An alternating current reverses direction periodically. The most common waveform is the sine wave, described by its frequency, amplitude and phase. The frequency is the number of cycles per second in hertz; the period is the time for one complete cycle.
Period and frequency
Period of a 50 Hz supply
Problem. What is the period of a 50 Hz sine wave?
Solution. T = 1 ÷ f = 1 ÷ 50 = 0.020 s = 20 ms.
Peak, peak-to-peak and RMS
The peak value is the maximum instantaneous magnitude. The peak-to-peak value is twice the peak for a symmetrical waveform. The RMS value is the equivalent steady value that would produce the same heating in a resistor; for a pure sine wave it is the peak divided by √2.
RMS from peak
Problem. A sine wave has a peak voltage of 325 V. What is its RMS voltage?
Solution. V_rms = 325 ÷ √2 = 325 ÷ 1.4142 = 229.8 V, approximately 230 V.
- Frequency is cycles per second; period is seconds per cycle.
- V_peak is the maximum instantaneous value.
- V_peak-to-peak is the full swing between the two extremes.
- V_rms is the heating-equivalent value; for a sine it is V_peak/√2.
- f = 1/T.
- V_rms = V_peak/√2 for a sine wave.
- RMS represents heating equivalence.
- The sine shortcut does not apply to distorted waveforms.
Worked example
A sine wave with 20 ms period has f = 1 ÷ 0.020 = 50 Hz. With a peak of 170 V, its RMS value is 170 ÷ 1.4142 = 120.2 V.