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Discussion Note: In the solution to the above problem, note that the impedance re ected into the primary coil is capacitive in nature, although the secondary circuit " is, by itself, inductive in nature, being 2 j9 ohms. This illustrates the fact that if Z2 is inductive, then the impedance re ected into the primary coil will be capacitive in " form. The reason for this can be seen from eq. (387), for Z2 R jX; thus !2 M 2 !2 M 2 " Zref R jX 2 R jX R X 2 " which is a capacitive type impedance. Or, if the net reactance of Z2 is capacitive in nature, then the impedance re ected into the primary coil will be inductive in nature. " Thus, for Z2 R jX, eq. (387) gives !2 M 2 !2 M 2 " Zref 2 R jX R jX R X 2 which is inductive in nature. Of course, if the secondary circuit is purely resistive (because of series resonance on the secondary side), the re ected impedance will also be a pure resistance. 188. (a) Let us rst calculate the reactances; thus XL1 XL2 4 105 60 10 6 24 ohms; XC 1=4 105 5 10 7 5 ohms; !M 20 ohms " Now let us look back to Fig. 222 and eq. (383). From Fig. 222 we see that Zb is the impedance connected to the terminals of the secondary coil. Thus, in our " problem here, Zb is the parallel combination of R and jXC , so that for Fig. 225 we have jXC R j50 " 2 j4 ohms Zb R jXC 10 j5 hence " " Z2 Zb JXL2 2 j20 ohms Now, putting all the known values into eq. (387) you should nd that " Zref 1:980 j19:80 ohms " Since there is no resistance in the primary circuit, Z1 jXL1 j24 ohms. Thus, by eq. (388), we have 28 "j j I1 1:980 j4:20 6:030 amperes; answer: (b) By eq. (389), " " j!M I !MjI j 120:6 " jI2 j " 1 " 1 p 6:000 amperes; answer: Z jZ 2 j 404 2 " " " First, ! 2f 105 radians/second; thus, the values of I1 ; I2 , and I3 are 40 " j5 A; I1 j8 40 " I2 4 A; 10 40 " I3 j10 A j4

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189. (a)

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" Next, the value of I4 is found by making use of eqs. (388), (387), and (371); thus 40 40 40 2:659 j3:713 A jXL1 Zref j12 5:099 j19:120 5:099 j7:120

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" " " " " Hence, I I1 I2 I3 I4 6:659 j8:713 , thus " jI j 10:966 amperes; answer: (b) (c) p " j j!MI4 !k L1 L2 j2:659 j3:713j 7:292 amperes; answer: j Is Z j2 j7:5j s The principle referred to here is that TRUE POWER is equal to the voltage

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times the component of current IN PHASE with the voltage. From the solution to (a), we have the vector diagram shown below. Thus, P 40 6:659 266:36 watts approx., answer. (d) Note that the value of can be found from the above gure. However, since " we re instructed to use Fig. 155, we must nd the value of Z R jX, and to do this we ll use the basic relationship " 40 " V Z " 2:2149 j2:8981 ohms 6:659 j8:713 I and hence, by Fig. 155, arctan X=R arctan 2:8981=2:2149 52:6118 and thus, by eq. (228), P 40 10:966 cos 52:6118 266:35 watts; answer: (e) As explained in section 8.5, there is no net energy loss in an ideal inductor or capacitor; that is, energy is removed from a network only through resistive elements which, in Fig. 226, are the devices having 10 ohms and 2 ohms of resistance. Since the current in the 10-ohm load is 4 amperes and the magnitude of the current in the 2-ohm load is 7.292 amperes, we have that P 4 2 10 7:292 2 2 266:35 watts; answer: As must be the case, all three procedures give the same value of P, after taking into account a slight round-o di erence.) 190. Since the two coils have equal inductances of L henrys, let us, for convenience, refer to them as the left-hand and right-hand (LH and RH) inductors. Next note that there are ve voltage drops to consider, as follows.

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