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> Because a sine wave is symmetrical, its area below the
horizontal axis is the same as its area above the axis;
> Thus, over a full cycle its net area is zero, independent
of frequency and phase angle.
» The area under the half-cycle is:
In Since dex = [-ta cos af = Un
2
lang = et = = 0.318ly
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n
7
(o)ACale. £209, (6)DCalone.¢=0¥ (Supine ac nd
Vog20¥ Vag Voge
> Since the average of a sine wave is zero, the average value of the combined
waveform will be its de component, E.
> However, peak voltages depend on both components as illustrated in (c). Since the average value of AC would directly provide you the answer zero.
Is this means No power is delivered to the load?
> While instantaneous, peak, and average values provide useful information about
a waveform, none of them truly represents the ability of the waveform to do
useful work.
Y Most ac voltmeters display the rms voltage
v The value measured by ac voltmeter for the voltage at the wall outlet in
homes is (120 Vac at North America & Japan , or 220 Vac at Egypt)
> The rms value relates de
toa load.
quantities with respect to the power delivered
A R
No
ince current is constant, power is constan E R
PR. Therefore, P is constant.
— =a PB:
2=P=rR fj (b) de Circa [Forth ac case:
A sketch of p(t) is shown in Figure, obtained by squaring values of current at
various points along the axis, then multiplying by R.
wi)
PO =PR nm
Um Sin @fPR = IPR sinter ” are
ete =
= ner| 5a — cos 200| 4
1k) = BR, Therefore, varies ey
(@) Cait
2
effective values for sinusoidal waveforms depend only on amplitude
It is important to note that these relationships hold only for sinusoidal
waveforms. However, the concept of effective value applies to all waveforms
_ [ares
fen = ==
root - mean - square 2
10s)
T234567
i eycle 4
@ ©)
FIGURE 15-66
Tess asers ae
Solution Square the curve, then apply Equati
yn 15-30. Thus,
9 <3) + GX 2)+ 4x
Ten 8
= fa =2.26A
