Homework10Key

Homework10Key - Irwin, Basic Engineering Circuit Analysis,...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

Ask a homework question - tutors are online