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Question #43878

Difficulty: 8/10 | 🟢 Free

The Question

A series LCR circuit is driven by an external alternating voltage source such that the steady-state current in the circuit is $I(t) = I_0 \sin(\omega t)$ . The circuit consists of an inductor, a capacitor, and a non-linear resistor. The voltage drop across the non-linear resistor is given by $V_R = k I|I|$ , where $k = 2 \text{ V/A}^2$ . The net reactance of the inductor and capacitor at the driving frequency $\omega$ is $X = X_L - X_C = 8 \text{ } \Omega$ . If the peak current is $I_0 = 4 \text{ A}$ , what is the peak voltage $V_{max}$ of the external source?

Options

A.
$40 \text{ V}$
B.
$32\sqrt{2} \text{ V}$
C.
$64 \text{ V}$
D.
$32\sqrt{5} \text{ V}$

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

Resistor Phase Invariance|Series LCR Impedance Vector|Dynamic Resistance

Why This Question Is Tricky

Common Mistakes

This question tests the boundary of the standard phasor method, which strictly requires Linear Time-Invariant (LTI) systems.

Students typically assume the resistor's phase invariance allows them to use the static impedance triangle $Z = \sqrt{R^2 + X^2}$ .

By introducing a non-linear resistor (dynamic resistance), the effective resistance fluctuates with the instantaneous current.

The trap lures students into calculating an 'effective' static resistance at peak current ( $R_{eff} = V_{R,max}/I_0 = 8 \text{ } \Omega$ ) and incorrectly applying the phasor sum $V_{max} = \sqrt{32^2 + 32^2} = 32\sqrt{2} \text{ V}$ .

The correct approach forces the abandonment of algebraic phasor addition in favor of maximizing the instantaneous non-linear Kirchhoff's voltage equation using calculus.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Alternating Current - Class 12
Concept	Resistor Phase Invariance
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free