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# Homework10Key - Irwin Basic Engineering Circuit Analysis...

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Unformatted text preview: Irwin, Basic Engineering Circuit Analysis, 9/E 3.3 Calculate the current in the induumr alum-'1] in Fig. Flt-1.3 if the unllagc input is; (a) Elli?) = CDHIIET-Ir + I”, {'3} 1'2“) : 5 5m {jar}: _ 90..) V Gm: tho uuh'wera in bath thus llrm‘ uml ['j'cqucncy' Llnrnains. L=1mH Figure FEE SOLUTION: (0») 1‘ ’9 V(t) L: ImH Mt)?- vmou: Mi) : :o J'Cosumt +qs°)ax I'm 4U): ~19” Am (3771+L450) Imam) J”) : 26'53 Slﬂ(377t+w5°) km) : 2653 (05 (3171— use) A Chapter 8: AC Steady-State Analysis Problem 8.8 2 lrwin, Basic Engineering Circuit Analysis, 9/E gm : 13-26 Cos (3711+4O") A 1‘: i326 <QO°A Problem 8.8 Chapter 8: AC Steady-State Analysis Irwin, Basic Engineering Circuit Analysis. 9/E 1 8.11 Find 1hr: ['i'r‘qucnuy'ﬂrimuiu inipmlmiuc. I. as; sham-n in Fig. PtH ]. Figure P3 .11 SOLUTION: Z _: 2.4331(2on 2mo° ~r 2L0“ 2‘: itinsUz _______—______——_____—_—_._—————— Chapter 8: AC Steady-State Anaiysis Problem 8.11 InNin, Basic Engineering Circuit Analysis, 9/E 346 Find it in the UL‘IWUE'K in Fig. P315. 211 1:1 _- q—Mr J’s—“£31; a“ "' (2:1 Figure P816 SOLUTION: Problem 8.16 Chapter 8: AC Steady-State Analysis 2 Inivin, Basic Engineering Circuit Analysis, 9/E I [0-7052 +O-S’2j] 7‘2 : 2-7058 +0-371J': 2'3724/6-99: __________________________——————————— Problem 8.16 Chapter 8: AC Steady-State Analysis Irwin, Basic Engineering Circuit Analysis, 9/E 3.43 Fiml 1hr: ‘v'L‘llk'lgC' ‘r' H-II'lD'Wﬂ in Fig. P343. 1 no girlf- v Figure P843 SOLUTION: looLo°v Chapter 8: AC Steady-State Analysis Problem 8.43 Irwin, Basic Engineering Circuit Analysis, 9/E 1 8.55 Fiml [J and 1’0 in this nclwnrk in Fig. P355. 12 £501.! 9 Figure P355 SOLUTION: VG” : «40° U0” : 6L igo°v Chapter 8: AC Steady-State Analysis Problem 8.55 2 Irwin. Basic Engineering Circuit Analysis, 9/E I, : il‘i V0 : "05+ VD” = IzLHS" +euxo° r39 :73- ex°v <7< H ____________—__—_—___—._—————-—-———-—————- Problem 8.55 Chapter 8: AC Steady-State Analysis Irwin, Basic Engineering Circuit Analysis, 9/E 1 85;! Find 1"? iii the circuit in Fig. FEET. 2 n 1'13“ H; Figlil‘ﬁ P35? SOLUTION: 2 ﬂ —j2 0 1'2 0 + / —'; I V 2 Q "0 i0": (1 ) (24q00):O‘EC[4H65‘/°A L4»J'2 V0”: 2( O'8QLii6-57‘): I'7Ciéli6’57°v 9‘0 : KID/+170” = (0-73426'57°-+i'7‘i 0'65)“ Chapter 8: AC Steady-State Analysis Problem 8.57 2 Irwin, Basic Engineering Circuit Analysis, 9/E V0 3 low 1 3s-OH°\/ ___________________.__._—__—————— Problem 8.57 Chapter 8: AC Steady-State Analysis ...
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