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Question #42851

Difficulty: 8/10 | 🟢 Free

The Question

A series LCR circuit consists of a resistor with resistance $R = 4 \, \Omega$ , an inductor with inductance $L = 10 \, \text{mH}$ , and a capacitor with capacitance $C = 2.5 \, \text{mF}$ . The circuit is driven by a time-varying voltage source given by $V(t) = 15\sqrt{2} \sin(100t) + 12\sqrt{2} \sin(200t)$ where $V$ is in volts and $t$ is in seconds. A hot-wire ammeter of negligible resistance is connected in series with the circuit. What is the reading of the ammeter?

Options

A.
$3\sqrt{2} \, \text{A}$
B.
$6 \, \text{A}$
C.
$6\sqrt{2} \, \text{A}$
D.
$\frac{3\sqrt{41}}{5} \, \text{A}$

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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

Superposition of Non-identical Waves|Series LCR Impedance Vector

Why This Question Is Tricky

Common Mistakes

This question tests the application of the Superposition Principle in AC circuits with frequency-dependent impedance.

A common trap is assuming the circuit has a single global impedance and dividing the total RMS voltage by it, or linearly adding the RMS currents ( $3 \, \text{A} + 3 \, \text{A} = 6 \, \text{A}$ ) instead of adding them in quadrature.

Students must recognize that impedance must be evaluated separately for each frequency component, and because the harmonic cross-terms average to zero over time, the total RMS current is the square root of the sum of the squares of the individual RMS currents.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Waves - Class 11 Part 2
Concept	Superposition of Non-identical Waves
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free