PrepRung

JEE Advanced 2019 Paper 1 · Physics · Electromagnetic Induction and Alternating Current · medium

A series circuit has R=10 Ω, L=0.10 H and C=100 muF connected to a variable-frequency source of constant rms emf 100 V.

One or more options correct

A series circuit has R=10 ΩR=10\ \Omega, L=0.10 HL=0.10\ \mathrm{H} and C=100 μFC=100\ \mu\mathrm{F} connected to a variable-frequency source of constant rms emf 100 V100\ \mathrm{V}. Which of the following statements is/are correct?
  1. A. The average power dissipated at resonance is 1000 W1000\ \mathrm{W}.
  2. B. The quality factor of the circuit is 1010.
  3. C. At an angular frequency 2ω02\omega_0 (ω0\omega_0 being the resonant value) the current leads the applied voltage.
  4. D. The half-power bandwidth is 50 rad s−150\ \mathrm{rad\,s^{-1}}.
Show answer and solution

Answer: A

At resonance XL=XCX_L=X_C, so the impedance equals RR and the rms current is 100/10=10 A100/10=10\ \mathrm{A}. The power is I2R=1000 WI^2R=1000\ \mathrm{W}, so (A) is true. The resonant frequency is ω0=1/LC=1/10−5=316 rad s−1\omega_0=1/\sqrt{LC}=1/\sqrt{10^{-5}}=316\ \mathrm{rad\,s^{-1}}, so Q=ω0L/R=31.6/10=3.16Q=\omega_0L/R=31.6/10=3.16, not 1010; (B) is false. At 2ω02\omega_0 we have XL=2ω0LX_L=2\omega_0L doubled and XCX_C halved, so XL>XCX_L>X_C and the circuit is inductive: the current lags the voltage, hence (C) is false. The half-power bandwidth of a series RLC circuit is Δω=R/L=100 rad s−1\Delta\omega=R/L=100\ \mathrm{rad\,s^{-1}}, not 5050, so (D) is false. Only (A) is correct.
Report a mistake

Original question written in the JEE Advanced 2019 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake