Power in alternating current
In DC, power is the product of voltage and current, and that is the end of the story. In AC there are three powers, and only one of them does work. The other two explain why a workshop pays more than it consumes and why cables are sized thicker than would seem necessary.
01Instantaneous power
The power at each instant is always p = v · i. What changes in AC is that v and i may be out of phase: there are moments when one is positive and the other negative, and then the instantaneous power is negative. That means the load is returning energy to the grid.
- Positive area: energy that flows from the grid to the load and is converted into heat, light or motion. It does not come back.
- Negative area: energy that the load had stored in a magnetic or electric field and returns to the grid. It goes back and forth fifty times per second.
- The average of p over one cycle is the active power: the only thing that is really consumed, and the only thing the energy meter measures.
02The three powers
The one that produces work: heat, light, motion. It is the one the utility bills for and the one that defines the size of the motor.
The one that goes back and forth between the grid and the magnetic fields of motors and transformers. It does no work, but it takes up room in the cables.
The product of the RMS values, ignoring phase. It is the one that sizes everything: conductors, transformers, breakers, the generator set.
Because what heats it and what saturates its core is the current, regardless of what phase it comes in. A 10 kVA transformer delivers 10 kW to a resistive load, but only 8 kW to a load with cos φ = 0.8. The manufacturer cannot know what will be connected to it: that is why it specifies the apparent power. The same goes for a generator set and for a UPS.
03Power factor: what makes it worse and what it costs
- Induction motors, especially running unloaded or lightly loaded: a motor at 25% load can have cos φ = 0.5.
- Transformers at no load.
- Welding equipment and induction furnaces.
- Magnetic ballasts of fluorescent tubes.
- More current for the same useful power.
- More losses in the cables: they grow with I², so they rise faster than the current.
- More voltage drop along the line, and motors that have trouble starting at the end of the installation.
- Less available capacity in the transformer and in the switchboard.
- Surcharge on the bill: in Argentina, below 0.85 the utilities apply a penalty.
A workshop consumes P = 20 kW on a 380 V three-phase line.
| cos φ = 0.75 | cos φ = 0.95 | |
|---|---|---|
| Apparent power S = P/cos φ | 26.67 kVA | 21.05 kVA |
| Line current I = S/(√3·V) | 40.5 A | 32.0 A |
| Cable losses (∝ I²) | 100% | 62% |
| Reactive power Q | 17.64 kvar | 6.57 kvar |
The same machine doing the same work, with 8.5 A less in each conductor and 38% lower losses in the wiring. That is the economic argument for compensation, even before talking about the fine.
04Compensation with capacitors
The idea is simple: if the inductive load asks for reactive power, we supply it from the same place instead of bringing it from the power plant. A capacitor delivers exactly the reactive power that an inductor consumes, because its current leads instead of lagging. Placed in parallel, they cancel.
Bring the 20 kW of the previous example from cos φ = 0.75 to 0.95:
- φ₁ = arccos 0.75 = 41.41° → tan φ₁ = 0.8819
- φ₂ = arccos 0.95 = 18.19° → tan φ₂ = 0.3287
- QC = 20 kW × (0.8819 − 0.3287) = 20 × 0.5532 = 11.06 kvar
- A commercial bank of 12.5 kvar is installed, which is the next step up.
If, instead, compensation is done single-phase, the required capacitance comes from QC = V²·2πf·C. For 1 kvar at 220 V: C = 1000 / (314.16 × 220²) = 65.8 µF. These are special capacitors for continuous AC duty, not just any motor-start capacitor.
- Individual: one capacitor per motor, connected together with it. It is the most effective because it relieves the whole installation, but it is the most expensive option.
- By groups: at the sub-panel that feeds several motors.
- Automatic central: a bank at the main switchboard with a controller that connects steps according to demand. It is the usual solution in industry.
Over-compensating leaves the installation capacitive, and that is just as bad as the original problem: the current rises again, the voltage tends to go up and there is a risk of resonance with the inductance of the grid, which can amplify harmonics. That is why banks are sized up to 0.95 and not up to 1, and why automatic ones connect steps one at a time instead of putting everything in at once.
If the capacitor stays connected to the terminals of a motor that is disconnected from the grid, the motor keeps spinning by inertia and the capacitor excites it: it behaves as a generator and overvoltages appear at its terminals. That is why, in individual compensation, the capacitor is always sized below the no-load current of the motor, and in installations with variable-speed drives you never compensate at the output of the drive.
05Measuring power properly
Multiplying the voltmeter reading by the ammeter reading gives the apparent power, not the active power. To measure P you need an instrument that takes phase into account.
| Instrument | What it measures | Detail |
|---|---|---|
| Electrodynamic wattmeter | P directly | Two coils: one for voltage and one for current. The torque is proportional to the instantaneous product, so the pointer shows the average: the active power (see analog measuring instruments). |
| Power factor meter | cos φ | Gives the power factor directly. Many switchboards have one permanently installed next to the ammeter. |
| Clamp meter with power function | P, S, Q and cos φ | It needs both things at once: clamping around the conductor and taking voltage with probes. Ordinary clamp meters measure current only. |
| Power analyzer | Everything, including harmonics | Logs for days and shows the consumption profile. It is the tool used to decide on a capacitor bank. |
| Energy meter | kWh and, in industry, kvarh | The utility’s meter. If it measures reactive energy, the penalty comes from there. |
In a three-wire three-phase system —with or without an accessible neutral, balanced or not— two wattmeters are enough to measure the total power: P = W₁ + W₂. It is a classic result (Blondel’s theorem) and is used all the time in practice. It also gives extra information: if the two readings are equal, cos φ = 1; if one reads zero, cos φ = 0.5; and if one reads negative, the power factor is less than 0.5. It is developed in three-phase circuits.
A switching power supply, a variable-speed drive or a cheap LED lamp does not draw sinusoidal current: it draws spikes. There the power factor has two components: the displacement one (the good old cos φ) and the distortion one, which comes from the harmonics. A capacitor corrects the first and does nothing about the second; worse still, it may resonate with the grid and amplify the harmonics.
That is what harmonic filters and power supplies with active power factor correction (PFC) are for, which is mandatory today in equipment above a certain power rating. An ordinary multimeter is not suitable for measuring these currents either: you need True RMS (AC measurements).
06In the lab
With a small single-phase motor, measure V, I and P with a wattmeter. Calculate S = V·I, cos φ = P/S and Q = √(S² − P²). Repeat unloaded and under load (braking it gently): you will see that the cos φ is very poor unloaded and improves as it is loaded. This is the central observation of the topic.
On the same motor, connect in parallel a continuous-duty capacitor of 4 to 8 µF. Note the line current before and after: it drops, while the active power measured with the wattmeter does not change. Try capacitors of different values and plot the current-versus-capacitance curve: it has a minimum, and past that point the current rises again. That minimum is the optimum compensation, and its existence is the practical demonstration of overcompensation.
With an isolation transformer and a small series shunt resistor, view the voltage and current of an inductive load simultaneously. Measure Δt between the zero crossings and calculate φ = 360°·Δt/T. Compare the cos φ obtained this way with the one given by the power factor meter. In X-Y mode, the resulting ellipse gives the same figure by another route.
A compensation capacitor retains dangerous charge after being disconnected. Continuous-duty ones have internal discharge resistors, but they take a minute or more. Before touching the terminals: switch off, wait and verify with the instrument that they are discharged.
07Common mistakes
| Symptom | Usual cause |
|---|---|
| V × I does not match what the wattmeter reads | That is correct: V×I is the apparent power. The difference is the power factor. |
| It was compensated and the current went up | Overcompensation: the installation became capacitive. Steps have to be removed. |
| The capacitor bank heats up or the protection trips | Harmonics from the installation resonating with the bank. A bank with detuning reactors is needed. |
| The plant’s cos φ is good by day and poor at night | Fixed compensation with a variable load: in the low-load hours there is excess capacitive reactive power. It is solved with an automatic bank. |
| A small motor has a terrible cos φ | It is working far below its rated power. Before compensating, check whether the motor is oversized. |
| It was compensated and the bill did not go down | The load is nonlinear: the problem is distortion, not displacement. Capacitors do not correct harmonics. |
| The current measurement differs depending on the instrument | Non-sinusoidal current measured with an average-responding, RMS-calibrated instrument. True RMS is needed. |
| When the motor is disconnected, its capacitor is left at high voltage | Self-excitation. The capacitor must be below the no-load current of the motor and have a discharge resistor. |
08Self-assessment
What does it mean for the instantaneous power to be negative during part of the cycle?
That at that instant the load returns to the grid the energy it had stored in its magnetic or electric field. That energy goes back and forth without producing work: it is the reactive power.
A motor draws 12 A at 220 V with cos φ = 0.8. What are S, P and Q?
S = 220 × 12 = 2640 VA. P = 2640 × 0.8 = 2112 W. Q = 2640 × sin(36.87°) = 2640 × 0.6 = 1584 var.
Why is a transformer rated in kVA?
Because what limits it is the current (heating of the winding) and the voltage (saturation of the core), regardless of the phase of the load. The active power it delivers will depend on the cos φ of whatever is connected to it, which the manufacturer cannot know.
With cos φ = 0.5, how much current flows compared with what would be needed at cos φ = 1?
Double, because I = P/(V·cos φ). And since losses go with I², the cable dissipates four times as much.
A 30 kW load with cos φ = 0.7 is to be brought to 0.92. How much reactive power has to be supplied?
tan(arccos 0.7) = 1.0202; tan(arccos 0.92) = 0.4260. QC = 30 × (1.0202 − 0.4260) = 30 × 0.5942 = 17.8 kvar.
Why is compensation not carried all the way to cos φ = 1?
Because the load varies and a fixed bank would leave the installation capacitive in the hours of low demand, with overvoltages and a risk of resonance with the harmonics of the grid. The target is 0.95, which also already avoids the penalty.
Does a capacitor correct the power factor of a switching power supply?
No. There the problem is the distortion factor: the current is not sinusoidal but peaky, and that is not corrected with capacitors. Harmonic filters or active correction (PFC) in the equipment itself are needed.
In the two-wattmeter method, one reads a negative value. What can be deduced?
That the power factor is less than 0.5. The total power is still the algebraic sum: P = W₁ + W₂, subtracting the negative reading.
Why does an unloaded motor have a poor power factor?
Because the magnetizing current —which is reactive— is practically the same with or without load, while the active component falls along with the work delivered. With little active and the same reactive, the cos φ plummets.
You want to supply 5 kvar at 380 V with a single-phase capacitor. What capacitance is needed?
C = Q/(2πf·V²) = 5000/(314.16 × 144,400) = 5000/45,365,000 = 1.102×10−4 F ≈ 110 µF.