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Question #42735

Difficulty: 8/10 | 🟢 Free

The Question

An ideal capacitor of capacitance $C = 40 \, \mu\text{F}$ is initially charged to a potential difference $V_0 = 100 \, \text{V}$ . At $t = 0$ , it is connected in series with an inductor $L = 1.0 \, \text{mH}$ and a resistor $R$ . The resistor has color bands in the sequence: Blue, Black, Gold, Silver. Assuming the nominal value of the resistance, calculate the exact root-mean-square (RMS) current $I_{\text{rms}}$ in the circuit over the first complete period $T$ of its damped oscillation. (Given: The period $T$ is defined as the time taken for the oscillatory part of the current to complete one full cycle).

Options

A.
$10 \sqrt{\frac{2}{3\pi} \left(1 - e^{-3\pi}\right)} \, \text{A}$
B.
$10 \sqrt{2} \, \text{A}$
C.
$10 \sqrt{\frac{2}{3\pi} \left(1 - e^{-1.5\pi}\right)} \, \text{A}$
D.
$20 \sqrt{\frac{1}{3\pi} \left(1 - e^{-3\pi}\right)} \, \text{A}$

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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

Color Code Logic|Instantaneous vs RMS vs Peak|LC Circuit Oscillations|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to bridge circuit theory, analytical calculus, and energy conservation.

The primary trap lies in the calculation of the RMS current for a damped oscillation.

Students often attempt to directly integrate the instantaneous power $P(t) = i^2(t) R$ , which is mathematically brutal and prone to errors (such as forgetting the factor of 1/2 from the average of sine squared, leading to Option D).

Others incorrectly assume the undamped RMS current $I_{rms} = I_{max}/\sqrt{2}$ (Option B).

The elegant JEE Advanced approach requires recognizing that at exactly one period $T$ , the current is zero, meaning all remaining energy is purely electrostatic.

By applying the Work-Energy Theorem globally over the period ( $\Delta U = I_{rms}^2 R T$ ), the complex integration is entirely bypassed, rewarding deep physical insight over brute-force mathematics.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Current Electricity - Class 12
Concept	Color Code Logic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free