A moving coil galvanometer,… - JEE Main 2019 Physics PYQ

JEE Main 2019 — Physics Electromagnetism

2019mcqhard

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 ${I}_{o}({I}_{0}>{I}_{g})$ by connecting a shunt resistance ${R}_{A}$ to it and (ii) into a voltmeter of range 0 to $V(V=G{I}_{0})$ by connecting a series resistance ${R}_{V}$ to it. Then,

Official previous-year question

Held on 12 Apr 2019 · Verified 6 Jul 2026.

Options

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

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Solution

For ammeter: $({I}_{0}-{I}_{G})\times {R}_{A}={I}_{G}\times G\Rightarrow {I}_{0}={I}_{G}\times \frac{(G+{R}_{A})}{{R}_{A}}$

For voltmeter: ${V}_{0}={I}_{G}(G+{R}_{V})$

$\because {V}_{0}={I}_{0}\times G$

$\Rightarrow {I}_{G}\times (G+{R}_{V})={I}_{G}\times \frac{(G+{R}_{A})\times G}{{R}_{A}}$

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

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

Again, $\frac{{R}_{A}}{{R}_{V}}=\frac{{I}_{G}\times G}{{I}_{0}-{I}_{G}}\times \frac{{I}_{G}}{({I}_{0}-{I}_{G})\times G}=\frac{{{I}_{G}}^{2}}{{({I}_{0}-{I}_{G})}^{2}}$

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

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

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

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

Solution

For ammeter: $({I}_{0}-{I}_{G})\times {R}_{A}={I}_{G}\times G\Rightarrow {I}_{0}={I}_{G}\times \frac{(G+{R}_{A})}{{R}_{A}}$

For voltmeter: ${V}_{0}={I}_{G}(G+{R}_{V})$

$\because {V}_{0}={I}_{0}\times G$

$\Rightarrow {I}_{G}\times (G+{R}_{V})={I}_{G}\times \frac{(G+{R}_{A})\times G}{{R}_{A}}$

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

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

Again, $\frac{{R}_{A}}{{R}_{V}}=\frac{{I}_{G}\times G}{{I}_{0}-{I}_{G}}\times \frac{{I}_{G}}{({I}_{0}-{I}_{G})\times G}=\frac{{{I}_{G}}^{2}}{{({I}_{0}-{I}_{G})}^{2}}$

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