Average AC Power
Strings (SiPjAjk) = S7P3A32 Base Sequence = 12735 String Sequence = 12735 - 3 - 32
Figure 8.7 shows a simple AC circuit. Figure 8.7(a) is the time domain circuit while figure 8.7(b) is its phasor form.
Given that the sinusoidal voltage and current of the circuit are as follows:
v(t) = Vcos(ωt); i(t) = Icos(ωt - θ); Determine:
(a) The average power of the circuit in the time domain
(b) The average power of the circuit in the frequency domain.
S7P3A32 (Force - Push).
Pj Problem of interest is of type force. Power and energy problems are force problems. Electric power is a force-push.
(a) General expression for electric power is:
p(t) = v(t)i(t)
So, p(t) = VIcos(ωt)cos(ωt - θ)----(1)
Using the following identities of trigonometry:
2cos2ωt - 1 = cos2(ωt) and cosωtsinωt = (sin2ωt)/2 .
Equation (1) reduces to:
p(t) = (VI/2)cos(θ) + (VI/2)cos(2ωt - θ) ----(2)
(a) The average power, Pav is obtained by integrating p(t) over one cycle of the sinusoidal signal and dividing by the period T, of the signal.
So, after substituting the expression for p(t) in equation (2) and integrating, we have:
Pav = (VI/2)cosθ
(b) In the frequency domain:
V(jω) = Vej0 and I(jω) = Ie-jθ
So, impedance, Z = (V/I)ejθ
So, I = (V/|Z|)ejθ
Hence, Pav = (V2/2)(1/|Z|)cosθ = (I2/2)(|Z|)cosθ
rms values are usually used for the voltage and current amplitudes.
Where, Vrms = V/√2 and Irms = I/√2.
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