Sinusoidal quantity
u(t) = Umax · sin(ω · t + φ)
| Quantity | Symbol | Unit |
|---|---|---|
| Peak value | Umax | V |
| Angular frequency | ω | rad/s |
| Initial phase | φ | rad |
Elec Class
learn electricity hands-on
Lesson · chapter 12 of 15
Alternating current changes direction every half-period: the mains voltage follows a sine wave that rises to 325 V, falls to −325 V and starts again 50 times a second. Its RMS value, Ueff = Umax / √2 = 230 V, is the one that heats as much as a 230 V DC voltage.
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 Umax = Ueff · √2
3The result
Peak value Umax325 V
A sinusoidal voltage of 230 V RMS rises to 325 V at each peak.
The definition, the key points to remember and everyday examples.
A sinusoidal quantity is written u(t) = Umax·sin(ωt + φ). Its period T (s) and frequency f = 1/T (Hz) relate to the angular frequency ω = 2πf (rad/s). The European mains is 50 Hz (T = 20 ms).
The RMS value Urms = Umax/√2 is the one producing the same heating in DC: AC voltmeters read it (230 V), while the oscilloscope shows Umax (325 V). Two sine waves of the same frequency are phase-shifted by φ; they are drawn as phasors.
Think of a saw being pushed and pulled: the blade goes back and forth without moving forward, and yet it cuts. In AC, electrons do the same: they oscillate on the spot, by much less than a millimeter, 50 times a second, and this back-and-forth is enough to make the lamp light up. Where it falls short: a hand pushes in jerks, whereas the voltage slows down and speeds up smoothly, following a sine wave.
| Quantity | Symbol | Unit |
|---|---|---|
| Peak value | Umax | V |
| Angular frequency | ω | rad/s |
| Initial phase | φ | rad |
| Quantity | Symbol | Unit |
|---|---|---|
| Frequency | f | Hz |
| Period | T | s |
| Quantity | Symbol | Unit |
|---|---|---|
| RMS value | Ueff | V |
| Peak value | Umax | V |
Four steps to solve a problem, then an example solved line by line.
RMS or peak value? Frequency, period or angular frequency? A plain "230 V" is an RMS value.
Umax = Ueff × √2, T = 1 / f, ω = 2π × f: write the relationship before substituting, with T in seconds.
u(t) = Umax · sin(ω·t + φ): it's the peak value, not the RMS value, that goes in front of the sine.
Calculator in radians for sin(ω·t), then check: an instantaneous value never exceeds Umax.
The mains supplies 230 V rms at 50 Hz. What are the peak voltage, the period and the angular frequency? Write u(t) and compute its value at t = 5 ms.
Umax = Urms · √2 = 230 × 1.414 = 325 V.
T = 1 / f = 1 / 50 = 20 ms; ω = 2π · f = 314 rad/s.
u(t) = 325 · sin(314 · t). At t = 5 ms (a quarter period): 314 × 0.005 = 1.57 rad = π/2, so u = 325 × 1 = 325 V, the peak.
Direct current always flows in the same direction (battery, USB port); alternating current changes direction every half-period (wall socket). The grid uses AC because transformers, which only work with AC, make it possible to transmit energy at very high voltage with few losses.
For a sine wave, divide the peak value by √2: Ueff = Umax / √2. A 17 V peak voltage gives 17 / 1.414 = 12 V RMS. For a square or triangle wave, the ratio is different.
f = 1 / T, with T in seconds. A period of 20 ms gives f = 1 / 0.02 = 50 Hz; a period of 2.5 ms gives 1 / 0.0025 = 400 Hz.
It's the rotational speed, in radians per second, of the phasor that represents the sine wave: ω = 2π × f. At 50 Hz, ω = 2π × 50 ≈ 314 rad/s. It appears in u(t) and in the reactances of coils and capacitors.
The two sine waves are offset by a quarter of a period: when one passes through its maximum, the other passes through zero. At 50 Hz, 90° corresponds to 5 ms. It's the offset between current and voltage in an ideal coil or capacitor.
About 650 V, from +325 V to −325 V. Oscilloscopes often display it as Vpp; don't confuse it with Umax (325 V) or with Ueff (230 V).
230 V is the rms value: a 230 V direct voltage would heat a resistor exactly as much as this sine wave. It is the only useful value for power calculations, and the one voltmeters display.
That the voltage completes 50 full cycles per second: one period lasts 20 ms, the voltage crosses zero 100 times a second. In North America the mains is 60 Hz.
From the simplest to the most complete, each problem comes with a step-by-step solution.
With a low-frequency signal generator and an oscilloscope, display a 6 V peak sine wave at 50 Hz, measure Umax and T on the screen, then compare with the reading of a multimeter set to AC (about 4.2 V).
Using Elec Class in class