True power in an RC series circuit: Is the real power dissipated given by P_R = I^2 * R, with the capacitor contributing zero average real power?
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ACorrect
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BIncorrect
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COnly true at DC
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DOnly true at the circuit's cutoff frequency
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EOnly true if R is very small
Answer
Correct Answer: Correct
Explanation
Introduction / Context:In AC circuits, resistors dissipate real power as heat, while ideal capacitors and inductors exchange reactive power with the source, averaging to zero over a complete cycle. Understanding this distinction is essential for power factor correction and thermal design.
Given Data / Assumptions:
- Sinusoidal steady state with RMS current I.
- Ideal capacitor with no dielectric loss (no ESR for first-order analysis).
- Series RC topology.
Concept / Approach:Real or true power is P = V_RMS * I_RMS * cos(phi). For an RC series, the phase angle phi satisfies cos(phi) = R / |Z|. Equivalently, the real power equals I^2 * R because only the resistor dissipates energy. The capacitor's voltage and current are 90 degrees out of phase, so its average power over a cycle is zero, though it stores and returns energy within each cycle.
Step-by-Step Solution:
Write Z = R − j * Xc and compute I = V / |Z|.Real power: P = I^2 * R (since the resistive part alone dissipates).Reactive exchange in C: Q_C = V * I * sin(phi) with zero average real power in an ideal capacitor.Therefore, P_R = I^2 * R is the correct expression.Verification / Alternative check:Wattmeter readings in lab confirm that power scales with the resistive element; adding series capacitance at fixed current does not increase true power, but changes phase and current at fixed voltage.
Why Other Options Are Wrong:
Incorrect: contradicts basic AC power relations.DC-only or cutoff-only qualifiers are unnecessary; the expression holds across frequencies for ideal components.Common Pitfalls:Confusing apparent power S = V * I with real power P; ignoring capacitor ESR which introduces small real-loss in practice.
Final Answer:Correct