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Question #43002

Difficulty: 8/10 | 🟢 Free

The Question

A capacitor of capacitance $C$ is initially charged to a voltage $V_0$ . At time $t=0$ , it is connected in series with an inductor of inductance $L$ and a resistor of resistance $R$ to form a closed loop. The system is weakly damped such that $R \ll \sqrt{\frac{L}{C}}$ . Which of the following expressions correctly represents the average power dissipated in the resistor over the time interval from $t=0$ to $t = \frac{L}{R}$ ?

Options

A.
$\frac{C V_0^2 R}{2L} (1 - e^{-1})$
B.
$\frac{C V_0^2 R}{2L}$
C.
$\frac{C V_0^2 R}{2L} (1 - e^{-1/2})$
D.
$\frac{C V_0^2 R}{L} (1 - e^{-2})$

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

LC Circuit Oscillations|Energy Partitioning|Average Power over Cycle

Why This Question Is Tricky

Common Mistakes

This question tests the distinction between steady-state RMS power and the time-averaged power of a decaying signal.

A common trap is assuming the circuit behaves like a steady AC circuit, leading students to blindly apply the steady-state formula $P_{avg} = I_{rms}^2 R = \frac{1}{2} I_0^2 R = \frac{C V_0^2 R}{2L}$ (Option B).

This ignores the exponential decay of the current envelope.

Another trap is confusing the amplitude decay constant ( $2L/R$ ) with the energy decay constant ( $L/R$ ), leading to Option C.

By requiring the calculation of average power via the macroscopic energy difference $\Delta U / \Delta t$ , the question forces students to reconcile the decaying envelope with the true energy-loss rate.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electromagnetic Induction - Class 12
Concept	LC Circuit Oscillations
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free