Journeyman Electrician Exam (NEC)General Electrical Knowledge and Plan ReadingMedium
A client's single-phase AC motor draws 15 A at 240 V with a power factor of 0.8 lagging. What is the approximate real power (P) consumed by the motor?
- A2.88 kW
- B5.40 kW
- C3.60 kW
- D4.32 kW
Show answer & explanationAnswer & explanation
Correct answer: A. 2.88 kW
For a single-phase AC circuit, real power (P) = V × I × PF, where V is voltage, I is current, and PF is power factor. P = 240 V × 15 A × 0.8 = 2880 W. Converting to kilowatts, P = 2.88 kW.
Why the other options are wrong
- B. This is an incorrect calculation, possibly dividing by 0.8 instead of multiplying.
- C. This would be the apparent power (S = V * I = 240V * 15A = 3600 VA = 3.60 kVA).
- D. This is an incorrect calculation, possibly multiplying by 1.2 or other error.
Single-Phase Real Power
Real power (P) in a single-phase AC circuit is the actual power consumed by the load and converted into useful work or heat, measured in Watts (W).
- Formula: P = V × I × PF, where V is voltage, I is current, and PF is power factor.
- Also known as active power or true power.
- Always less than or equal to apparent power (S).
Memory trick: P.I.F. – Power Is Fun, with PF.