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# I have a doubt, kindly clarify. - Current Electricity - JEE Main

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,

• Option 1)

$R_{A}R_{V}=G^{2}\left ( \frac{I_{0}-I_{g}}{I_{g}} \right )$ and $\frac{R_{A}}{R_{V}}=\left ( \frac{I_{g}}{(I_{0}-I_{g})} \right )^{2}$

• Option 2)

$R_{A}R_{V}=G^{2}$ and $\frac{R_{A}}{R_{V}}=\frac{I_{g}}{I_{0}-I_{g}}^{2}$

• Option 3)

$R_{A}R_{V}=G^{2}\left ( \frac{I_{g}}{I_{0}-I_{g}} \right )$ and $\frac{R_{A}}{R_{V}}=\left ( \frac{I_{0}-I_{g}}{I_{g}} \right )^{2}$

• Option 4)

$\frac{R_{A}}{R_{V}}=G^{2}\frac{R_{A}}{R_{V}}=\frac{I_{g}}{(I_{0}-I_{g})}$

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Conversion of galvanometer into voltameter -

Connected a large Resistance $R$  in series

- wherein

$(I_{o}-I_{G})R_{A}=I_{G}G$             $V_{0}=I_{G}(G+R_{A})$

$V_{0}=I_{0}G$

$R_{A}=\frac{I_{G}G}{I_{o}-I_{G}}$

$I_{0}G=I_{G}(G+R_{V})$

$\frac{I_{0}G}{I_{G}}-G=R_{V}$

$R_{V}=\frac{G(I_{0}-I_{G})}{I_{G}}$

$\frac{R_{A}}{R_{V}}=\frac{I_{G}G}{I_{0}-I_{G}}\times \frac{I_{G}}{G(I_{0}-I_{G})}=\left ( \frac{I_{G}}{I_{0}-I_{G}} \right )^{2}$

$R_{A}R_{V}=\frac{I_{G}G}{I_{0}-I_{G}}\times \frac{G(I_{0}-I_{G})}{I_{G}}=G^{2}$

Option 1)

$R_{A}R_{V}=G^{2}\left ( \frac{I_{0}-I_{g}}{I_{g}} \right )$ and $\frac{R_{A}}{R_{V}}=\left ( \frac{I_{g}}{(I_{0}-I_{g})} \right )^{2}$

Option 2)

$R_{A}R_{V}=G^{2}$ and $\frac{R_{A}}{R_{V}}=\frac{I_{g}}{I_{0}-I_{g}}^{2}$

Option 3)

$R_{A}R_{V}=G^{2}\left ( \frac{I_{g}}{I_{0}-I_{g}} \right )$ and $\frac{R_{A}}{R_{V}}=\left ( \frac{I_{0}-I_{g}}{I_{g}} \right )^{2}$

Option 4)

$\frac{R_{A}}{R_{V}}=G^{2}\frac{R_{A}}{R_{V}}=\frac{I_{g}}{(I_{0}-I_{g})}$

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