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Homework2 - P3245 a2an = —[3x(2x10‘3 =...

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Unformatted text preview: P3245 a2an = —[3x(2x10‘3)] = —6><10‘3 = —5 mar leA = —[—tx(1x10'3}] = 7x10'3 = 7'111W (checked using LNAP 8516.902) P3.2—T Pu. = +[2x(1x10'3)]= 2:-<1'D_3 = 2 mw P3... = + [s >< (—2 x 10‘3 = —6><10_3 = —6 mar [checked using LNAP 3.316.502) P3241 The subscripts suggest a Iiluuberiug cf the scrui‘ces. Apply K‘s-’1 to get 1:1: v2 +1L='j +1:g —vﬁ i1 and v1 d0 not adhere ta the passive caus‘entieu. sen p1=i1v1=f1[v3 +1:S +L'9 —1=5] is the pau-‘er supplied by source 1. Next. apply KC'L ta get i: = —(i‘1+i4] i. and v2 dc not adhere ten the passive convention. so P! =T2V2 =_|:!-1 ‘H-ilv': is the power supplied by source 2.Next, apply K171. to get v-=v,_—|:v,+vu:| : 1'3 and v3 adhere to the passive convention, so P: =_’-3V3 =43 [vs—[vs'l'vsﬂ is the power supplied by source 3. Next, apply K171. to get v, = v3 + v, + vN 1', and v , do not adhere to the passive convention, so Pd =14“ =14 live +Vs +VH] is the power supplied by source 4. Next, apply EC]. to get ’53&3 =1? —:',=:' —[—|::'l+:',}:|.=:'1+z'3+r, 1'5 and v1 adhere to the passive convention, so pi =—r', v, = —|:J"| +z'3 +17, :lvj is the power supplied by source 3. Next, apply EC]. to get 1', =I'_,—|:a'J +13} 1', and v6 adhere to the passive convention, so Pa = 45% =‘l1'7— [171+ 1': is the power supplied by source 6. Next, apply K171. to get V-I. = —'l-'¢_ I', and v, adhere to the passive convention, so _ —J-'T [I—vn ] = 1'7 1"eh = 4.1“? F7 is the power supplied by source Next: apply KCI. to get I" and 11“ do not adhere to the passive convention, so p, =1}thH =|:—r'_,:I‘LrH -I'*VH is the power supplied by source 8. Finally, apply KCL to get 1;, = r, +r3 1'9 and v,J adhere to the passive convention, so Ps- =_!"J vu = _|:__::| ‘H-silvu is the power supplied by source 9. U (Check: 2}?" = [l .j I’ll-21 —18+U—12—vu=ﬂ :a vu=—sov and :- P3.3—T All of the elements are connected in series. Replace the series voltage sources with a single equivalent voltage having voltage 12+20—18=14V. Replace the series 15 ﬂ. 5 Q and 20 ﬂ resistors by a single equivalent resistance of 15+5+20=40£1 40 Q By voltage division (checked: LNAP 6.59.504) P3.3—8 Use voltage division to get 13% 10 ](120)=20v M 10+50_ ' =0.2(20)=4A e i I. The power supplied by the dependent source is given by p = (llﬂﬁa = 480 1N (checked: LNAP 6 21 O4) P3.4_4 Ctu'rent division: 3 t = —6 =—.t A 1 15+al ] 3 i' =— —6 =—3A 2 3+3( ) t =r'1—.t"E =+1 P3.4—8 All of the elements of this circuit are connected in parallel. Replace the parallel current sources by a single equivalent 2 — 0.5 + 1.5 = 3 A current source. Replace the parallel 12 £1 and 6 .0 12 x 6 12+6 resistors by a single = 4 (1 resistor. 3 A 3 By current division 4 12 I=[—)3 =—=l "’14 A 3+4 * (checked: LNAP 6 9 O4) P3.6—2 6;; 1,. an 3.. 4L: 24V 3-5 a R =4+— =6 U 1 3+6 — (in i — i+l+i :>R 44:.) Then R — 3+R 404:1 R 12 6 5 P h 2 P '— P (a) KCL: I'_._+2 =r'1 and —24+6i3+R2i1= 0 => —24+6(f1—2}+1O.4i1= 0 2;; i1=—=".195A :> 1=1=f1R2=2.2(10.4)=22.33V 1 i3 — 4:- 4n 9' '~,—— 2.195 =O.878A. W? ‘“ l+l+i( ] 5")“; 8 3n 4. 6 6 12 “2 6n v3 =(O.ST8](6} = 5.3 v _ 6 1 e 3': i=0.585A 2) P=3i"=1.03W H 3 3+62 3 Fiﬁ—19 (a) R1=1UII(3U+1U]=SQ R1=4+(13||9]=1U§1 R3=|5||(5+5)=4Q (b) I=1A F1=3VJ=3=4V (ﬂ 1’4=— 10 3=—2V 10+3O r5—— 9 12—115; 9+18 3 l _ 1’?=—18[—g]=+6\' 4 l -——=—-A I6 12 3 (checked: LNAP 6 6 D4) P4.2—2 KC'L at node 1: L1*: ‘1 +—+1=UI :> 5v —v 2—20 20 5 1 2 KC'L at node 2: v1_v2 v2_vs +2: => —1’+31’7—2v =40 20 10 1 - 3 KC‘L at node 3: v —v v 2 3 3 = - :5= = 10 +1 15 2:: 3L2 13 3O Solving gives v1 = 2 V. v: = 30 V and v; = 24 V. (checked Ming LNAP 8 13 02} P4.2—7 Apply KC L at node a to get v —‘|’ v v 1'_ i5+ " b: b b '5 ' lID=£+£ :> i,=4A 2 S 8 2 8 8 ‘ (checked: LNAP 6.521.504) 14.3-4 Apply KCL to the supemcde: va+8 [‘Iw'a+3)-12 va —12 ea + + + = O 500 125 250 500 Solving yield-s v0 = 4 v (checked using LNAP 8.513.502} P435," 4k!) Apply KCL at nodes 1 and 2 to get 10—vl v1 121—191 = + 1000 5000 5000 10—1}, ‘91—'93 v3 _ + = 4000 5000 2000 :: 231,-1 41:2 =150 2} -41’1+19‘r'3 = 50 Solving. eg. using MATLAB. gives 23 —3 1:1 150 _ _ _ = _ 2) v1 = 3.06 V and 1.11 = 4.12 V —4 19 v2 50 Then = —= 7 _ ’J ih=11 12 =—"{:”5 4'1-=0.538111A 5000 5000 Apply KCL at the top node to get ‘r' —10 v} —10 T _ ’J_ g; =1_+-_=ﬂ+ﬂ=_441mg1 1000 4000 1000 4000 (checked: LNAP 5.531.504) P4.3—11 Label the node voltages: Express the voltage source voltages 1'11 terms of the node voltages: v1 —v2 = S and v5 = —28 Apply KC L to the supernode corresponding to the 8-\-" source: 1, -1, 1, 1, _1, “ﬂ. 2— 1 6+ 2: 2 3: ‘ 3 :3 => —3v1+3v5—28v2+24v3=48 16 12 3 6 Apply KC L at node 6 to get “ti—"1 "5 +—=0 => 5v1—91=5=160 16 20 2+ Apply KC L at node 3 to get "2 _ "3 "2 ‘1’3 "3 ""4 + = => —15v +131: —3v =0 6 3 10 1 3‘ 4 Apply KC L at node 4 to get 1.- _-l: 1: _-l:_ 3+ 3 4 = 4 3 z) 21o=—?v3+17’v4—10vj 10 7 Solving. cg. using MATLAB. gives —1 0 D U 0' 1’1 8 V1 —3.5 O 0 O D l 0 1’1 —28 1»; —16.5 —3 —23 24 D U 3 ‘Ir'g 48 1’} —15.5 = z; = 5 0 0 D U —9 1’4 160 V4 —10.5 0 —15 1s —3 0 0 v5 0 v5 —28 O 0 —T’ 1? —10 0 v5 210 vs —22.5 (checked: PSpice 6512.504) P4.4_4 Apply KCL to the supernode of the CC VS to get 12—10 14—10 1 4 + 2 —E+ib=0 2) ib=—2A Next 10—12 1 rt, = = —— —2 v 4 2 => : =—1=4 — r in =1: —14 —— A I (checked using LNAP 8514.502) P4.4—? Label the node voltages: V? First. 1:: = 10 V. due to the independent voltage source. Next. express Va and is. the controlling voltage and current of the dependent sources. in terms of the node voltages: and Next. express :1, and Reva. the controlled voltages of the dependent sources. in terms of the node voltages: _ 9.23 —10 83b =v1 —1»‘3 :> 8 =v1—v3 S and 3»; =v1 => 3{v1—10}=v1 => v1=15 v So 1=}—1{}=15—v3 => v3=125V Xext 2.5— 1===15—10=5V and ib=%=0.3125A Finally. apply KC L to the top node to get v s a =T‘“+;b = §+ 0.3125 =2.s125 A 4. a. ...
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Homework2 - P3245 a2an = —[3x(2x10‘3 =...

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