JEE Main 2022 · Physics · Electromagnetic Induction and Alternating Current · easy

A series LCR circuit with R=60 Ω, L=0.5 H and C=250 muF is connected to an AC source of rms voltage 200 V and angular frequency 200 rad s^-1.

Single correct option

A series LCR circuit with

R=60\ \Omega

L=0.5\ \mathrm{H}

and

C=250\ \mu\mathrm{F}

is connected to an AC source of rms voltage

200\ \mathrm{V}

and angular frequency

200\ \mathrm{rad\,s^{-1}}

. The average power dissipated in the circuit is

Show answer and solution

Answer: (B) $240\ \mathrm{W}$

X_L=\omega L=200\times0.5=100\ \Omega

and

X_C=\dfrac{1}{\omega C}=\dfrac{1}{200\times250\times10^{-6}}=20\ \Omega

. Impedance

Z=\sqrt{60^{2}+(100-20)^{2}}=\sqrt{3600+6400}=100\ \Omega

I_{rms}=200/100=2\ \mathrm{A}

and

P=I_{rms}^{2}R=4\times60=240\ \mathrm{W}

. Equivalently

P=VI\cos\phi=200\times2\times0.6=240\ \mathrm{W}

. Using

V^{2}/R=667\ \mathrm{W}

is the resonance value and is wrong here.

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