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CHAPTER 8 Reactance and Impedance
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Fig. 148
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Equation (224) corresponds to the dc case of eq. (62) for Fig. 55 in section 4.5. If we now apply to Fig. 148 the same basic procedure and steps as we applied to Fig. 55, we get the ac equivalent of eq. (63), thus " " " " " " Y V Y2 V2 Yn Vn " Vo 1 1 " " " Y1 Y2 Yn 225
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" " which is Millman s theorem for the ac case of Fig. 148, where Y 1=Z . As examples,
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" Z jXC " Y 1 jXC
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" Z jXL 1 " Y jXL " Y
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" Y j=XC j!C
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XL !L; XC 1=!C Problem 132 In Fig. 149, the values of the network components are given in ohms, microhenrys, and microfarads. The frequency is 100,000 rad/sec. Find the ac voltmeter reading between point a and ground. (We ve numbered the branches from left to right, 1 through 5, as shown.) Use Millman s theorem.
Fig. 149
CHAPTER 8 Reactance and Impedance
Real and Apparent Power. Power Factor
Electrical POWER is the RATE at which energy is being expended in a circuit, and is measured in WATTS. In section 5.5 we found that the AVERAGE POWER, P, produced in a purely RESISTIVE ac circuit is equal to P VI watts 226
where V and I are the rms values of voltage and current. It must be understood, however, that eq. (226) is correct only if the load is a pure RESISTANCE of R ohms. This is because in a purely resistive circuit the current and voltage waves are completely IN PHASE with each other, so that the current never reverses direction before the voltage does, and vice versa. If the current wave is not completely in phase with the voltage wave, then the average power P is less than the value given by eq. (226) and is then given by eq. (117) in Chap. 5, which is repeated below as eq. (227) (using  instead of ); thus P VI cos  watts 227 * in which  is the PHASE ANGLE between the current and voltage waves, where V and I are the magnitudes of the rms values. The situation is represented in vector form in Fig. 150.
Fig. 150
" In the gure, note that I cos  is the component of I that is IN PHASE WITH THE " , and thus, in accordance with eq. (227), we see that TRUE POWER P, is VOLTAGE V " equal to the voltage V times the component of current that is IN PHASE with V . Thus we have the important fact that, in alternating-current work. TRUE POWER is equal to the RMS VOLTAGE times the rms component of current that is IN PHASE with the voltage. Consider rst the case of an ideal inductor or capacitor.{ In either case, the voltage drop across, and the current through, are 908 out of phase, and thus the true power expended in an ideal reactor would be zero, since by eq. (227) P VI cos 908 0
* cos  is called the power factor. { An ideal inductor or capacitor would have zero resistance and thus zero losses. If an actual coil or capacitor does have appreciable resistance, such resistance can be taken into account by assuming an equal resistance to be in series with an ideal coil or capacitor.
CHAPTER 8 Reactance and Impedance
This is because inductors and capacitors store energy in their magnetic and electric elds during one-half of the cycle, then return the stored energy to the circuit during the other half of the cycle. Thus an ideal inductor or capacitor would be a totally lossless (wattless) device (see section 7.1). The di erence in the instantaneous power relationships in a pure resistance and in a pure reactance, such as a capacitor, is illustrated graphically in Figs. 151 and 152 and the following discussion.
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