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Question #43082

Difficulty: 8/10 | 🟢 Free

The Question

A series LCR circuit consists of a resistor $R = 12 \, \Omega$ , an inductor $L = 90 \, \text{mH}$ , and a capacitor $C = 2.5 \, \text{mF}$ . The circuit is driven by an alternating voltage source given by $v(t) = 260 \sin(100 t) + 100 \sin(200 t)$ , where $v(t)$ is in volts and $t$ is in seconds. Calculate the total average power dissipated in the circuit.

Options

A.
$2400 \, \text{W}$
B.
$2550 \, \text{W}$
C.
$2850 \, \text{W}$
D.
$5100 \, \text{W}$

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

Average Power over Cycle|Superposition Principle|Fourier-style Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's understanding of average power in AC circuits subjected to non-sinusoidal harmonic injection.

A common misconception is attempting to calculate a single overall $V_{rms}$ and applying the standard $P = V_{rms} I_{rms} \cos(\phi)$ formula, failing to realize that impedance and phase angle are frequency-dependent.

Furthermore, students often ignore the orthogonality of trigonometric functions, which dictates that average power must be calculated independently for each harmonic frequency and then summed.

The traps include ignoring the power factor for individual harmonics (yielding 2850 W), forgetting the RMS factor of 1/2 (yielding 5100 W), or completely neglecting the higher-order harmonic's contribution (yielding 2400 W).

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Alternating Current - Class 12
Concept	Average Power over Cycle
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free