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Alternating Current

JEE Main level, numerical answer, medium

116 of 190

Question 116 of 190

A sinusoidal voltage of peak value $250V$ is applied to a series $LCR$ circuit, in which $R=8\Omega ,L=24mH$ and $C=60\mu F$ . The value of power dissipated at resonant condition is $x$ $kW$ . The value of $x$ to the nearest integer is _______.

Show answer and solution

Answer: 4

At resonance power,

P=\dfrac{{\left({V}_{rms}\right)}^{2}}{R}

P=\dfrac{{\left(250/\sqrt{2}\right)}^{2}}{8}=3906.25W

\approx 4kW

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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More Physics practice

JEE Main Physics practice question on Alternating Current (AC Circuits), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.

A sinusoidal voltage of peak value 250V250V250V is applied to a series LCRLCRLCR circuit, in which R=8Ω,L=24mHR=8\Omega ,L=24mHR=8Ω,L=24mH and C=60μFC=60\mu FC=60μF. The value of power dissipated at resonant condition is xxx kWkWkW. The value of xxx to the nearest integer is _______.

JEE Main question paper PDF

More Physics practice

A sinusoidal voltage of peak value $250V$ is applied to a series $LCR$ circuit, in which $R=8\Omega ,L=24mH$ and $C=60\mu F$ . The value of power dissipated at resonant condition is $x$ $kW$ . The value of $x$ to the nearest integer is _______.