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Question #43081

Difficulty: 8/10 | 🟢 Free

The Question

A series circuit consists of an ideal capacitor of capacitance CC, an inductor of inductance LL, and a radiation resistance RradR_{rad} (accounting for electromagnetic energy radiated by the system). The circuit is driven by an alternating voltage source V(t)=V0sin⁡(ωt)V(t) = V_0 \sin(\omega t). The driving frequency is set to ω=2ω0\omega = 2\omega_0, where ω0=1LC\omega_0 = \frac{1}{\sqrt{LC}} is the natural undamped frequency of the LCLC combination. At this driving frequency, the steady-state current lags the source voltage by a phase angle of π3\frac{\pi}{3} rad. If the peak voltage across the capacitor is measured to be VCV_C, what is the average power PavgP_{avg} dissipated as radiation?

Options

  • A.

    3VC2CL\sqrt{3} V_C^2 \sqrt{\frac{C}{L}}

  • B.

    23VC2CL2\sqrt{3} V_C^2 \sqrt{\frac{C}{L}}

  • C.

    33VC2CL\frac{\sqrt{3}}{3} V_C^2 \sqrt{\frac{C}{L}}

  • D.

    34VC2CL\frac{\sqrt{3}}{4} V_C^2 \sqrt{\frac{C}{L}}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Average Power over Cycle|LC Oscillations Energy Cycle|Bandwidth in RLC

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to transition from ideal LC circuits to non-conservative driven RLC systems, specifically focusing on radiation damping.

The primary trap lies in the calculation of average power.

Students often incorrectly assume Pavg=VrmsIrmsP_{avg} = V_{rms} I_{rms} without accounting for the power factor (cos⁡ϕ\cos\phi), which leads to Option B.

Alternatively, they might incorrectly apply the capacitor's voltage directly to the resistor (Option C) or fail to use the driven frequency ω\omega instead of the natural frequency ω0\omega_0 when relating VCV_C to I0I_0 (Option D).

The correct approach requires determining the hidden radiation resistance via the phase lag and using the local current-voltage relationship at the capacitor to find the true dissipated power.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterAlternating Current - Class 12
ConceptAverage Power over Cycle
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree