NEET UG 2016 — Physics Electromagnetism

2016mcqhard

An inductor $20 \text{mH}$ , a capacitor $50 \mu F$ and a resistor $40 \Omega$ are connected in series across a source of emf $V=10\mathrm{sin}⁡340 t$ . The power loss in $\text{AC}$ . circuit is:

Official previous-year question

Held on 30 Apr 2016 · Verified 9 Jul 2026.

Options

A
$0.51 \text{W}$
B
$0.67 \text{W}$
C
$0.76 \text{W}$
D
$0.89 \text{W}$

Did you get this right?

Solution

Given : $\text{R=40Ω , L= 20 mH , C=50 μF}$

Comparing the given equation for voltage with general equation $V={V}_{0}\mathrm{sin}\omega t$

$\omega =340\mathrm{rad}{s}^{-1}$ and peak voltage ${V}_{0}=10V$

Inductive reactance ${X}_{L}$ $\omega L=340\times 20\times {10}^{-3}=68\times {10}^{-1}=6.8\Omega$

Capacitive reactance ${X}_{C}$

$\frac{1}{\omega C}=\frac{1}{340\times 50\times {10}^{-6}}=\frac{{10}^{4}}{34\times 5}=\frac{2}{34}\times {10}^{3}$

$=0.0588\times {10}^{3}=58.82\Omega$

$\text{ Z=}\sqrt{{({X}_{L}-{X}_{C})}^{2}{\text{+R}}^{2}}\text{⇒Z}=\sqrt{{(\omega L-\frac{1}{\omega C})}^{2}+{R}^{2}}$

$\text{ Z=}\sqrt{{(58.8-6.8)}^{2}\text{+}{(40)}^{2} }\text{ ⇒ Z}=\sqrt{2704+1600}\approx 65.6\Omega$

${I}_{\mathrm{rms}}=\frac{{V}_{\mathrm{rms}}}{\text{Z}}=\frac{10}{65.6\times \sqrt{2}} ({V}_{\mathrm{rms}}=\frac{{V}_{0}}{\sqrt{2}})$

Power $={E}_{\mathrm{rms}}{I}_{\mathrm{rms}}\mathrm{cos}\phi ={I}_{\mathrm{rms}}^{2}R$

$(\because \mathrm{cos}\phi =\frac{R}{Z},{V}_{\mathrm{rms}}={I}_{\mathrm{rms}}Z)$

$=\frac{100\times 40}{{(65.6)}^{2}\times 2}=\frac{2000}{{(65.6)}^{2}}$

$=0.51 W$

Your note

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NEET UG 2016 — Physics Electromagnetism

2016mcqhard

An inductor $20 \text{mH}$ , a capacitor $50 \mu F$ and a resistor $40 \Omega$ are connected in series across a source of emf $V=10\mathrm{sin}⁡340 t$ . The power loss in $\text{AC}$ . circuit is:

Official previous-year question

Held on 30 Apr 2016 · Verified 9 Jul 2026.

Options

A
$0.51 \text{W}$
B
$0.67 \text{W}$
C
$0.76 \text{W}$
D
$0.89 \text{W}$

Did you get this right?

Solution

Given : $\text{R=40Ω , L= 20 mH , C=50 μF}$

Comparing the given equation for voltage with general equation $V={V}_{0}\mathrm{sin}\omega t$

$\omega =340\mathrm{rad}{s}^{-1}$ and peak voltage ${V}_{0}=10V$

Inductive reactance ${X}_{L}$ $\omega L=340\times 20\times {10}^{-3}=68\times {10}^{-1}=6.8\Omega$

Capacitive reactance ${X}_{C}$

$\frac{1}{\omega C}=\frac{1}{340\times 50\times {10}^{-6}}=\frac{{10}^{4}}{34\times 5}=\frac{2}{34}\times {10}^{3}$

$=0.0588\times {10}^{3}=58.82\Omega$

$\text{ Z=}\sqrt{{({X}_{L}-{X}_{C})}^{2}{\text{+R}}^{2}}\text{⇒Z}=\sqrt{{(\omega L-\frac{1}{\omega C})}^{2}+{R}^{2}}$

$\text{ Z=}\sqrt{{(58.8-6.8)}^{2}\text{+}{(40)}^{2} }\text{ ⇒ Z}=\sqrt{2704+1600}\approx 65.6\Omega$

${I}_{\mathrm{rms}}=\frac{{V}_{\mathrm{rms}}}{\text{Z}}=\frac{10}{65.6\times \sqrt{2}} ({V}_{\mathrm{rms}}=\frac{{V}_{0}}{\sqrt{2}})$

Power $={E}_{\mathrm{rms}}{I}_{\mathrm{rms}}\mathrm{cos}\phi ={I}_{\mathrm{rms}}^{2}R$

$(\because \mathrm{cos}\phi =\frac{R}{Z},{V}_{\mathrm{rms}}={I}_{\mathrm{rms}}Z)$

$=\frac{100\times 40}{{(65.6)}^{2}\times 2}=\frac{2000}{{(65.6)}^{2}}$

$=0.51 W$

Your note

NEET UG Electromagnetism in other years

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Work through every NEET UG Electromagnetism PYQ, year by year.

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NEET UG 2016 — Physics Electromagnetism

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Solution

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NEET UG Electromagnetism in other years

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NEET UG 2016 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

NEET UG Electromagnetism in other years

More NEET UG Electromagnetism questions

Other Physics topics for NEET UG

Mechanics

Modern Physics

Thermodynamics

Optics

Waves & Oscillations

Keep practicing Electromagnetism