Reactive power Q for the load with P = 831 W and S = 960 VA is approximately?

Reactive power Q for the load with P = 831 W and S = 960 VA is approximately?

• A. Approximately 481 VAR
• B. Approximately 960 VAR
• C. Approximately 831 VAR
• D. Approximately 120 VAR

Answer: (A)

Think of P, Q, and S as a right triangle: real power P is the along-the-velocity leg, reactive power Q is the perpendicular leg, and apparent power S is the hypotenuse. They’re related by P^2 + Q^2 = S^2. So with P = 831 W and S = 960 VA, the reactive power is Q = sqrt(S^2 − P^2) = sqrt(960^2 − 831^2). Calculating: 960^2 = 921,600 and 831^2 = 690,561, so the difference is 231,039. sqrt(231,039) ≈ 481 VAR. Thus reactive power is about 481 VAR. This also aligns with the power factor: cos(phi) = P/S ≈ 831/960 ≈ 0.866, giving a phase angle around 30°, and Q = S sin(phi) ≈ 960 × 0.5 ≈ 480 VAR. The other options don’t fit the P^2 + Q^2 = S^2 relationship.

Think of P, Q, and S as a right triangle: real power P is the along-the-velocity leg, reactive power Q is the perpendicular leg, and apparent power S is the hypotenuse. They’re related by P^2 + Q^2 = S^2. So with P = 831 W and S = 960 VA, the reactive power is Q = sqrt(S^2 − P^2) = sqrt(960^2 − 831^2).

Calculating: 960^2 = 921,600 and 831^2 = 690,561, so the difference is 231,039. sqrt(231,039) ≈ 481 VAR.

Thus reactive power is about 481 VAR. This also aligns with the power factor: cos(phi) = P/S ≈ 831/960 ≈ 0.866, giving a phase angle around 30°, and Q = S sin(phi) ≈ 960 × 0.5 ≈ 480 VAR. The other options don’t fit the P^2 + Q^2 = S^2 relationship.