Ideal capacitor power consideration in AC circuits: Evaluate the statement: “In a purely capacitive AC circuit (ideal capacitor, no resistive losses), the energy converted to heat is proportional to the circuit current.” Choose the most accurate assessment.

Difficulty: Easy

Correct Answer: Invalid

Explanation:


Introduction / Context:
Reactive components (ideal capacitors and inductors) in AC circuits exchange energy with the source but, ideally, do not dissipate real power as heat. This distinction between reactive power and real power is foundational for power factor, filter design, and AC analysis.


Given Data / Assumptions:

  • Purely capacitive AC circuit with an ideal capacitor.
  • No series resistance or dielectric loss is considered (lossless assumption).
  • Steady sinusoidal operation.


Concept / Approach:
Instantaneous power in a pure capacitor alternates between positive and negative as the capacitor charges and discharges. The average over a full cycle is zero because the current leads the voltage by 90 degrees. Real power P_avg = V_rms * I_rms * cos(phi). For a pure capacitor, phi = +90 degrees, so cos(phi) = 0 and P_avg = 0 W. Therefore, no average energy is converted to heat in the ideal case, even though current can be large. Any heating in real components comes from non-ideal losses (equivalent series resistance or dielectric losses), not from the ideal reactive behavior itself.


Step-by-Step Solution:

Recall that for a pure capacitor, current leads voltage by 90 degrees.Real power formula: P_avg = V_rms * I_rms * cos(phi).Set phi = 90 degrees ⇒ cos(phi) = 0 ⇒ P_avg = 0 W.Therefore, the statement about heat proportional to current is incorrect in the ideal case.


Verification / Alternative check:
Compute instantaneous power p(t) = v(t) * i(t) for v = Vp sin(ωt) and i = C dv/dt = ωC Vp cos(ωt). Over one period, the average of sin(ωt)cos(ωt) is zero, confirming zero average real power.


Why Other Options Are Wrong:

  • Valid / applies only at resonance / true for low frequencies only: These either contradict the zero real power result or introduce conditions (like resonance) that still do not create real power in an ideal, lossless capacitor.


Common Pitfalls:
Equating high reactive current with heating; ignoring that only resistive components dissipate real power; forgetting the role of phase angle in P_avg calculation.


Final Answer:
Invalid

More Questions from RC Circuits

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion