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CHAPTER 8 Reactance and Impedance
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In using the foregoing equations, all we need do is write each impedance in the form of a complex number R j!L, and then apply the algebra of complex numbers, remembering j 2 1. Problem 118 In Fig. 133, the resistance and inductance values are in ohms and henrys.
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Fig. 133
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" Given that the applied reference voltage V is 60 volts rms, and that ! 2f 100 rad/sec, nd* " (a) impedance ZT seen by generator, " (b) generator current IT , (c) (d) (e) (f ) (g) phase angle  between generator voltage and generator current, " I1 " I2 " I3
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verify that the sum of the answers to (d), (e), (f ) equals the answer to (b). " (h) using V V =08 60 volts rms as the reference vector, make a rough sketch of the vector diagram showing the answers to (d) through (g). Problem 119 The load on a generator of V =08 volts rms consists of a resistance of R ohms in parallel with a coil of inductance L henrys, as shown in Fig. 134.
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Fig. 134
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" Using eq. (209), write the equation for the generator current IT . Write the nal 0 00 answer in the rectangular form I jI .
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* We are here assuming there is NO MAGNETIC COUPLING between the two coils; that is, the magnetic eld of each coil is con ned to that coil only. The important case where this is not true is studied in Chap. 10.
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CHAPTER 8 Reactance and Impedance
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Let us pause here, just a moment, to again point out the close correspondence between the procedures used in dc circuit analysis and steady-state ac circuit analysis. Consider, for example, eqs. (32) and (34) in section 2.6; note that, to convert these two " equations to the ac case (eqs. (207) and (209)), all we need do is replace the Rs with Z s " R j!L) and apply the algebra of complex numbers, remembering that (in which Z j 2 1. In the same way, the treatment of series-parallel ac networks exactly parallels the " treatment of dc networks in section 2.7, with the Rs replaced by Z s. Likewise, the basic dc procedure of loop current analysis, explained in section 4.4, applies also to ac circuit analysis, remembering, of course, that j 2 1. In this regard, " consider the example of Fig. 135, containing two ac generators of known voltages V1 and "2 and three unknown loop currents, as shown. V
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Fig. 135
All the quantities in Fig. 135 denote complex numbers representing both the rms vector " values of voltage and current and the passive network components (the Z s). In this regard, using eq. (158) in Chap. 6, the two generator voltages can be put in the complex rectangular form; thus " V1 V1 =08 V1 cos 0 j sin 0 V1 1 j0 V1 and " V2 V2 =8 V2 cos  j sin  V20 jV200 In the example of Fig. 135, let us assume that the voltages and impedances are given, and that the problem is to nd the unknown values of the currents. In doing this, the voltage and current arrows shown in the gure are used in conjunction with KIRCHHOFF s VOLTAGE LAW, following the same basic procedure outlined for dc networks in section 4.4, except now we re dealing with complex (vector) quantities instead of scalar "" quantities. Also, the VOLTAGE DROPS will now be of the form ZI instead of RI as in the dc case. Thus, upon applying the rules of section 4.4, we have that the equations for the ac network of Fig. 135 are " " " " " " Z2 I2 0 I3 V1 Z1 Z2 I1 " " " " " " " " Z2 I1 Z2 Z3 Z4 I2 Z4 I3 0 " " " " " " " 0 I1 Z4 I2 Z4 Z5 I3 V2 " In regard to Fig. 135, it should be noted that the value of voltage V2 must be given with " respect to the reference voltage V1 V1 =08. Fundamentally, this is done by remembering that the voltages are sinusoidal waves of the same frequency, the wave of V2 being  degrees out of phase with V1 . (See discussion given with eq. (108) in section 5.6.) Lastly, to solve the resulting simultaneous equations for any particular value of current it will generally be easiest to use the method of determinants.
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