21.1 Characteristics of alternating currents
- Syllabus
- 9702–2028–2029
- Topic
- 21.1
- Level
- A2
Learning objectives
21.1.1The terms period, frequency and peak value as applied to an alternating• understand and use the terms period, frequency and peak value as applied to an alternating current or voltage21.1.2Equations of the form x = x0 sin ωt representing a sinusoidally alternating• use equations of the form x = x0 sin ωt representing a sinusoidally alternating current or voltage21.1.3The fact that the mean power in a resistive load is half the maximum power• recall and use the fact that the mean power in a resistive load is half the maximum power for a sinusoidal alternating current21.1.4Root-mean-square (r.m.s.) and peak values and recall and use I r.m.s. = I0 /• distinguish between root-mean-square (r.m.s.) and peak values and recall and use I r.m.s. = I0 / sqrt(2) and Vr.m.s. = V0 / sqrt(2) for a sinusoidal alternating current
An alternating quantity is described by period, frequency, angular frequency and peak value
An alternating voltage or current varies periodically, with period T, frequency f=1/T, angular frequency ω=2πf and peak magnitude I₀ or V₀.
Read peak from the centre line to an extreme, not peak-to-peak, and distinguish the waveform’s cycle time from its angular frequency.
A 50 Hz mains waveform has T=20 ms and ω=100π rad s⁻¹.
The mean of a symmetric AC waveform can be zero while its heating effect is not zero.
Use x=x0 sin(ωt+φ) for sinusoidal alternating current or voltage
x=x0sin(ωt+φ),whereω=2πf=2π/T
| Parameter | Controls |
|---|---|
| x0 | peak magnitude |
| ω | cycle rate / period |
| φ | value and direction at t=0 |
| For φ=0 | t=0 | T/4 | T/2 | 3T/4 | T |
|---|---|---|---|---|---|
| x | 0 rising | +x0 | 0 falling | -x0 | 0 rising |
For v=12 sin(100πt) V, V0=12 V, f=50 Hz and T=0.020 s. At t=5.0 ms=T/4, v=+12 V; at 10 ms, v=0 and falling.
A sinusoid of peak 18 V that starts at +18 V is represented conveniently by v=18 cosωt, or equivalently v=18 sin(ωt+π/2).
Use angular frequency ω—not f—inside the trigonometric phase unless a factor 2π is included. x is instantaneous, x0 is peak, and neither is automatically an rms value.
For a sinusoidal current in a resistor, mean power is half the maximum instantaneous power
With i=I₀sinωt and fixed resistance R, instantaneous power i²R varies from zero to I₀²R; its cycle average is ½I₀²R.
Use average over a complete cycle and distinguish peak power from mean heating power.
If peak current doubles, mean resistive power quadruples because it depends on I₀².
A zero mean current does not imply zero mean power, since power depends on current squared.
RMS values give the DC-equivalent heating effect: I_rms=I₀/√2 and V_rms=V₀/√2
For a sinusoidal AC signal, rms current or voltage produces the same mean power in a resistor as a DC value: I_rms=I₀/√2 and V_rms=V₀/√2.
Use rms values in P=VI or P=I²R for resistive loads, and specify whether a quoted AC value is rms or peak.
A 10 A peak sinusoidal current has I_rms≈7.07 A.
RMS is not the arithmetic average of a symmetric waveform, which is zero.