JEE Main 2019 — Physics Electromagnetism

2019mcqmedium

A series $AC$ circuit containing an inductor $(20 \mathrm{mH}),$ a capacitor $(120\text{ μ}F)$ and a resistor $(60 \Omega )$ is driven by an $AC$ source of $24 V/50 \mathrm{Hz}.$ The energy dissipated in the circuit in $60 s$ is:

Official previous-year question

Held on 9 Jan 2019 · Verified 6 Jul 2026.

Options

A
$5.17\times {10}^{2}J$
B
$3.39\times {10}^{3}J$
C
$2.26\times {10}^{3}J$
D
$5.65\times {10}^{2}J$

Did you get this right?

Solution

(Power factor), ${P}_{f}=\mathrm{cos}⁡\text{ϕ}=\frac{R}{Z}$

Where $R,Z$ are the resistance and impedance respectively.

(Energy dissipated) $E=\frac{{V}_{rms}^{2}}{Z}\times \mathrm{cos}\text{ϕ}\times t$

$\Rightarrow E=\frac{{V}_{rms}^{2}}{Z}\times \frac{R}{Z}\times t$

$\Rightarrow E=\frac{{V}_{rms}^{2}}{{Z}^{2}}\times R\times t$

${V}_{rms},t$ are the RMS voltage and time respectively.

We know that the Impedance of a LCR circuit is given by the formula,

${Z}^{2}={R}^{2}+{(\text{ω}L-\frac{1}{\text{ω}C})}^{2} \Rightarrow {Z}^{2}={60}^{2}+{(2\text{π}\times 50\times 20\times {10}^{-3}-(\frac{1}{2\text{π}\times 50\times 120\times {10}^{-6}}))}^{2}$

Where $\text{ω},L,C$ are the angular frequency, inductance and capacitance in the circuit respectively.

$\Rightarrow {Z}^{2}={60}^{2}+{(2\text{π}-\frac{1000}{12\text{π}})}^{2}$

$\Rightarrow {Z}^{2}=4009.7$

$\therefore E=\frac{{24}^{2}}{4009.7}\times 60\times 60J$

$= 517.14 J$

$=5.17\times {10}^{2}J$

Your note

JEE Main Electromagnetism in other years

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JEE Main 2019 — Physics Electromagnetism

2019mcqmedium

Official previous-year question

Held on 9 Jan 2019 · Verified 6 Jul 2026.

Options

A
$5.17\times {10}^{2}J$
B
$3.39\times {10}^{3}J$
C
$2.26\times {10}^{3}J$
D
$5.65\times {10}^{2}J$

Did you get this right?

Solution

(Power factor), ${P}_{f}=\mathrm{cos}⁡\text{ϕ}=\frac{R}{Z}$

Where $R,Z$ are the resistance and impedance respectively.

(Energy dissipated) $E=\frac{{V}_{rms}^{2}}{Z}\times \mathrm{cos}\text{ϕ}\times t$

$\Rightarrow E=\frac{{V}_{rms}^{2}}{Z}\times \frac{R}{Z}\times t$

$\Rightarrow E=\frac{{V}_{rms}^{2}}{{Z}^{2}}\times R\times t$

${V}_{rms},t$ are the RMS voltage and time respectively.

We know that the Impedance of a LCR circuit is given by the formula,

${Z}^{2}={R}^{2}+{(\text{ω}L-\frac{1}{\text{ω}C})}^{2} \Rightarrow {Z}^{2}={60}^{2}+{(2\text{π}\times 50\times 20\times {10}^{-3}-(\frac{1}{2\text{π}\times 50\times 120\times {10}^{-6}}))}^{2}$

Where $\text{ω},L,C$ are the angular frequency, inductance and capacitance in the circuit respectively.

$\Rightarrow {Z}^{2}={60}^{2}+{(2\text{π}-\frac{1000}{12\text{π}})}^{2}$

$\Rightarrow {Z}^{2}=4009.7$

$\therefore E=\frac{{24}^{2}}{4009.7}\times 60\times 60J$

$= 517.14 J$

$=5.17\times {10}^{2}J$

Your note

JEE Main Electromagnetism in other years

Keep practicing Electromagnetism

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

Browse all Physics topics

JEE Main 2019 — Physics Electromagnetism

Options

Did you get this right?

Solution

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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