# Power factor, efficiency and horsepower calculations

> True power in watts equals apparent power in volt-amperes times the power factor, so kW = kVA × PF. Efficiency is output divided by input, and one horsepower of motor output equals 746 watts, so a motor's input watts are hp × 746 ÷ efficiency.

Exams: Journeyman Electrician, Master Electrician. Electrical theory.

## Formula

    True power (W) = Apparent power (VA) × PF
    Efficiency (%) = Output (W) ÷ Input (W) × 100
    Motor output (W) = hp × 746

## Variables

- **W**: True (real) power (W)
- **VA**: Apparent power, E × I (× 1.732 for three-phase) (VA)
- **PF**: Power factor, W ÷ VA, as a decimal
- **Eff**: Efficiency, output ÷ input, as a decimal
- **hp**: Motor output horsepower (1 hp = 746 W) (hp)

## Open your 2026 NEC to

- 430.6(A)(1): Size motor conductors and protection from the table full-load current, not a nameplate or calculated current
- Table 430.250: Three-phase motor full-load currents, to compare with a calculated current

Motors, transformers and ballasts are inductive loads. Their current is out of step with the voltage, so voltage times current (apparent power, in VA) is more than the power that actually does work (true power, in watts). Power factor is the ratio of the two, W ÷ VA, a number between 0 and 1 that is often written as a percent.

Efficiency compares what comes out of a machine to what goes in. A motor's horsepower rating is its mechanical output, and each horsepower is 746 W. Because of losses in the windings and bearings, the electrical input is always larger: input watts = hp × 746 ÷ efficiency.

Put together, a three-phase motor draws I = hp × 746 ÷ (E × 1.732 × Eff × PF). Exam questions may ask for this current. Remember that for sizing motor conductors and protection, 430.6(A)(1) sends you to the full-load current tables instead.

## Worked example

A balanced three-phase load on a 480 V system draws 181 A per phase at a power factor of 0.90. What is the true power in kW?

1. Apparent power: VA = E × I × 1.732 = 480 × 181 × 1.732 = 150,476 VA.
2. True power: W = VA × PF = 150,476 × 0.90 = 135,428 W.
3. Convert to kilowatts: 135,428 ÷ 1,000 = 135.43 kW.

**Answer: 135.43 kW**

## How to solve it on the exam

1. Identify what the question gives: watts (true power), volt-amperes (apparent power) or horsepower (motor output).
2. Convert horsepower to watts at 746 W per hp, and kW or kVA to W or VA by multiplying by 1,000.
3. Write power factor and efficiency as decimals (85% = 0.85).
4. Divide by efficiency to go from output to input; multiply by power factor to go from VA to watts, or divide to go from watts to VA.
5. For three-phase, include 1.732 with the voltage, as in any three-phase formula.

## Common mistakes

- Multiplying by efficiency when finding input power. Input is always larger than output, so divide.
- Leaving out the power factor, which gives kVA instead of kW.
- Using 1,000 W per horsepower instead of 746 W.
- Using a calculated or nameplate current to size motor branch-circuit conductors, when 430.6(A)(1) calls for the table full-load current.
- Putting a percentage (85) into a formula instead of a decimal (0.85).

## Exam tips

- Power factor and efficiency are never more than 1. If your input watts come out smaller than the output, you multiplied instead of divided.
- Write 1 hp = 746 W next to the power wheel on your exam notepad.

## Questions people ask

### How do you convert kVA to kW?

Multiply kVA by the power factor: kW = kVA × PF. A 100 kVA load at a 0.8 power factor uses 80 kW of true power.

### How many watts are in one horsepower?

One horsepower equals 746 watts. That is the motor's output; its electrical input is larger because no motor is 100% efficient.

### How do you calculate motor efficiency?

Divide output watts by input watts and multiply by 100. A 10 hp motor (7,460 W output) that draws 8,500 W has an efficiency of 7,460 ÷ 8,500 × 100 = 87.8%.

### What is the difference between true power and apparent power?

True power, in watts, is the power that does work. Apparent power, in volt-amperes, is voltage times current. With inductive loads such as motors, apparent power is larger, and the ratio between the two is the power factor.

## Sources

- [TDLR: Electrician examination information](https://www.tdlr.texas.gov/electricians/elecexam.htm)
- [PSI: Texas electrician Candidate Information Bulletin](https://test-takers.psiexams.com/api/content/bulletin/1682)

Interactive practice: https://texaselectricianexam.com/calculations/power-factor-efficiency/