If the apparent power S is 50 kVA and the power factor PF is 0.9, what is the real power P?

Unlock all questions

This demo includes only 20 questions. Upgrade to access hundreds of questions, flashcards, exam simulations, and disable ads.

Full question bankExam simulationsFlashcards
From $24.99Unlock all

Study for the Ontario Power Generation (OPG) Orange 1 Test. Ace your exam with multiple choice questions and detailed explanations. Prepare confidently!

Multiple Choice

If the apparent power S is 50 kVA and the power factor PF is 0.9, what is the real power P?

Explanation:
Real power is the portion of apparent power that does work. It’s found by multiplying the apparent power by the power factor: P = S × PF. With S = 50 kVA and PF = 0.9, the real power is 50 × 0.9 = 45 kW. So the answer is 45 kW. There is also reactive power, Q, which satisfies S² = P² + Q². Here Q ≈ sqrt(50² − 45²) ≈ 21.8 kVAR, indicating the presence of reactive power when PF is less than 1. The other options would require PF values outside 0–1 (or exceed the given S), so they don’t match the given PF.

Real power is the portion of apparent power that does work. It’s found by multiplying the apparent power by the power factor: P = S × PF. With S = 50 kVA and PF = 0.9, the real power is 50 × 0.9 = 45 kW. So the answer is 45 kW.

There is also reactive power, Q, which satisfies S² = P² + Q². Here Q ≈ sqrt(50² − 45²) ≈ 21.8 kVAR, indicating the presence of reactive power when PF is less than 1.

The other options would require PF values outside 0–1 (or exceed the given S), so they don’t match the given PF.

More Multiple Choice Questions
Subscribe

Get the latest from Examzify

You can unsubscribe at any time. Read our privacy policy