JEE Advanced 2018 Paper 1 · Physics · Electromagnetic Induction and Alternating Current · medium

A series LCR circuit has R=8.0 Ω, L=0.20 H and C=50 muF, and is driven by a source of rms voltage 40 V whose angular frequency is set to 0.90 times…

Numerical answer

A series LCR circuit has

R=8.0\ \Omega

L=0.20\ \mathrm{H}

and

C=50\ \mu\mathrm{F}

, and is driven by a source of rms voltage

40\ \mathrm{V}

whose angular frequency is set to

0.90

times the resonant angular frequency of the circuit. Find the rms potential difference, in volts, across the capacitor. Give the answer correct to two decimal places.

Show answer and solution

Answer: 180.59

The resonant angular frequency is

\omega_0=1/\sqrt{LC}=1/\sqrt{0.2\times50\times10^{-6}}=316.23\ \mathrm{rad\,s^{-1}}

, so the drive is

\omega=0.9\omega_0=284.60\ \mathrm{rad\,s^{-1}}

. The reactances are

X_L=\omega L=56.92\ \Omega

and

X_C=1/(\omega C)=70.28\ \Omega

. The impedance is

Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{64+178.6}=15.57\ \Omega

. The rms current is

I=40/15.57=2.570\ \mathrm{A}

, and the rms voltage across the capacitor is

V_C=IX_C=2.570\times70.28=180.59\ \mathrm{V}

. This exceeds the source voltage several times, since the circuit has a high quality factor and the drive is near resonance.

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