In an RC AC circuit, what is the correct expression for true (real) power PR?

Difficulty: Easy

Correct Answer: I^2 * R

Explanation:


Introduction:
True (real) power is the portion of power converted to heat or useful work. In an RC AC circuit, only the resistor dissipates real power; the capacitor exchanges energy with the source but does not dissipate it (ideal case). This item checks recognition of the correct algebraic form for PR.


Given Data / Assumptions:

  • Series or parallel RC under sinusoidal steady state.
  • Ideal capacitor with zero loss.
  • R is the only dissipative element.


Concept / Approach:
The resistor's real power is PR = I_rms^2 * R. Alternative forms include PR = V_R(rms)^2 / R. Apparent power S = V_rms * I_rms, while reactive power Q is associated with the capacitor. The product VS * I does not generally equal real power when there is phase shift between voltage and current.


Step-by-Step Solution:

Identify the dissipative element: the resistor R.Use PR = I^2 * R (rms quantities).Capacitor contributes Q, not P, because current and voltage are 90° out of phase for the reactive branch.


Verification / Alternative check:
Compute P = V * I * cos(phi). For a series RC, cos(phi) = R / Z. With I the same through both components, PR = I^2 * R, which matches P = V * I * cos(phi).


Why Other Options Are Wrong:

  • VS × I: This is apparent power when there is a phase shift; not necessarily real power.
  • VT × RT: Not a standard power expression; has unit inconsistencies.


Common Pitfalls:
Equating V * I with real power regardless of phase, or neglecting that only resistors dissipate real power in the ideal model.


Final Answer:
I^2 * R.

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