hw04solution

hw04solution - P 4.12 ’01 *— 3 111 ~ ’02 _ 4 10 80...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: P 4.12 ’01 *— 3 111 ~ ’02 _ 4 10 80 " '2) H71 1) w3+ 280 1+—51"-=0 so ~vl+17v2=240 Solving, 01 = 100 V; v2 = 20 V 0 so 29111 - 1153 = 2880 P 4.18 [a] —~>ia 38.5V , ’U ’U — ’02 ‘220+Iblé+ 125 =0 so 501—802—l—40320 v2 # “1 + £2— + ’02 _ v3 so —8111 + 1302 ~ 4'03 :2 0 25 200 50 v3 —— v2 03 - 52}, 03 —- 38.5 _ 50 + 5 + 20 — Solving, 711 = —50 V; 02 = ——30 V; 113 2 2.5 V 112 —r 113 _ ~30 — 2.5 0 so 051 — 4712 + 29123 = 192.5 [b] 20: 50 —————50 =—0.65A — '0 2. —r—. 7:321», 51 z 5 o( 065)=1.15A 5 5 385—25 ‘ =——-———= . A $9 20 18 Either-2mm Calculate Zpdev because we don’t know if the dependent sources are developing or absorbing power. Likewise for the independent Source. P225 = "—2730’01 == —2(-0.65)(—50) == -—65 W(dev) 1255, = 55053 = 5(——0.65)(1.15) z ~3.7375 W(dev) 299 = «38.5(18) = ~69.30 W(dev) Zpdev = 69.3 + 65 + 3.7375 = 138.0375 W 2500 900 400 2 2 2 100 + 200 + 25 + (0.65) (50)+ (1.15) 5+(1v8) (20) = 138.0375 W Zpdev = Ems : 138.0375 w E? g H P 4.37 ~80 + 31i1 - 161'2 *- 7i3 = 0 ~16i1 + 272'2 —— 42'3 x 0 ——7i1 ~— 42'2 + 3123 + 2415 = 0 Solving, 2'1 = 3.5 A p89 : (3-5>2(8> = 98 W P 4.46 [a] 20 SD Mesh equations: ——50 + 6i1 — 4132 + QiA == 0 —9'£A — 42'1 + 292'2 — 2023 = 0 Constraint equations: DA = 221 43 = 17 A; 7:3 = ~1.7’UA; M = Z.2; Solving, i1 = —5 A; 9A = _1() V 944 = 9(16) = 144 V ia=i2—7§1=21A ib=i2—7$32—1 A vb = 202']3 = —20 V Psov = —50’£1 = 250 W (absorbing) p9“ = 44924) = —(21)(144) = “3024 w (delivering) puv = —1.7vAvb = igvb = (17)(~—20) 2 —340 W (delivering) [b] ZPdeV = 3024 + 340 = 3364 W ZPdis = 250 + (—5)2(2) + (21)2(4)+(15)2(5) + (4)2(20) = 3364 W P 4.59 [a] Apply source transformations to both current sources to get 2.31:?) 2.10:0 1m 13 . SUI loé" T4 . 21! 13.8 + 4.2 ' = W = 3 1" 2700 + 2300 + 1000 mA The node voltage equations: _3 ’01 '01 —- ’U2 _ 6x10 +230 + 2700 _ 0 “2 +7)?”1 —4.2>< 10”3 = 0 1000 2700 Place these equations in standard form: 1 1 1 13 “1 (2700 + 2300) + "2 (M 2700) ’" ‘6 X 10 1 1 1 _3 m (—2700) + 1’2(1000 + 2700) _ 4‘2 X 10 Solving, 721 = ~6.9 V; 722 = 1.2 V . __ ’02 ~ ’01 1° _ 2700 = 3 mA ...
View Full Document

This note was uploaded on 07/29/2008 for the course ECE 376 taught by Professor Milenkovic during the Summer '06 term at Wisconsin.

Page1 / 7

hw04solution - P 4.12 ’01 *— 3 111 ~ ’02 _ 4 10 80...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online