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Question #43158

Difficulty: 8/10 | 🟢 Free

The Question

A series LCR circuit with resistance RR (in Ω\Omega), inductance LL (in H), and capacitance CC (in F) is driven by a superposition of two AC voltage sources: V(t)=V0cos⁡(ω1t)+V0cos⁡(ω2t)V(t) = V_0 \cos(\omega_1 t) + V_0 \cos(\omega_2 t), where V0V_0 is the voltage amplitude (in V) and tt is time (in s). The angular frequencies (in rad/s) are symmetrically offset from the circuit's resonant frequency ω0=1/LC\omega_0 = 1/\sqrt{LC} such that ω1=ω0+Δω\omega_1 = \omega_0 + \Delta \omega and ω2=ω0−Δω\omega_2 = \omega_0 - \Delta \omega, where Δω≪ω0\Delta \omega \ll \omega_0. The total voltage V(t)V(t) exhibits a beat pattern with an envelope that periodically reaches a maximum. The resulting steady-state current I(t)I(t) also exhibits a beat pattern. What is the time lag Δt\Delta t of the maximum of the current envelope relative to the nearest maximum of the voltage envelope?

Options

  • A.

    1Δωtan⁡−1(2LΔωR) s\frac{1}{\Delta \omega} \tan^{-1}\left(\frac{2L\Delta \omega}{R}\right) \text{ s}

  • B.

    1ω0tan⁡−1(2LΔωR) s\frac{1}{\omega_0} \tan^{-1}\left(\frac{2L\Delta \omega}{R}\right) \text{ s}

  • C.

    1Δωtan⁡−1(Lω0R) s\frac{1}{\Delta \omega} \tan^{-1}\left(\frac{L\omega_0}{R}\right) \text{ s}

  • D.

    0 s0 \text{ s}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Frequency Sensitivity of Z|Beats and Frequency Difference

Why This Question Is Tricky

Common Mistakes

This question tests the student's understanding of AC circuit analysis and the superposition principle in the presence of multiple driving frequencies (beats).

The primary trap is assuming that because the average driving frequency is the resonant frequency ω0\omega_0, the circuit behaves purely resistively, leading to the incorrect conclusion that the current envelope is perfectly in phase with the voltage envelope (Option D).

Students often fail to recognize that the complex impedance Z(ω)Z(\omega) must be evaluated independently for each frequency component.

Because the impedance is inductive at ω1\omega_1 and capacitive at ω2\omega_2, the individual current components acquire equal and opposite phase shifts.

When these components superpose, the phase shift manifests as a macroscopic time delay in the beat envelope, rather than a phase shift of the high-frequency carrier.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterAlternating Current - Class 12
ConceptFrequency Sensitivity of Z
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree