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HW2-solutions

# HW2-solutions - 3-3(a 1 1 —-1 — —— — 1500 3000...

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Unformatted text preview: 3-3 (a) 1 1 —-1 — + —— _._._...._ _ — 1500 3000 3000 [VA] _ 20-10 3 — 40.10 3 i ;+;+———1—— v3 —20.10"3 3000 3000 6000 1000 + 1000 + VI (5) 43 mil ' 1 1 —1 “ 1 — + — m — 3 — 3 [VA] _ 1500 3000 3000 20.10 — 40.10 VB —_1 —1 + 1 + —————-—l —2o.10‘ 3 3000 3000 6000 1000 + 1000 VA ~30 VB _ _ VA — VB = vx:=———— vx=——15 1x: 1 1x2: v3 —30 2 3000 3-12 Assign mesh currents as shown (a) by KCL i2 -— i1 = is Super Mesh: I R2 R3 (R1 + R2)-i1 + (R3 + R4 + R5)-i2 = v5 Kr (b) 111;: 200 112:: 500 R3 := 60 R4 2: 240 R5 := 200 vs := 15 is := 0.05 vs R5 _ —(is-R3 + is-R4 + is-Rs — vs) . (is-R1 + 13.112 + vs) i: 43.333 >< 10' 3 1 := —————————————~—-—-—-——— 1 := ._,_...._......_— _ = 1 (R1+R2+R3+R4+R5) 2 (R1+R2+R3+R4+R5) 12 4.167X10-2 ix := —i1 ix = 3.333 x 10' 3 vx ;= R4-(i2) 0x = 10 (c) Ptotal := (R3 + R4 + R5)-(i2)2 + (R1 + R2)-i12 ptotal = 0.917 w 3-19 Writing one mesh equation (2000 + 4000 + 80000912; + 40001210’ 3 = 40 solving for iA 1A ;= w0.052.10‘ 3 VA := —3000-iA VA = 38.856 VB := iA + 12.10“ 3)4000 v3 = 28.572 Checking {2000 + 4000 + 8000).“ + 4000-(12-10_ 3) = 49.998 <—-checks the mesh equation 349 With the 20 V and 1 mA sources turned off 3-34 The input-output reiaﬂonship is of the form: From the given data we have Assume k1::1 k2:=l k3 :=1 Given 0 = MD + 112-4 + k3-(——4) 1.5 = {(1-2 + k2!) + k3-(2) 0.5 RINIZ=15103 + 1 1 1 1 3 + 3 + _.._ 5-10 15-10 104 i _“ 10 4 S" RN is] =5.641x 10 . __ 10'4 , 101-“ W 451 (5.103)"1 + (15103)“ 1 + 10‘ 4 £01: 1.538x 10‘4 V0 = k1-Vs1 + kz-Vsz + ks-Vss 2 = 1:12 + k2-4 + k3-0 Fiml(k1,k2,k3) :- 025 -:—-Values of KS in the input-output reiationship 0.25 ...
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