FCC Exam Question: 3-8A5

How do you compute true power (power dissipated in the circuit) in a circuit where AC voltage and current are out of phase?

A. Multiply RMS voltage times RMS current.
B. Subtract apparent power from the power factor.
C. Divide apparent power by the power factor.
D. Multiply apparent power times the power factor.
Correct Answer: D

Explanation: In AC circuits where voltage and current are out of phase, we distinguish between *apparent power* and *true power*. * **Apparent power** is the product of RMS voltage and RMS current (V_RMS * I_RMS), measured in Volt-Amperes (VA). It represents the total power flowing in the circuit. * **True power** (or real power) is the actual power dissipated by the circuit's resistive components, doing useful work, measured in Watts (W). * The **power factor** is a value between 0 and 1, representing the cosine of the phase angle between voltage and current. It indicates how much of the apparent power is true power. To compute the true power, you must account for the phase difference. Multiplying the apparent power by the power factor directly gives you the true power dissipated. This relationship is often expressed as P_true = P_apparent × PF. Let's look at the other options: A) Multiplying RMS voltage times RMS current calculates *apparent power*, not true power when they are out of phase. B) Subtracting apparent power from the power factor is mathematically incorrect and doesn't yield a meaningful result. C) Dividing apparent power by the power factor would give a value greater than or equal to apparent power, which is incorrect because true power can never exceed apparent power.

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Includes Elements 1, 3, 6, 7R, 8, and 9.