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Hence " VR 74:300 cos 38:5128 j sin 38:5128 58:138 j46:265 volts and " VC 59:126 cos 51:4888 j sin 51:4888 36:817 j46:265 volts thus " " VR VC 94:955 j0 95=08 approx:; the applied reference voltage; answer: 122. (a) Since the value of each capacitance is 0:12 10 6 farads, eq. (222) becomes " I 75 j3 36 5 105 0:12 10 6 37:5 18 j25
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which, upon rationalizing (multiplying numerator and denominator by conjugate of denominator), becomes " " I 0:7113 j0:9879; thus jI j 1:2173 A; answer: (b) " Here, I a jb amperes, where a and b have the values found in part (a); thus, from the vector diagram to the right,  arctan b=a arctan (c) 123. (a) 0:9879 54:2468; answer 0:7113
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" " jVC j jI j 1=!C 1:2173 1=0:06 20:288 V, answer. First, in complex notation, for 0:25 mF capacitor; j 1=!C j40 ohms; for 0:32 mF capacitor; j 1=!C j31:25 ohms: " Here we re asked to nd the total (generator) current IT ; let us do this in two di erent ways, as follows. " 60 V " " FIRST WAY: Since IT " " , where ZT is the total impedance ZT ZT seen by the generator, let us begin by making use of eq. (207); thus, noting that 1= j40 j=40, 1 j 1 1 40 j32 1 "T 40 32 16 j31:25 1280 16 j31:25 Z
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hence 60 3 60 " IT " 5 j4 16 j31:25 ZT 8 which, after rationalizing the fraction, becomes 3 60 16 j31:25 " 2:6539 j3:0212 A approx: IT 5 j4 8 1232:5625 " which is a leading current of magnitude jIT j 4:021 A, answer. SECOND WAY: Let us now apply the loop current procedure, explained in connection with Fig. 135 and problem 120. Using the same current notation " " and reference directions as in Fig. 135 (denoting I1 by IT ), the three simultaneous equations for Fig. 140 are " j40IT " 0I T " j2IT " 0I T " j40I2 " 0I3 60 " 32I3 0 " " j40IT 32 j40 I2
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" " 32I2 48 j31:25 I3 0 " j2I2 " 0I3 3 " 4I3 0
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" " 32I2 48 j31:25 I3 0
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The value of the denominator determinant D is therefore equal to       1   1 0 0 1 0           j2 j5  j5 4 j5 4 4 D j2 4         0 32 48 j31:25    0 32 48 j31:25 8 31:25 j16    3 j2 0       0 4 j5 4     0 32 48 j31:25  3 4 j5 48 j31:25 128 " IT 8 31:25 j16 8 31:25 j16 Hence " j IT j  arctan 3j92:25 j365j 4:021 A; answer: 8j31:25 j16j 3 92:25 j365 8 31:25 j16
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3:0212 48:7038 approx., answer. 2:6539
124. Note: To save space, we ve generally rounded calculator values o to ve decimal places.
for 35 microhenry coil; for 65 microhenry coil; for 0:22 microfarad cap:;
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First, ! 2f 2 28 103 1:75929 105 rad/sec. Hence the reactances are j! 35 10 6 j6:15752 ohms; j! 65 10 6 j11:43539 ohms; j=! 22 10 8 j25:83687 ohms;
total resistance 2 4 6 ohms; a real number: " Let ZT be the total impedance seen by the generator. Since Fig. 141 is a series " circuit, the REAL PART of ZT is the sum of all the real components and the IMAGINARY PART is the sum of all the imaginary components; hence, for the case of Fig. 141 we have " ZT 6 j8:24396 thus, by Ohm s law, " I 20 20 6 j8:24396 1:15426 j1:58594 A; leading: 6 j8:24396 103:96288
Therefore, since the reactance of the 65 microhenry coil is j11:43539 ohms, and " since the voltage drop across the coil is equal to the coil reactance times the current I , we have that " Va 1:15426 j1:58594 j11:43539 thus " Va 18:13584 j13:19941 volts approx:; answer: The above answer is in terms of rectangular coordinates, expressed by a complex number lying in the second quadrant. To put the same answer in terms of polar coordinates, note that     "a j 22:43063 and  arctan13:19941 36:058; approx: jV 18:13584 Since  actually ends in the second quadrant, we have that  180 36:05 143:958 Thus, in polar coordinates the answer is " Va 22:43=143:958 as shown to the right. Thus an ac voltmeter* connected from point a to ground would read 22.43 volts, which, you ll note, is greater than the generator voltage of 20 volts. This is the result of the phenomenon of series resonance, which we take up in section 8.6. 125. First, reactance of inductor j!L j 106 25 10 6 j25 ohms and reactance of capacitor j=!C j= 106 5 10 8 j20 ohms
* Be reminded that ac meters are normally calibrated to read magnitudes of rms voltages and currents.
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