JEE PHYSICS MCQ: ALTERNATING CURRENT

1 Medium

1 of 1

Application +4 | -1

An L-C-R series circuit with a resistance of

100 \Omega

is connected to

200 \mathrm{~V}

(AC source) and angular frequency

300 \, \mathrm{rad} / \mathrm{s}

. When only the capacitor is removed, then the current lags behind the voltage by

60^{\circ}

. When only the inductor is removed the current leads the voltage by

60^{\circ}

. The average power dissipated in original L-C-R circuit is

View answer and explanation

Correct option: D

400 W

Option D (400 W) is correct because at resonance, the impedance equals the resistance, and power is calculated as V²/R.

More Questions from ALTERNATING CURRENT

Practice All Chapter MCQs →

R=300 \, \Omega

C=25 \, \mu \mathrm{F}, \varepsilon_0=50 \mathrm{~V}

v=50 / \pi \, \mathrm{Hz}

R=300 \Omega, L=0.9 \mathrm{~H}, C=2.0

\mu F

\omega=1000 \mathrm{~rad} / \mathrm{sec}

L-C \cdot R, \mathrm{AC}

L-C-R, \mathrm{AC}

L-C-R

\mathrm{AC}

L

\frac{\pi}{4}

C

\frac{\pi}{4}

{X_L}

R