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Question #43060

Difficulty: 8/10 | 🟢 Free

The Question

A voltage source $V(t) = V_0 (1 + \cos \omega t)$ is connected across a parallel combination of an ideal capacitor of capacitance $C$ and a real inductor having inductance $L$ and resistance $R$ . The circuit has been operating for a long time such that steady-state conditions are established. Given that $\omega L = R$ and $\omega C = 1/R$ , find the total heat dissipated in the circuit during the time interval from $t=0$ to $t=\pi/\omega$ .

Options

A.
$\frac{V_0^2}{\omega R} \left( \frac{5\pi}{4} + 2 \right)$
B.
$\frac{V_0^2}{\omega R} \left( \frac{5\pi}{4} \right)$
C.
$\frac{V_0^2}{\omega R} \left( \frac{5\pi}{4} - 1 \right)$
D.
$\frac{V_0^2}{\omega R} \left( \frac{5\pi}{4} + 2\sqrt{2} \right)$

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

Wattless Current|LC Oscillations Energy Cycle|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the precise application of Joule heating in a circuit with superimposed DC and AC components over a non-integer number of full cycles.

A common trap is erroneously assuming that the cross-product term $2 I_{DC} I_{AC}$ in the power integral $P = I^2 R$ averages to zero, which is only true for complete cycles, not the half-cycle specified.

Additionally, students might incorrectly equate the total heat dissipated to the net energy delivered by the voltage source, failing to account for the non-zero change in stored magnetic energy ( $\Delta E_L$ ) in the inductor over this specific interval.

Finally, ignoring the $\pi/4$ phase lag introduced by the inductor's resistance leads to an incorrect evaluation of the cross-term integral.

Quick Facts About This Question

Question Information

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