In an RC AC circuit, what is the correct expression for true (real) power PR?
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AVS * I
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BVS × I
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CI^2 * R
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DVT × RT
Answer
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.