CSLB C-10 Electrical ContractorElectrical Theory and CalculationsMedium
A 208V, 3-phase pure resistive load draws 30 amps. What is the approximate total real power consumed by the load?
- A12.5 kW
- B6.24 kW
- C7.2 kW
- D10.8 kW
Show answer & explanationAnswer & explanation
Correct answer: D. 10.8 kW
For a pure resistive 3-phase load, the power factor (PF) is 1. The real power (P) is calculated as P = √3 × V × I × PF. So, P = √3 × 208V × 30A × 1 = 1.732 × 208V × 30A = 10795.2 W, which is approximately 10.8 kW.
Why the other options are wrong
- A. This could be a miscalculation or an incorrect application of the square root of 3.
- B. This would be V*I for single phase, or a calculation error.
- C. This would be if the power factor was less than 1, implying reactive power.
Three-Phase Real Power (Resistive)
The total real power consumed by a three-phase resistive load, where the power factor is effectively 1, calculated using P = √3 × V × I.
- PF = 1 for purely resistive loads.
- P = √3 × V × I (simplified due to PF=1).
- Represents all energy converted to heat or work.
Memory trick: Resistive means Power Factor is One, so just Root-3 V I.