IRWIN 9e 8_112

IRWIN 9e 8_112 - |n~in, 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: |n~in, Basic Engineering Circuit Analysis, 9/E 8.112 Um: Hanna‘s theorem m fiml 1} in the netwurk in Fig. PH] 12. xi)?- A Figure P8112 SOLUTION: 4&A 10 i : 32:1, Kvu lCih JIFi'xH Mic) JrJI 11:0 Chapter 8: AC Steady-State Analysis Problem 8.112 2 Irwin, Basic Engineering Circuit Analysis, 9/E TL": +J'(L1£O°—i5¢)‘l‘ff5¢+jl—fzto ~f. w“ (H1!) "1’1 -l C r—Mffgg =— 94432:“ T, = HLO" A CHJi) i. H: Hi)?“ : 566 L415" kcu 29} + ic = "I: v”: 'J‘j: : ~J'CHLO°~ Tu.) ~12 (wound-r i; -350 ~‘I; +(m' 2) {SC : uqo" OH.) Ii H.401; .1 5-66 1-HT “’1; —l ( H31) T36 : 8046’ ~— 11 : 293 L435" A \l Tgc, 2'33 ALISOA Problem 8.112 Chapter 8: AC Steady-State Analysis Irwin, Basic Engineering Circuit Analysis, 9/E 1f). Chapter 8: AC Steady-State Analysis Problem 8.112 4 Irwin. Basic Engineering Circuit Analysis. 9/E vac : 5'44 446° +(—!~jl)(5’Z—90°) VOL: 5-65 msov 2H : ML: Mi= 2mm it 2-373 /. (45° .. ‘L : ; ( Zqu" )(2-33Lw50):2'5217r6°/\ /o |+2L90 R70 : M’Z‘SZLW‘S“) VD: 2'53 L7"6° V _______________________.__—_————————————— Problem 8.112 Chapter 8: AC Steady-State Analysis ...
View Full Document

Ask a homework question - tutors are online