = G^{2} $\left(\dfrac{{I}_{0} - {I}_{g}}{Ig}\right)$ and $\dfrac{RA}{RV} = \left(\dfrac{Ig}{{I}_{0} - {I}_{g}}\right)$ 2
B ${R}_{A}{R}_{V} = G^{2} and$ \dfrac{RA}{Rv} = {I}_{g} \left({I}_{0} - {I}_{g}\right)
C ${R}_{A}{R}_{v} = G^{2} and&$ \dfrac{RA}{Rv} = \left(\dfrac{Ig}{{I}_{0} - {I}_{g}}\right)2
D{R}_{A}{R}_{V} = G^{2} $\left(\dfrac{Ig}{{I}_{0} - {I}_{g}}\right)$ and $\dfrac{RA}{RV} = \left(\dfrac{{I}_{0} - {I}_{g}}{Ig}\right)$ 2

C. ${R}_{A}{R}_{v} = G^{2} and&$ \dfrac{RA}{Rv} = \left(\dfrac{Ig}{{I}_{0} - {I}_{g}}\right)2

$( {I}_{0} - {I}_{g} ) {R}_{4} = {I}_{g}G\text{and}{V}_{0} = {I}_{0}\left(G + {R}_{v}\right)$

${I}_{0} = \dfrac{{I}_{0}\left(G + {R}_{A}\right)}{{R}_{A}}{V}_{0} = {I}_{0}\left(G + {R}_{V}\right)$

${V}_{0} = {I}_{0}G ( \text{given} ) ⇒ {I}_{0}\left(G + {R}_{V}\right) = {I}_{g}\dfrac{\left(G + {R}_{A}\right)G}{{R}_{A}}$

${R}_{A}G + {R}_{V}{R}_{A} = {G}^{2} + {R}_{A}G$

${R}_{V}{R}_{A} = {G}^{2}$

${R}_{A} = \dfrac{{I}_{g}G}{{I}_{0} - {I}_{g}}; {R}_{V} = \dfrac{\left({I}_{0} - {I}_{g}\right)G}{{I}_{g}}$

$\dfrac{{R}_{A}}{{R}_{V}} = {I}_{g}^{2}{\left({I}_{0} - {I}_{g}\right)}^{2}$