For a phasor voltage V and current I with power factor cos(phi), which set of equations correctly defines real, reactive, and apparent power and the power factor?

• A. P = VI; Q = VI sin(phi); S = VI*; PF = cos(phi)
• B. P = VI cos(phi); Q = VI sin(phi); S = VI*; PF = cos(phi)
• C. P = VI cos(phi); Q = VI sin(phi); S = VI; PF = sin(phi)
• D. P = VI cos(phi); Q = VI sin(phi); S = VI*; PF = sin(phi)

Answer: (B)

Power in an AC circuit splits into three parts that relate to how the voltage and current line up in time. The real (active) power is the portion that actually does work, and it depends on how much the current overlaps the voltage in phase: P = V I cos(phi). The reactive power is the part that exchanges energy between the source and the reactive elements (it oscillates with the cycle), and it depends on the sine of the phase angle: Q = V I sin(phi). The apparent power is the overall power flow, the product of the rms voltage and current, giving a magnitude S = V I. When you treat the voltages and currents as phasors, the complex power is S = V I*, whose real part is P and whose imaginary part is related to Q, and the power factor is PF = P / S = cos(phi). So the best set of relations uses P = V I cos(phi), Q = V I sin(phi), and S represented as V I* (with PF = cos(phi)). The other options mix in an incorrect dependence (like missing the cos(phi) in real power or assigning PF as sin(phi)) or use the wrong form for the complex power.