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Question #43332

Difficulty: 8/10 | 🟢 Free

The Question

An ideal voltage source outputs a periodic non-sinusoidal signal given by $V(t) = 120 \left[ \sin(100 t) + \frac{1}{2}\sin(200 t) + \frac{1}{3}\sin(300 t) \right] \text{ V}$ where $t$ is in seconds. The source is connected across a pure capacitor of capacitance $C = 50 \text{ \mu F}$ . What is the root-mean-square (RMS) value of the total current $I_{rms}$ flowing through the circuit?

Options

A.
$0.3\sqrt{6} \text{ A}$
B.
$0.35\sqrt{2} \text{ A}$
C.
$0.6\sqrt{3} \text{ A}$
D.
$0.3\sqrt{2} \text{ A}$

Check Your Answer

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

Capacitive Reactance Dependency|Principle of Superposition|Fourier-like Harmonic Summation

Why This Question Is Tricky

Common Mistakes

This question tests the student's understanding of capacitive reactance in the presence of non-sinusoidal periodic inputs.

A common misconception is to calculate the total RMS voltage ( $V_{rms} = 70\sqrt{2} \text{ V}$ ) and divide it by the capacitive reactance of the fundamental frequency ( $X_C = 200 \text{ \Omega}$ ), which incorrectly yields $0.35\sqrt{2} \text{ A}$ .

Students fail to recognize that a capacitor acts as a high-pass filter; its reactance $X_C = 1/(n\omega C)$ decreases for higher harmonics.

Consequently, even though the voltage amplitude of the $n$ -th harmonic decreases as $1/n$ , the current amplitude remains constant, significantly increasing the true RMS current.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Alternating Current - Class 12
Concept	Capacitive Reactance Dependency
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free