ECE109 HW 4 SOLUTIONS

# ECE109 HW 4 SOLUTIONS - 7-Β ~ β wv ~-ββw.mββ...

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Unformatted text preview: , 7-Β» ~ β wv _, ~-ββw.mββ wvβ-ββ β25:) 500 + o o :\ .r mο¬f V2 100!) V2 5200:) 8 JV , 112 β 1!] In ' " 5β . 1), 7" > - 712 r _ 0 ~ 21, 4 , ~4 ~~ β ~ LO so am - 81:2 + 4123 ββ ' β)0 I β1 712 - 1)] 717 7β2 β '01 ~- Β»-- -~βΒ«β-βββ- so -811 +130 -β 4113 = 0 2:, ' 200 β 50 1 2 ._ , f" . _ ; "r 7 "1-33 + 33-7333 + 3β β8 ββ 0 so 001 β 402 + 29113 = 192.5 50 0 20 Solving, 7), .-~ Β«50 V; Hg : ~30 V; '03 2: 2.5 V β βU2 ~ 113 ' '30 β 2.5 β)1 7.β 7 1 ~~Eο¬a~ 77 'β 50 =' ~β06:β) A yrβrβocr 1-; ~ 11β5-12β : β βΓ© β) 21.15 A 88.5 β 9 5 s = 1.8 A " 20 Calculate Emβ because we donβt know if the dependent sources are developing or almorbing power. Likewise for the independent source. p2,, = 41001 = -2(--0.65)(β50) = ~65 W(dev) pa, = 52'02'3 = 5(β0.65)(1.15) = β3.7375 W(dev) pa = β38.5(1.8) = β69.30 W(dev) 2de = 69.3 + 65 + 3.7375 = 138.0375 W CHECK 2500 900 400 Elm. = β-- β'β β 2 2 2 100 + 200 + 25 + (0.65) (50) + (1.15) 5 + (1.8) (20) = 138.0375 W Dd" z 2%. = 138.0375 W .zmβewm) to 25 A .Cβ pm, = 7523,13, = β375 w The dependent voltage source delivers 375 W m the circuit. @ P4.20[a]β5+13β5-+v1;v2=0 so dulβ302+0iA=75 vzβm v2 v2 v2+5iA_ .. _ 5 +30+m+ 3O β0 so -60;+11v2+mAβ0 iA=Ulgv2 so vlβvz-fn'A=0 " f Solving,v1=3OV; v2=15V; iA=3A; io=10;1Β°=1A Pa}; = (β15)(1) 2 ~15 W(del) PM = -5(30) = -150 W(del) 2 2 2 [b]. 21"β = (1105)β + 93% + (11%) + (3)2(5) + (1)2(30) = 165 w 2pm = 2m. = 165 w P 4.25 + "o β --_βββ The two node voltage equations are: β10+3"β+vβ"βc = 0 = 0 6 2 21);; uc β vb vc β 24 3 + 2 + 4 The constraint equation for the dependent source is: 0A = Uh Place these equations in standard form: 1 1 l vb ('6 + 5) + βCC-β5) + UA(0) = 10 1 1 l 2 24 βbβ? + βc (a + z) + βA (a) = I vb(1) + vc(0) + UA(β1) = 0 Solving,vb=18V,vc=4V,vA=18V, andvo=24βvc=20V " 250113 2 44V! β 2.0 β2β; 4β 2/113 β900 WWJβL/Kbl) SMVβO; ,25c(Zeβ\rz) β2. quv' β-ZO\/& 1r ZlV3βSβOO β4300 1- (4.111, β21003;?β β2ββ; (3) noβ) Wβsym W3 (9,) (#4 C?) : -β: 'Uβ; 1);: >βS'(Tβi. 4.1)} (.L-β\ W Mtg βE 5(3\ (MA C L13 0 β32.00 -: HquSβql) +V3] ββ 211,0 v1 +2Jv3 _ββ2.0o: Xrl<HL{[βVB>]"Z_\'O\) *651β73 ββ6300 ': _ g β . β31 V2+26Β©V3 (1,) WβLwe <1Β» (6): 90Β° : g'Vβzβm V3 (5) Let ig be the current delivered by the 20 V source, then . _ 20 +(2o.25) + 20 β 10 β-ββ-β =3.12 A 19 2 1 0 5 Pg (deliVered) = 20(30. 125) = 602.5 W a" 3:25): EiyβWi} } 80V _β p ~80 + 311] β 162'; - 71:; =β 0 ~16i1+ 271': β 453 = 0 _7,-1_ 4i; + 311'3 + 242β2 = 0 51mm. 1β. = l 2_ Solving, 2'; = 3.5 A Fan = (3.502(8) = 98 W P 4.38 250 ,. 660 = 302'1 β 10;; β 152β3 202},5 = ~10i1 + 602'2 - 5013 0 = Β«2152'1 β 502β2 + 9023 Q=QβQ Solving, 2'; = 42 A; 2'2 = 27 A; is = 22 A; i4, = 5 A 29iΒ’_=100 V pm = ~100i2 = β100(27) = β2700 w p29β (developed) = 2700 W The mesh current equation for the right mesh is: 3300(1'1 β 0.008) + 65002β1 + 2000'1 β 0.003) = 0 Solving, 10,0002'1 = 28 i1 = 2.8 mA Then, it: = 2'1 β 0.008 = β5.2 mA [b] v0 = (0.008)(980) β (β0.0052)(3300) = 25 v P8mA = β(25)(0.008) = ~200 mW Thus, the 8 mA source delivers 200 mW [c] 200iA = 200(β0.0052) = β1.04 V = 2002201 = (β1.04)(0.0028) = β2.912 mW The dependent source delivers 2.912 mW 10 = 18711 β 16i2 0 = β16βi1 + 28i2 + 42}; 4 = 8:}; Solving, i1 = 1 A; i2 = 0.5 A; it; = 0.5 A v0 = 16(2β1 β i2) = 16(0.5) = 8 V [13] pa}: = 4iAi2 = (4)(0.5)(0-5) = 1 W (8-105) βPuA (deliver) = -β1 W 0 = β75i; + 351'; + 1501'; (supermesh) i3 - i2 = 4-3β! β 52) Solving, i] = 4.6 A; i: = 5.7 A; i3 = 0.97 A i.=i2=5.7A; ib=i1=4.6A ic=i3=0.97A; id=i1~i2= β1.1 A i.=Β£1β2'3=3.63A {b1 10Β’β, +vo+ 25(1'2 β i1) = 0 '. u, = ~57 β- 27.5 = β-84.5 V I pug = βv.,(4.3id) = β(β84.5) (4.3)(β1.1) = β399.685 W(dev) onov = -200(4.6) = -β920 W(dev) 21):!" = 1319.685W 21%, = (5.7)210+(1.1)2(25)+(0.97)2100+(4.6)2(10).+ (353)2(50) = 1319.685 W 2P5" = 21%;. = 1319.685 W 400's β i1)+ 10(48 β Β£2) + 35(44 β 2'2) + 150 = 0 350;Β». βi4)+10(Β’, βi3)+15id = o 31). =i3βi4; 0:41β43 I'd =1}; i1 = 30 A Solving, 0', =30 A; i2=8A; i3=24 A; i4=6A (β=30β24=6A; ib=8β24=β16A; ic=8β6=2A; 23:615.; ia=iΒ’+id=6+2=8A [b] v. = 401'. = 240 V; vb = 150 β 352'c = 80 V Pam = β3011., = β30(240) = β7200 W (gen) pm, = 155415: = 15(6)(8) = 720 W (diss) p34, = 3400,, = 3(0)(80) = 1440 w (diss) leV = 1502}; = 150(6) = 900 W (diss) p409 = (6)9(40) = 1440 w (diss) pm = (β16)β(10) = 2560 w (diss) 17350 = (2)2(35) = 140 w (diss) 2PM = 7200 w 231;. = 720 + 1440 + 900 + 1440 + 2560 + 140 = 7200 W .ο¬ - Van}; .1- " u N.β 4).. .4 β.-.β;J'Y'T;1c'-, '12! . , 1) _ β "34ββ β1'" v, WA ο¬rmr vars) ο¬rmβ . - _, Mc~~β _. ~. β7 .7 a . β1 t p β \ β In , I > I?!β / 4.: 5 ' (If. 3'27" 1 β (I 1 4/4 β13'β βI .β~_ _β _ 7 ,_ _. 1 ' L1". 11β. '1 any"): 112β.Β» a amt" β\$331301β? β5β" 12:1: 1 ' - β . I . - aβ: _ . wg.βz:: areΒ» L any; rapt}; "I, 12!β β-βw 11Β»: 5 β12β): I?!β g β.376 '/ .5!Β» 3r 3 rm; tmsiβmrn m as, damage- wxr-βz βΓ©βaβο¬ x; a 317- 94113β5β3 57-4 m 7:. '3β 34991426 7/ ο¬ab: 3β53 ββ3 -S+\'2_ : \w v H n Β«as = C(%)-q(3)β5(5): 23-31: \1β4 \$ ...
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