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JEE Main 2024 · Physics · Electromagnetic Induction and Alternating Current · medium

In a series LCR circuit driven by an ac source of fixed rms voltage and angular frequency ω, the reactances are R=30 Ω, XL=40 Ω and XC=80 Ω. If only…

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In a series LCR circuit driven by an ac source of fixed rms voltage and angular frequency ω\omega, the reactances are R=30 ΩR=30\ \Omega, XL=40 ΩX_L=40\ \Omega and XC=80 ΩX_C=80\ \Omega. If only the source frequency is varied until the circuit is in resonance, the ratio of the average power at resonance to the average power at the original frequency is
  1. A. 259\dfrac{25}{9}
  2. B. 169\dfrac{16}{9}
  3. C. 53\dfrac{5}{3}
  4. D. 925\dfrac{9}{25}
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Answer: (A) 259\dfrac{25}{9}

Original impedance: Z=302+(80−40)2=50 ΩZ=\sqrt{30^2+(80-40)^2}=50\ \Omega. Average power P=V2RZ2=V2R⋅R2Z2P=\dfrac{V^2R}{Z^2}=\dfrac{V^2}{R}\cdot\dfrac{R^2}{Z^2}. At resonance Z=RZ=R, so Pres=V2/RP_{res}=V^2/R. Hence PresP0=Z2R2=2500900=259\dfrac{P_{res}}{P_0}=\dfrac{Z^2}{R^2}=\dfrac{2500}{900}=\dfrac{25}{9}. (The resonant frequency is 2 ω\sqrt2\,\omega since XC/XL=2X_C/X_L=2, but it is not needed.) Using Z/RZ/R without squaring gives 5/35/3.
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