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

Did you get this right?

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

Your note

JEE Main Electromagnetism in other years

Electromagnetism 2026 Electromagnetism 2025 Electromagnetism 2024 Electromagnetism 2023 Electromagnetism 2022 Electromagnetism 2021 Electromagnetism 2020 Electromagnetism 2018 Electromagnetism 2017 Electromagnetism 2016 Electromagnetism 2015 Electromagnetism 2014

Keep practicing Electromagnetism

Work through every JEE Main Electromagnetism PYQ, year by year.

Browse all Physics topics

Options

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

JEE Main 2019 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism

JEE Main 2019 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism