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 this load?
- A18.7 kW
- B6.24 kW
- C10.8 kW
- D14.4 kW
Show answer & explanationAnswer & explanation
Correct answer: C. 10.8 kW
For a 3-phase pure resistive load, the real power (P) is calculated as P = √3 * V_L * I_L * Power Factor. Since it's purely resistive, the Power Factor (PF) is 1. So, P = 1.732 * 208V * 30A = 10798.56 watts, or approximately 10.8 kW.
Why the other options are wrong
- A. This is significantly too high, indicating a major calculation error or incorrect formula.
- B. This could be a miscalculation, perhaps omitting the √3 factor or an incorrect power factor.
- D. This value is too high, possibly from an incorrect power factor or calculation error.
Three-Phase Real Power (Resistive)
Real power (P) in a three-phase purely resistive circuit is calculated as the product of the square root of 3, line voltage, line current, and a power factor of 1 (since it's purely resistive).
- Formula: P = √3 * V_L * I_L * PF (where PF = 1 for purely resistive).
- Measured in watts (W) or kilowatts (kW).
- V_L is line-to-line voltage, I_L is line current.
- Purely resistive loads have a power factor of 1.
Memory trick: Root three Volts Amps Power Factor, that's the real power factor.