The three powers
S = U · I
P = U · I · cos φ
Q = U · I · sin φ
P = U · I · cos φ
Q = U · I · sin φ
| Quantity | Symbol | Unit |
|---|---|---|
| Apparent power | S | VA |
| Active power | P | W |
| Reactive power | Q | var |
Elec Class
learn electricity hands-on
Lesson · chapter 14 of 15
In AC, the power actually used is the active power P = U × I × cos φ, in watts. The power factor cos φ, between 0 and 1, tells you what share of the product U × I is really turned into heat, light or motion.
The result is worked out before your eyes, line by line, just like on the board.
1I'm looking for
2I enter the known values
Formula used P = U · I · cos φ
3The result
Active power P920 W
At 230 V, an appliance drawing 5 A with a cos φ of {cos φ} uses 920 W of active power.
The definition, the key points to remember and everyday examples.
In AC we distinguish the apparent power S = U·I (VA), the active power P = U·I·cos φ (W, the only one producing work or heat) and the reactive power Q = U·I·sin φ (var, exchanged with coils and capacitors). They form a triangle: S² = P² + Q².
The power factor cos φ = P/S measures the efficiency of the installation: a low cos φ requires more current for the same useful power, hence more line losses. It is corrected by adding capacitors in parallel.
In a glass of soda, there's the drink (active power P, the part that quenches your thirst) and the foam (reactive power Q, which takes up room without giving you anything). The whole glass is the apparent power S: that's what the grid has to fill. cos φ is the share of drink in the glass. Where it falls short: the foam doesn't simply add to the drink, P and Q combine as a triangle (S² = P² + Q²).
| Quantity | Symbol | Unit |
|---|---|---|
| Apparent power | S | VA |
| Active power | P | W |
| Reactive power | Q | var |
| Quantity | Symbol | Unit |
|---|---|---|
| Power factor | cos φ | - |
| Quantity | Symbol | Unit |
|---|---|---|
| Capacitance to add | C | F |
| Phase shifts before / after | φ₁, φ₂ | rad |
Four steps to solve a problem, then an example solved line by line.
S = U × I with the RMS values, in VA: it's the longest side of the triangle.
sin φ = √(1 − cos² φ): cos φ = 0.8 gives sin φ = 0.6.
P = S × cos φ in W, Q = S × sin φ in var, then check that P² + Q² = S².
Calculate tan φ before and after, then C = P × (tan φ₁ − tan φ₂) / (U² × ω); the result comes out in farads, often a few tens of µF.
A single-phase motor draws 5 A at 230 V with cos φ = 0.8. Find S, P and Q. What capacitance should be placed in parallel to raise the power factor to 0.95 (50 Hz)?
S = U · I = 230 × 5 = 1 150 VA; P = S · cos φ = 1 150 × 0.8 = 920 W; Q = S · sin φ = 1 150 × 0.6 = 690 var (check: 920² + 690² = 1 150²).
tan φ₁ = 0.75 (cos φ = 0.8); tan φ₂ = 0.329 (cos φ = 0.95).
C = P · (tan φ₁ − tan φ₂) / (U² · ω) = 920 × 0.421 / (230² × 314) = 387 / 16.6 × 10⁶ = 23 µF.
Active power P (W) is turned into heat, light or work; reactive power Q (var) goes back and forth between the grid and the coils or capacitors; apparent power S (VA), equal to U × I, is what the cables and the transformer have to handle.
Divide the active power by the apparent power: cos φ = P / (U × I). A 1,500 W appliance drawing 8 A at 230 V has cos φ = 1,500 / 1,840 ≈ 0.82.
I = P / (U × cos φ), with P the electrical power drawn. A motor drawing 1,840 W at 230 V with cos φ = 0.8 takes 1,840 / (230 × 0.8) = 10 A. Careful: the rating plate gives the useful mechanical power; divide it by the efficiency first.
So as not to confuse it with active power: var stands for "volt-ampere reactive". It really is volts multiplied by amps, like the VA of apparent power, but this power does no work.
1 for a resistor (current and voltage in phase), 0 for an ideal coil or capacitor (90° phase shift). A real appliance is somewhere in between, for example 0.8 for a motor at full load.
No: for a household, only active energy, in kWh, is billed. cos φ matters for businesses, whose contract may charge for reactive energy above a threshold.
For the same useful power P, a low cos φ requires a larger current (I = P / (U·cos φ)), hence line losses and oversized cables the utility pays for. Beyond a threshold it bills the reactive power.
The active energy, in kWh: the integral of P over time. A reactive energy meter (kvarh) is only installed at large consumers.
From the simplest to the most complete, each problem comes with a step-by-step solution.
In the lab, measure U, I and P with the wattmeter on an iron-core coil, calculate cos φ = P / (U × I), then add a capacitor in parallel: the line current drops while P stays the same.
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