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Hence (a) " I 95 , thus 30 j37 95 " jI j q 1:994 amperes; answer: 2 30 37 2 (b) (c) By eq. (203),  arctan 37=30 50:9658 approx:; answer: Let L 0 ( L prime ) denote the 25 mH inductor; then V !L 0 I 103 25 10 3 1:994 49:85 volts rms; answer: 118. (a) " " " Setting Z1 8 ohms; Z2 j12 ohms; Z3 4 j6 ohms in eq. (207), you should nd that 1 5 j24 "T 24 3 j2 Z thus 24 3 j2 " ZT 5 j24 which, upon rationalizing (section 6.3), becomes " ZT 2:516 j2:476 ohms approx:; answer: (b) By eq. (206), " IT 60 2:516 j2:476
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which, after rationalizing, becomes 60 2:516 j2:476 " 12:115 j11:922 amperes; answer: IT 12:461 (c) From part (b) we have the vector diagram:
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thus  arctan
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11:922 44:548 approx: 12:115
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" " meaning that IT lags V by approximately 44.548, answer. (In polar form, " IT 16:997= 44:548:
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(d) (e) (f)
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60 " 7:50 amperes, answer. (In polar form, 7:50=08 I1 8 60 " j5 amperes, answer. (In polar form, 5= 908 I2 j12 " I3 60 30 which, upon rationalizing, becomes 4 j6 2 j3 30 2 j3 " I3 4:615 j6:923 amperes; answer: 13 In polar form; 8:320= 56:318 (g) By eq. (206), " IT 7:50 j5:00 4:615 j6:923 12:115 j11:923 amperes; which does check: (h) Relative to the reference voltage 60=08, the vector diagram can be drawn as follows:
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The basic graphical procedures is as follows. " " " First nd the vector sum of I1 and I2 , then combine this vector with the I3 vector ". to get the nal resultant vector IT " " " 119. For this particular case Z1 R; Z2 j!L , eq. (209) becomes V V =08 real number V jR!L " ZT R j!L thus V V R j!L " IT " jR!L ZT Now, to express the current in the rectangular form I 0 jI 00 , all we need do is multiply the numerator and denominator by j, thus getting   " V jR !L V 1 j 1 ; answer: IT R!L R !L
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" " 120. Drawing the loop currents in the manner of Fig. 135 (denoting I1 by IT ), the three simultaneous equations are " " " 8I T 8I 2 0I3 60 " " " 8 j12 I2 j12I3 0 8IT " " " 0I T j12I2 4 j18 I3 0 We next must nd the value of the denominator determinant D, thus       1   8 1 0 8 0         j3  D  8 8 j12 j12  8 4 2  2 2 j3      0  0 j6 2 j9  j12 4 j18  The easiest way, now, is to multiply each element of the rst row by 2 and add the result to the corresponding element of the second row, thus getting     1 1 0      1 1     D 64  0 j3 j3  j192  j6 2 j9  192 3 j2    0 j6 2 j9  hence  1   60 4 2  0  0 8 2 j3       2 j9  0 j3
" IT
j6 192 3 j2
480 5 j24 192 3 j2
thus, rationalizing, " IT 12:115 j11:923 amperes A ; answer; as in problem 118 b : 121. (a) Here, ! 2f 2:5133 104 rad/sec, C 5 10 7 farads; thus, 1 1 79:577 ohms !C 2:5133 104 5 10 7 thus, by eq. (219), " jZ j hence, by eq. (220), " jI j (b) By eq. (221),  arctan 0:7958 38:5128; answer: (c) Ac meters read magnitude of rms values; hence, voltmeter reading across R 0:743 100 74:3 volts, answer. " (d) Voltmeter reading across C 1=!C jI j 79:577 0:743 59:126 volts, answer. The details of the arrival of the full answer are as follows. Using rms values (instead of peak values), Fig. 138 becomes, on the complex plane, for the above values, 95:00 0:743 A; answer: 127:799 p 16332:50 127:799 ohms; approx:
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