Library / Electrical / ELE-38
Electrical
Apparent power in a single-phase circuit
Product of the RMS voltage and current: the total loading the source and cables must carry.
- Ref
- ELE-38
- Rev
- v1
- Field
- Electrical
- Result
- |S| [VA]
The apparent power is |S| = VrmsIrms = √(P² + Q²), in volt-amperes (VA). It is the product the cables, transformers and generators must be rated for, because they must carry the full current whatever the power factor. Real power P is the part that does work and reactive power Q the part that oscillates; the three form a right-angled triangle.
Product of the RMS voltage and current: the total loading the source and cables must carry. It combines the real power P and the reactive power Q at right angles, and it is measured in VA.
Calculator
- Real power, V I cos φ995.9 W
- Reactive power, V I sin φ575 var
- Power factor, cos φ0.866
- Check, √(P² + Q²)1,150 VA
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- A 230 V, 5 A supply is carrying 1,150 VA, of which 995.9 W is real power at a power factor of 0.866.
- A poor power factor does not reduce the current: the same 5 A flows and heats the cable whether or not the power is real, so equipment is rated in VA, not W.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (VA) | Local step | Change in result |
|---|---|---|---|---|---|---|
| Irms RMS current | 0.1 to 100 A | 23 | 23,000 | 22,980 | +10% | +10 % |
| Vrms RMS voltage | 1 to 1,000 V | 5 | 5,000 | 4,995 | +10% | +10 % |
| φ Phase of voltage minus current (positive for an inductive load) | −90 to 90 ° | 1,150 | 1,150 | 0 | +3.6 ° | 0 % |
Sweep of RMS current
| RMS current (A) | Apparent power (VA) |
|---|---|
| 0.1 | 23 |
| 0.2371 | 54.54 |
| 0.5623 | 129.3 |
| 1.334 | 306.7 |
| 3.162 | 727.3 |
| 7.499 | 1,725 |
| 17.78 | 4,090 |
| 42.17 | 9,699 |
| 100 | 23,000 |
Sweep one variable
Variables and how each one contributes
Apparent power
What the equation gives you. Everything else feeds it.
RMS voltage
The RMS voltage, in volts. Raises the apparent power in direct proportion. Raising it by 10 V (to 240 V) moves the result by +4.35 %, from 1,150 to 1,200 VA.
RMS current
The RMS current, in amps. Raises the apparent power in direct proportion. Raising it by 0.5 A (to 5.5 A) moves the result by +10 %, from 1,150 to 1,265 VA.
Phase of voltage minus current (positive for an inductive load)
The phase angle between voltage and current, in degrees. It does not change the apparent power. Has no effect: the apparent power is the product of the two RMS values, whatever the phase. Raising it by 5° (to 35°) moves the result by +0 %, from 1,150 to 1,150 VA.
Worked example with the values above
- Multiply the RMS values: 230 V × 5 A.
- This does not depend on the phase angle (here 30°).
- |S| = 1,150 VA.
Assumptions and limits
- Sinusoidal voltage and current at a single frequency.
- Single-phase circuit; for three-phase multiply by √3 with line values.
Common mistakes
- Quoting a generator or transformer rating in W when it is in VA.
- Adding apparent powers of different loads arithmetically: add the real and reactive parts separately.
Statistics and uncertainty
The RMS voltage and current are each measured to 1 % or better, so the apparent power is known to about 1.5 %, much better than the real power, which also needs the phase angle.
Where it is used
- Rating transformers, generators, cables and UPS systems.
- Checking the current drawn by a load at a known power.
Maths you need
- RMS values: effective values of an AC quantity
- Power triangle: real, reactive and apparent power as a right triangle