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

Info iconThis preview shows pages 1–10. 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

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

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

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

View Full DocumentRight Arrow Icon
Background image of page 10
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

Page1 / 10

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

This preview shows document pages 1 - 10. Sign up to view the full document.

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