Which expression correctly relates apparent power S to real P and reactive Q?

Which expression correctly relates apparent power S to real P and reactive Q?

• A. S = √(P^2 + Q^2)
• B. S = P × Q
• C. S = P / cosφ
• D. S = V × I

Answer: (A)

In AC power, real power P (watts) and reactive power Q (volt-amps reactive) combine to form apparent power S (volt-amperes). They relate like a right triangle: P is the horizontal leg, Q is the vertical leg, and S is the hypotenuse. By the Pythagorean theorem, the square of the hypotenuse equals the sum of the squares of the legs, so S^2 = P^2 + Q^2. Therefore S = sqrt(P^2 + Q^2). This directly ties S to both P and Q. For example, if P = 300 W and Q = 400 VAR, then S = sqrt(300^2 + 400^2) = 500 VA. The other expressions don’t directly relate S to P and Q in this context: multiplying P and Q gives a quantity with incompatible units; S = P / cosφ involves the phase angle φ and isn’t a direct P–Q relationship, though cosφ = P/S would tie them if φ is known; and S = V × I describes apparent power in terms of voltage and current, not P and Q.

In AC power, real power P (watts) and reactive power Q (volt-amps reactive) combine to form apparent power S (volt-amperes). They relate like a right triangle: P is the horizontal leg, Q is the vertical leg, and S is the hypotenuse. By the Pythagorean theorem, the square of the hypotenuse equals the sum of the squares of the legs, so S^2 = P^2 + Q^2. Therefore S = sqrt(P^2 + Q^2).

This directly ties S to both P and Q. For example, if P = 300 W and Q = 400 VAR, then S = sqrt(300^2 + 400^2) = 500 VA.

The other expressions don’t directly relate S to P and Q in this context: multiplying P and Q gives a quantity with incompatible units; S = P / cosφ involves the phase angle φ and isn’t a direct P–Q relationship, though cosφ = P/S would tie them if φ is known; and S = V × I describes apparent power in terms of voltage and current, not P and Q.