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JEE Main 202120 JulyEvening Shift

Exam: JEE Main
Year: 2021
Date: 20 July
Shift: Evening Shift
Subject: Physics
Chapter: Alternating Current

JEE Main 2021 · 20 July · Evening Shift · Physics · Alternating Current

A series LCR circuit of R = 5Ω, L = 20mH and C = 0 . 5 is connected...

A series LCR circuit of

R = 5Ω, L = 20mH

and

C = 0 . 5\muF

is connected across an

AC

supply of

250V,

having variable frequency. The power dissipated at resonance condition is ________

\times {10}^{2}W .

Correct answer

0

Step-by-step solution

${X}_{L} = {X}_{C}$ (due to resonance)

$Z = R$ so ${i}_{rms} = \dfrac{V}{Z} = \dfrac{V}{R}$

$\dfrac{{V}^{2}}{R} = \dfrac{250 \times 250}{5} = 125 \times {10}^{2}W$

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