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b If C1 = C2 = C, RL = RS = 600 , and 1/ LC = R/L = 1/RC = 2,000 , plot |Vo (j )/VS (j )| in dB versus frequency (logarithmic scale) in the range 100 Hz f 10,000 Hz
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640 The T lter of the circuit of Figure P640 is a
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low-pass lter that may be used to pass signals to the woofer portion of a speaker a Find the frequency response Vo (j )/VS (j ) b If L1 = L2 = L, RS = RL = 600 , and 1/ LC = R/L = 1/RC = 2,000 , plot |Vo (j )/VS (j )| in dB versus frequency (logarithmic scale) in the range 100 Hz f 10,000 Hz
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Figure P639
AC Power
he aim of this chapter is to introduce the student to simple AC power calculations, and to the generation and distribution of electric power The chapter builds on the material developed in 4 namely, phasors and complex impedance and paves the way for the material on electric machines in s 16, 17, and 18 The chapter starts with the de nition of AC average and complex power and illustrates the computation of the power absorbed by a complex load; special attention is paid to the calculation of the power factor, and to power factor correction The next subject is a brief discussion of ideal transformers and of maximum power transfer This is followed by an introduction to three-phase power The chapter ends with a discussion of electric power generation and distribution Upon completing this chapter, you should have mastered the following basic concepts:
Calculation of real and reactive power for a complex load Operation of ideal transformers Impedance matching and maximum power transfer Basic notions of residential circuit wiring, including grounding and safety Con guration of electric power distribution networks
7
AC Power
POWER IN AC CIRCUITS
The objective of this section is to introduce the notion of AC power As already mentioned in 4, 50- or 60-Hz AC power constitutes the most common form of electric power; in this section, the phasor notation developed in 4 will be employed to analyze the power absorbed by both resistive and complex loads
Instantaneous and Average Power From 4, you already know that when a linear electric circuit is excited by a sinusoidal source, all voltages and currents in the circuit are also sinusoids of the same frequency as that of the excitation source Figure 71 depicts the general form of a linear AC circuit The most general expressions for the voltage and current delivered to an arbitrary load are as follows: v(t) = V cos( t V ) i(t) = I cos( t I ) (71)
i(t)
v(t) + ~
AC circuit v(t) = V cos( t uV) i(t) = I cos( t uI) I = Ie ju V = Ve juV
+ ~
Z= e j(u)
where V and I are the peak amplitudes of the sinusoidal voltage and current, respectively, and V and I are their phase angles Two such waveforms are plotted in Figure 72, with unit amplitude and with phase angles V = /6 and I = /3 From here on, let us assume that the reference phase angle of the voltage source, V , is zero, and let I =
Voltage waveforms for unity amplitude, zero deg voltage phase angle and 60 deg current phase angle Voltage Current
1 AC circuit in phasor form 08 06
Figure 71 Circuit for illustration of AC power
Volts, amps
04 02 0 02 04 06 08 1 0 001 002 003 004 005 006 007 008 009 Time (s) 01
Figure 72 Current and voltage waveforms for illustration of AC power
Since the instantaneous power dissipated by a circuit element is given by the product of the instantaneous voltage and current, it is possible to obtain a general expression for the power dissipated by an AC circuit element: p(t) = v(t)i(t) = V I cos( t) cos( t ) (72)
Part I
Circuits
Equation 72 can be further simpli ed with the aid of trigonometric identities to yield VI VI (73) cos( ) + cos(2 t ) 2 2 where is the difference in phase between voltage and current Equation 73 illustrates how the instantaneous power dissipated by an AC circuit element is equal to the sum of an average component, 1 V I cos( ), plus a sinusoidal component, 2 1 V I cos(2 t ), oscillating at a frequency double that of the original source 2 frequency The instantaneous and average power are plotted in Figure 73 for the signals of Figure 72 The average power corresponding to the voltage and current signals of equation 71 can be obtained by integrating the instantaneous power over one cycle of the sinusoidal signal Let T = 2 / represent one cycle of the sinusoidal signals Then the average power, Pav , is given by the integral of the instantaneous power, p(t), over one cycle: p(t) = Pav = 1 T
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