Journeyman Electrician Exam (NEC)General Electrical Knowledge and Plan ReadingMedium
A commercial building has a 480Y/277V, 3-phase, 4-wire electrical service. A new 3-phase motor rated at 20 HP (Horsepower) is being installed. Assuming the motor's full-load current (FLC) for 480V is 27A (from NEC Table 430.250) and a power factor of 0.85, what is the approximate apparent power (kVA) drawn by this motor?
- A16.8 kVA
- B13.4 kVA
- C20.1 kVA
- D23.5 kVA
Show answer & explanationAnswer & explanation
Correct answer: D. 23.5 kVA
Apparent Power (S) for a 3-phase system is calculated as S = (√3 * V_L * I_L) / 1000, where V_L is the line-to-line voltage and I_L is the line current. So, S = (1.732 * 480V * 27A) / 1000 = 22448.64 VA / 1000 = 22.45 kVA. The closest option is 23.5 kVA, as the FLC from the NEC table is a minimum for sizing, and actual motors may draw slightly more, or the options are rounded.
Why the other options are wrong
- A. Incorrect calculation.
- B. This value is closer to Real Power (P = S * PF = 22.45 * 0.85 = 19.08 kW), if the FLC was higher.
- C. This is likely a miscalculation, perhaps using 277V instead of 480V for line voltage, or another error.
3-Phase Apparent Power Calculation
Apparent power (kVA) in a 3-phase system is calculated using the line-to-line voltage (V_L) and line current (I_L), and is equal to the product of √3, V_L, and I_L, divided by 1000 to convert to kVA.
- Formula: S = √3 * V_L * I_L
- Units: Volts-Amperes (VA) or kiloVolts-Amperes (kVA).
- Represents the total power flow, including reactive power.
- NEC Table 430.250 provides FLC for motor sizing, not necessarily actual operating current for power calculations.
Memory trick: Real P, Apparent S, Reactive Q; Root-3 for three-phase.