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:
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:
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:
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
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