Question

A moving coil galvanometer, having a resistance G, produces full scale deflection when a current $\mathrm{I}_{\mathrm{g}}$ flows through it. This galvanometer can be converted into (i) an ammeter of range 0 to $\mathrm{I}_0$ ($\mathrm{I}_0 > \mathrm{I}_{\mathrm{g}}$) by connecting a shunt resistance $R_A$ to it and (ii) into a voltmeter of range 0 to $\mathrm{V}$ ($\mathrm{V} = \mathrm{GI}_0$) by connecting a series resistance $R_V$ to it. Then,

Accepted Answer

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

Answer

(A) $\mathrm{R}_{\mathrm{A}} \mathrm{R}_{\mathrm{V}}=\mathrm{G}^{2}$ and $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{V}}}=\frac{\mathrm{I}_{\mathrm{g}}}{(\mathrm{I}_{0}-\mathrm{I}_{\mathrm{g}})}$

Answer

(C) $\mathrm{R}_{\mathrm{A}} \mathrm{R}_{\mathrm{V}}=\mathrm{G}^{2}\left(\frac{\mathrm{I}_{\mathrm{g}}}{\mathrm{I}_{0}-\mathrm{I}_{\mathrm{g}}}\right)$ and $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{V}}}=\left(\frac{\mathrm{I}_{0}-\mathrm{I}_{\mathrm{g}}}{\mathrm{I}_{\mathrm{g}}}\right)^{2}$

Answer

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

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Current Electricity

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