Homework3

# Homework3 - P4.4—1t‘i Express the controlling voltage...

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Unformatted text preview: P4.4—1t‘i Express the controlling voltage and current of the dependent sources in terms of the node voltages: and Express the dependent voltage source voltage in terms of the node voltages: v3 —v3 =Ava =At=1 so vu—t' Ti— —15 A: 4 3 = ( )=4v.tv 124 22.5 Apply KC L to the supernode corresponding to the dependent voltage source 1| _‘Ll 1| _1_s_ _ _ 5—w’)‘ - 1 1+ 3 4:15 :> 510+—1 “ :23. :> 31:209 R1 R1 R1 50 Apply KC L at node 4: 1-‘ —1-‘ V. V —1-‘ — 7—22.: 2".5 — 3—22.: 3 4:_4+3 3 4 :3 —13‘ ﬂ: ‘ Jug—1" " :- B=2.5A-’A R2 R3 R3 50 2G 50 P-‘l.4-—1'er va—v 21—12 v —3 3.121: ' 1= =5o and 12,: 1 = =4o 2—0.5 1.5 _ 125—2 40.75 b. The power supplied by the voltage source is 12(05 +1.25 — 2] = —3 7W . The power supplied by the 1.2 S-A current source is l.25(—3—12) = —18.T5 1W. The pou—‘er supplied by the 0.5-A current source is —G.5(21) = —l'[}.5 ‘W. The power supplied by the 2-H current source is 2(21—[—3}}=43 W. P4.4—18 12— —1. 1:1— ( )=1.666A 3 and f2=g=l4A 4 1"?_1"’ 96—12 1’ _1I33 3-31: _ .1: =65} and R1=_ 3 =—=3_93:4Q 3—31 2—2.4 :1—2 1.666—2 b. The power eupplied by the voltage aouree is 12(2.4+1.66 — 2] = 24.? W. The power supplied by the current source is 2(9.6—(—1.33]) = 21.9 Vi". Mesh Equation: 11168111 : Bil + l [fl—i2} + 1'1]: ID 111E511: : BEE—fl} + 4 (fl—i3} = Cl 11168113 : —l{]+ 4 (ﬁg—i3) + 5i} = '3 Solving: q i=t2 2:: i=—1'—'_’=—D.194 A 50 QllSDU (1:509 40 ﬂ+60 9:100 f2 and 100 Q+30 Q+(80 ﬁll 560 ﬂ)=200 Q so the simpliﬁed circuit is 21H} [1 HM] Q |2V 8‘; The mesh equations are 200i1+50[i1—i2)—12 = o 100i2+8—50[i1—i2]=0 250 —50 I1 _ 12 3 2'1 _ 0.04 4.0 150 i; —s :5 43.04 The power supplied by The 12 V source is 1211 =12(0.04) = 0.48 1N. The power supplied by The 8 V source is —Si3 = — (—0.04) = 0.32 1W. The power absorbed by The 30 .0 resistor is 1'12 (513)2(304): [3(3): 0.043 "W. 01' P4.ﬁ—3 30E] Expreaa tha current some: Clu‘rent aa a function of the 111133.11 Clll'l‘EllTE: 1'1—322 = 43.5 ::>:'1 = :12 43.5. Apply KVL to tha aupernleall: 3Ui1+2|3i3+10= n 2: snug—n.5}+2m2 II | ,_. c: mfg—15 .131 II | ._. ('3' U hr II n i1 =-.4 A and v1 = 2:21;} = _ I; Fiﬁ—T i2=—3A 21—1125 :> il—(—3)=5 :> i1=2A e 2(i3—i1)+4i3+R{i3—i2)=0 3A 14 A 2 2[—1—2)+4[—1)+R{—1—(—3))=O => R=5§1 P4.ﬁ—9 Label the mesh currents: Express the clu'rent source currents in terms of the mesh currents 1'3 =11 and if =t'3 —i3 Apply KVL to the supermesh cun‘espconding to the current source with current i},- to get 4p} —i1)+vz +11[f3 —i4)+2i_, = 0 2 4e 43;; —16i3 +12t4 =1=z Substituting i1 =ix and 3'2 = f} —t", gives 43'}:—2[f3—iy]—16i3+11i4=v2 => —181'3+12i4 “(lg—4:3 Apply KVL to mesh 4 to get _ . . . 3. Sc- } vz+1iv—4fx —18 —]+12i4=vz+2iF—4ix :> i4=—' _ l l l _ 1_ 10:13—r4= w’14=—£t:z+1—<rv+ﬁtx Sc: (1:3 E321 and c:—i P433 Express 1'3 and 11,. the controlling voltage and current of the dependent sources. in terms of the mesh cun‘ents Next express 20 fb and 3 13,. the controlled voltages of the dependent sou1‘ees._ in terms of the mesh cun‘ents 20 fb 2—20 f1 and 3 ~.-a 2150243) Apply KVL to the meshes 10 —35 15 :1 0 0 45 —s :2 = 0 o 10 —10 f3 —10 Solving. e.g. using MATLAB. gives f1: —1.25 A1363 2 +0125 A! and :3 :+1.125 A P4.T-3 Label the mesh etu‘rents: Express ii. the controlling current of the C'CCS. in terms of the mesh currents l” fa : Ei3 _ Ii1 9 (i3 9 m V EXPI'ES‘: 2 fat. the controlled current of the C C CS. in ternn of the ineah currenta: 60 ﬂ 4U El 21] ﬂ il—t'g =2 ta = 2(3'3 —i1} => 3 i1 —:'3 —2 i3 =0 Apply KVL to the raupermesh corresponding to the {CC 3: 80(i1—i3)+40[i3 —i3}+60 i3 + 20 i1: 0 => 100i1+100i3 420;; = 0 Apply KVL to mesh 3 10+4D(i3 —i3)+30(i3 —i1}= 0 2: -30 1'1—40 i3 +1201; = —10 These three equation-s can be written in matrix form 3 —1 —2 i1 0 100 100 —120 i1 = ID —80 —40 120 1'3 —10 Solving. e.g. uaing MATLAB. gives i1: —D.2 A. 3': = —'D.l A and 1'3 = —O.25 A Apply KVL to 1116811 2 to get vb +4I::i(t2 —i3]+6{)i2 = t] :» vb =—4ﬂ[—ﬂ.1—[—D.15)}—6D[—ﬂJ]=CJ v 50 the power aupplied by the dependent murce if. p = vb [2:53] = '3 “W . [ht-.111 Express the current aouree current 111 terms of the 111e1h CIHTEIITS: Apply KVL ten the supertnesh em‘respondhtg ten the current muree to get 1(13 —11)+3+314 +1113 = +1 :1» —311+913 +314 = —3 Applyr KVL to 111e1h 1 to get 15+4(—13]+3[11—13)+211 :11 :1» 511—113 = —15 Applyr KVL to mesh 2 ten get 112 —3—4[—13) = 11 :1 112 +413 = 3 Solving. e.g using MATLAE. git-res {It 11 —1 1 :1 1 1'1 —5 —3 +1 9 3 :1 —3 13 3 = :5 = 5 11 J 1 :3 —15 :3 —2 {It 2 4 {It 14 8 I4 —1 ...
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Homework3 - P4.4—1t‘i Express the controlling voltage...

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