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Question #43058

Difficulty: 8/10 | 🟢 Free

The Question

A series R−L−CR-L-C circuit has a resistance R=12 ΩR = 12\ \Omega. At an angular frequency ω0\omega_0, the inductive reactance is XL=413 ΩX_L = \frac{41}{3}\ \Omega and the capacitive reactance is XC=683 ΩX_C = \frac{68}{3}\ \Omega. The circuit is driven by a composite voltage source given by V(t)=V0sin⁡(ω0t)+V0sin⁡(2ω0t)V(t) = V_0 \sin(\omega_0 t) + V_0 \sin(2\omega_0 t). Calculate the effective power factor of the circuit, defined as pf=PavgVrmsIrmspf = \frac{P_{avg}}{V_{rms} I_{rms}}, where PavgP_{avg} is the average power dissipated, and VrmsV_{rms} and IrmsI_{rms} are the root-mean-square voltage and current of the source.

Options

  • A.

    12\frac{1}{\sqrt{2}}

  • B.

    710\frac{7}{10}

  • C.

    11

  • D.

    45\frac{4}{5}

Check Your Answer

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

Power Factor Definition|Phase Constant (Epoch)|Superposition of SHMs

Why This Question Is Tricky

Common Mistakes

This question tests the application of superposition in AC circuits with multiple frequencies.

The lethal trap occurs when students attempt to calculate an 'average' phase angle or use standard phasor addition for non-identical frequencies.

Many will incorrectly assume the effective power factor is the arithmetic mean of the power factors at ω0\omega_0 (cos⁡ϕ1=12/15=0.8\cos\phi_1 = 12/15 = 0.8) and 2ω02\omega_0 (cos⁡ϕ2=12/20=0.6\cos\phi_2 = 12/20 = 0.6), yielding 0.70.7.

Others might incorrectly calculate Vrms=V0/2V_{rms} = V_0/\sqrt{2} by treating the maximum amplitude of the composite wave as V0V_0, failing to recognize that orthogonal functions sum in quadrature (Vrms2=V1rms2+V2rms2V_{rms}^2 = V_{1rms}^2 + V_{2rms}^2).

The correct approach requires abandoning the single-frequency impedance triangle and reverting to the fundamental definitions of RMS values and average power via instantaneous superposition.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterAlternating Current - Class 12
ConceptPower Factor Definition
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree