Alternating currents
In AC, Ohm’s law still holds, but it has to be extended: inductors and capacitors oppose the flow of current without dissipating energy and shift the current in phase relative to the voltage. From this come impedance, resonance and the power factor.
01Where the sine wave comes from
A loop rotating at constant speed inside a uniform magnetic field generates, by Faraday’s law, an EMF that follows exactly the sine function. It is not a design choice: it is a geometric consequence of the rotation.
The characteristic values (peak, peak-to-peak, average and RMS) are covered in measurements in alternating current. Here they are taken as known, and any voltage or current written without qualification is RMS.
Because of the transformer: it only works with a changing flux. Being able to step the voltage up for transmission (less current, lower I²R losses) and step it down for consumption is what made large-scale distribution viable. That was the “war of the currents” between Tesla-Westinghouse (AC) and Edison (DC) in the late 19th century, and AC won it for this one reason.
02Resistor, inductor and capacitor in AC
All three oppose the flow of current, but in very different ways.
Both are measured in ohms and both limit the current, but a resistance converts energy into heat while a reactance stores it and returns it to the source in the next quarter cycle. An ideal inductor and an ideal capacitor consume no average power, even though current flows through them. That distinction is what gives rise to reactive power and to the power factor.
03Phasors and impedance
Adding out-of-phase sine waves point by point is impractical. The solution is to represent each quantity as a rotating vector (phasor): its length is the RMS value and its angle is the phase. Since they all rotate at the same speed, they are drawn frozen and added as vectors.
R = 100 Ω, L = 0.5 H, C = 20 µF, supplied with 220 V at 50 Hz.
- XL = 2π × 50 × 0.5 = 157 Ω
- XC = 1/(2π × 50 × 20×10−6) = 159 Ω
- X = XL − XC = −2 Ω (slightly capacitive)
- Z = √(100² + 2²) = 100.02 Ω
- I = 220 / 100.02 = 2.20 A
- φ = arctan(−2/100) = −1.1° — practically in phase
The circuit is almost at resonance: the two reactances cancel and the impedance reduces to the resistance. With the inductor alone, the current would be 220/157 = 1.4 A; with both, it rises to 2.2 A. Adding a component in series increased the current, which is impossible in DC and perfectly normal in AC.
In a series RLC circuit it is perfectly possible for the voltage measured across the inductor to be 350 V and that across the capacitor also 350 V, with only 220 V of supply. There is no contradiction: they are in antiphase and cancel each other. What adds vectorially is VR with (VL − VC), and that does give 220 V.
04Resonance
Since XL increases with frequency and XC decreases, there is a frequency at which they are equal and cancel. This is the resonant frequency.
- The impedance is minimum and equal to R.
- The current is maximum.
- Voltage and current in phase (φ = 0).
- Across L and across C, voltages Q times larger than the supply voltage appear.
Used to pass a frequency: the input of a radio receiver.
- The impedance is maximum.
- The line current is minimum.
- A large current circulates inside the L-C loop.
Used to reject a frequency or as a selective load of an oscillator or of a radio-frequency amplifier.
A tank circuit with L = 250 µH and a variable capacitor. To tune to 1000 kHz:
With R = 10 Ω of winding resistance: XL = 1571 Ω, Q = 157, and the bandwidth comes out to 1000 kHz / 157 = 6.4 kHz. Enough to separate one AM station from the next, which is 10 kHz away. That is exactly the function of the input circuit of a radio, and it is taken up again in Telecommunications, in Year 6 — where the Morse code and the phonetic alphabet and amplitude modulation also come in.
In series resonance with a high Q, voltages Q times the supply voltage appear across the inductor and the capacitor. With 220 V and Q = 20, that is 4400 V across components rated for 220. It is a real cause of destruction of power factor correction capacitors when the installation accidentally goes into resonance with the grid inductance.
05Power in AC
Since the current can be out of phase with the voltage, the product V × I is no longer the power actually consumed. Three kinds of power must be distinguished.
| Power | Symbol | Unit | What it is |
|---|---|---|---|
| Active | P | W | The power that is actually transformed into work or heat. It is the one the utility charges for in a home. |
| Reactive | Q | VAr | The power that goes back and forth between the source and the inductors or capacitors. It is not consumed, but it flows through the wires. |
| Apparent | S | VA | The product V × I. It determines the cross-section of the conductors and the size of the transformer. |
A transformer or a generator set heats up because of the current, not because of the active power. If the load has cos φ = 0.5, to deliver 5 kW you need a 10 kVA unit. That is why UPSs, generators and transformers are sold by their apparent power: it is the only one that says how much current they will have to handle.
A single-phase 220 V motor drawing 8 A with cos φ = 0.72.
- S = 220 × 8 = 1760 VA
- P = 1760 × 0.72 = 1267 W (the useful part)
- φ = arccos(0.72) = 43.9°; Q = 1760 × sin 43.9° = 1221 VAr
That is: the wires carry 8 A, but only 72% of that current does work. The rest goes back and forth heating the conductors without producing anything useful.
06Power factor correction
Almost all industrial loads are inductive (motors, transformers, ballasts), so the current lags and cos φ is low. Since a capacitor produces the opposite phase shift, connecting capacitors in parallel with the load compensates the reactive power.
P = 1267 W, V = 220 V, f = 50 Hz.
- φ₁ = 43.9° → tan φ₁ = 0.963
- φ₂ = arccos(0.95) = 18.2° → tan φ₂ = 0.329
- C = 1267 × (0.963 − 0.329) / (2π × 50 × 220²) = 52.8 µF
Result: with the capacitor connected, the new line current is I = P/(V·cos φ) = 1267/(220 × 0.95) = 6.06 A, compared with the original 8 A. The same useful power with 24% less current: thinner wires, lower losses and less penalty on the bill.
The capacitor must be of the type made specifically for correction (self-healing) and rated for at least 400 V working voltage, never an electrolytic.
Overcompensating (going past cos φ = 1) is as bad as not compensating: the installation becomes capacitive, the current rises again and there is a risk of resonance with the grid inductance. That is why it is corrected to 0.95 to 0.98 and not to 1, and large installations use capacitor banks with automatic step switching. The topic is taken up again in Electrical Technology II and in Industrial Electronics I.
07In the lab
With the function generator at different frequencies, measure the current flowing through an inductor (from the drop across a small series resistor) and calculate XL = V/I. Plot XL as a function of f: you should get a straight line through the origin. Repeat with a capacitor: XC as a function of f gives a hyperbola, and XC as a function of 1/f, a straight line.
Series RL circuit driven by the generator. CH1 on the input, CH2 across the resistor (which shows the current, because in R they are in phase). Measure Δt between the zero crossings and calculate φ. Compare with arctan(XL/R). Repeat with RC and verify that the sign reverses.
Series RLC with R = 100 Ω, L = 10 mH and C = 100 nF (theoretical f₀ = 5033 Hz). With constant input amplitude, sweep the frequency from 1 to 20 kHz, noting the voltage across R (proportional to the current). Plot it and find the peak. Measure the bandwidth between the points at 0.707 of the maximum and calculate Q = f₀/BW. Compare with Q = XL/R.
Also measure the voltage across L and across C at resonance: they will be several times larger than the input. This is the check of the resonance overvoltage.
On an inductive load (the primary of a small transformer at no load, or a ballast), measure V, I and P (with a wattmeter) and calculate cos φ. Connect a 400 V polyester capacitor in parallel and measure the current again: it should drop without the active power changing. Try two or three capacitors and find the one that reduces the current the most.
Work at 220 V: only with the teacher, with all wiring insulated and unplugging for every change.
08Common mistakes
| Symptom or confusion | Clarification |
|---|---|
| Z was calculated as R + XL − XC | They are 90° apart: they add vectorially, with Pythagoras. Adding them arithmetically gives a value that is far too large and wrong. |
| The voltages measured in a series RLC add up to more than the input | That is correct: VL and VC are in antiphase and partly cancel. |
| Power is calculated as V × I | That is the apparent power in VA. The active power is V·I·cos φ. |
| The correction capacitor exploded | An electrolytic (which is a DC type) or one with insufficient voltage rating was used. It must be made specifically for AC and rated 400 V or more. |
| After correcting, the current went up instead of down | Overcompensation: cos φ = 1 was exceeded and the installation became capacitive. |
| The measured resonant frequency does not match | The real inductor has parasitic capacitance and the capacitor has lead inductance. Also, the typical tolerance of an inductor is 10 to 20%. |
| A motor-start capacitor was connected to DC | In DC XC = ∞: nothing flows. And if it is a bipolar electrolytic, it degrades. |
09Self-assessment
What is XL of a 100 mH inductor at 50 Hz? And at 5 kHz?
At 50 Hz: XL = 2π × 50 × 0.1 = 31.4 Ω.
At 5 kHz: 2π × 5000 × 0.1 = 3140 Ω. One hundred times more, because the frequency is one hundred times
higher.
Why does a reactance not dissipate power if current flows through it?
Because during a quarter cycle it stores energy (in the magnetic or electric field) and during the next it returns it to the source. The average of the instantaneous power over a complete cycle is zero. A resistance, by contrast, converts it irreversibly into heat.
A series circuit with R = 60 Ω, XL = 100 Ω and XC = 20 Ω at 220 V. What are Z, I and φ?
X = 100 − 20 = 80 Ω. Z = √(60² + 80²) = 100 Ω. I = 220/100 = 2.2 A. φ = arctan(80/60) = 53.1° (inductive: the current lags).
What happens to the impedance of a series RLC circuit at resonance?
The reactances cancel and Z = R, which is the minimum possible value. The current is maximum and in phase with the voltage.
L = 40 mH and C = 250 nF. What is the resonant frequency?
f₀ = 1/(2π√(0.04 × 250×10−9)) = 1/(2π√(10−8)) = 1/(2π × 10−4) = 1592 Hz.
A piece of equipment draws 10 A at 220 V with cos φ = 0.6. What are the active, reactive and apparent powers?
S = 220 × 10 = 2200 VA. P = 2200 × 0.6 = 1320 W. φ = 53.1°, Q = 2200 × 0.8 = 1760 VAr. Check: √(1320² + 1760²) = 2200 ✓.
Why are industries charged for reactive energy?
Because even though it does no work, it flows through the whole grid: it forces conductors, transformers and generators to be sized larger, and it produces real I²R losses in the line. The penalty is the economic incentive for each user to compensate their own reactive power.
Can the voltage across a capacitor be greater than the supply voltage?
Yes, in a circuit at series resonance: across L and across C, voltages Q times larger than the input appear. They are in antiphase, so they cancel each other and do not contradict Kirchhoff’s law. It is a real and potentially destructive phenomenon.
Why is a transformer rated in kVA and not in kW?
Because its limit is the current it can handle without overheating, and that depends on the apparent power S = V·I. How much of that power is active depends on the cos φ of the load, which the transformer manufacturer does not know.