HW4 - ’Ul ’01—’U2.4 ——— 2 P49 2 125 25...

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Unformatted text preview: ’Ul ’01—’U2 . .4 ———- 2 P49 2 +125+ 25 0 (“2-01 +L2 +———v2 —3.2=0 25 250 375 Solving, v1 = 25 V; 1);; z 90 V CHECK: - (25V _ P1259 ~ 125 —» 5 W 90—252 P259=(———-r—*)*==169W 20 (90)2 = = 2.4 W P2500 250 3 (90)2 = z 21. W P3759 375 6 P2.4A = (25)(2.4) 2* 60 W Zpabs = 5 + 169 + 32.4 + 21.6 + 60 = 288 w 229.18, = (90)(3.2) = 288 w (CHECKS) P 4.11 [a] ’02—?11 U2 712*?)3 6 + ’5 + 12 2 0 ’03 + ’03 — ’02 ’03 — ’Ul _ 1 + 12 + 24 _ 0 In standard form: 1 1 1 1 v1 (1 + 6 + E) “2(7) “3171) ' 125 <—£>+v<-1-+-1—+i>+ (A) ~o “1 6 2 6 2 12 v3 12 ’ 1 1 1 1 1 ”1(‘§21‘)+“2(‘fi)+”3(1 + 1—2 + “22) = ‘125 Solving, v1 = 101.24 V; 112 = 10.66 V; 123 = -—106.57 V 125— — Thus, 11: v1=23.76A 7J4=v16v2=15A . ’U , ’U —-’U_ 222—22—=5.33A r15: 212 3:9.77A 2 ._ ig=vg+15=1843A 15:”124'U328fi6A [b] :Pdev == 1257:1 + 12513 = 5273.09 w ZPdis 2 13(1) + 13(2) + 13(1) + 13(6) + i§(12) + 13(24) = 5273.09 W P 4.14 [a] ’01 ’01 — ’01 — ’Uz 56 —|— ————5—-— + 2.5 = 0 so 31111 — 20112 + 0123 = 6400 v2#”1 v2_v3+12.8=0 so *2v1+3112“v3=—64 2.5 5 v 12 ~12 5:34. 35 2~12,8=0 so 0v1—112+303=64 Solving,’u1 = 380 V; 02 == 269 V; 123 = 111 V, [m%=%&£@=m1 Pg(del) = (640)(52) = 33,280 W ’Uo +’Uo+5’iA 110—80: P 4.17 «3 . + 200 10 + 20 20 [a] Solving, '00 z: 50 V . _’U0+5’iA [b] st " 10 iA = (50 — 80)/20 = «1.5 A -. ids = 4.25 A; 5% = —-7.5 V : pds z (~52'A)(ids) = 31.875 W [c] P3A = ~3v0 2 —3(50) 2 *150 W (del) Psov = SOiA 2 80(—1.5) == “120 W (del) 210561: 150 + 120 = 270 W CHECK: 192009 = 2500/200 = 12.5 W 29209 = (80 — 50)2/20 = 900/20 2 45 W P109 = (425)2(10) = 180.625 W Zpdgss = 31.875 + 180.625 + 12.5 + 45 = 270 w P 4.25 The two node voltage equations are: —10+3)3+vb_v° 6 2 20A+vC—vb+vC—24 _ 0 3 2 4 — = 0 The constraint equation for the dependent source is: 71A = vb Place these equations in standard form: 1 1 1 ’01, + + UGO—E) + ’UA(0) 7—" 10 (_£) + + + — “b 2 1": 2 4 “A 3 '” 4 1110(1) + 110(0) + 1)A(—1) 2: 0 Solving, vb 2 18 V, we 2 4 V,’UA = 18 V, and v0 = 24 —— 416 = 20 V P 4.29 —————————————————————————————————————————— supernode _ _ _ _ _. Node equations: ’01 ’01 ~ 713 — ’Ug ’03 20 + 2 + 4 + '8—0 + 3.125’UA =2 0 '02 '02 —— v3 412 —~ 20 _ 40 + 4 + 1 “ 0 Constraint equations: ’UA = -* ’02 ’01 - = ’03 73¢, a 112/40 Solving, 111 = —20.25 V; v2 = 10 V; 113 = —29 V Let i9 be the current delivered by the 20 V source, then 20 — (20.25) 20 — 10 = ~————— + 2'9 2 = 30.125 A pg (delivered) 2 20(30.125) = 602.5 W V , P 4.41 10:2 89 5mm /\\V 409% mm 600 2 642'1 *- 40z'2 — l4i3 —-400 = ~40’i1 + 50i2 —— 22'3 ~12 = i3 Solving, i1 = 2.9 A; i2 = W616 A; 2'3 = —12 A [a] v12A :: 2(12 — 6.16) + 14(12 + 2.9) = 220.28 V P12A = ~42va == ~12(220.Z8) = ~2643.36 W Therefore7 the 12 A source delivers 2643.36 W. [b] MOW = 400(—6.16) = —2464 W Peoov = “60011 = —600(2.9) = ~1740 W Therefore, the total power delivered is 2643.36+ 2464+ 1740 : 6847.36 W [a] Epigsism = (2.9)2(10) + (616)2(8) + (906)2(40) + (14.9)2(14) + (584)2(2) Zpabs = 6847-36 W = Zpdel (CHECKS) P 4.42 [a] The mesh current equation for the right mesh is: 3300(2‘1 — 0.008) + 65002} + 200(2‘1 —- 0.008) = 0 Solving, 10,0002} = 28 i1 = 2.8 mA Then, Q = i1 —— 0.008 = —5.2 mA [b] '00 = (0.008) (980) —— (~0.0052) (3300) = 25 V pgmA = *(25)(0.008) : —-200 mW Thus, the 8 mA source delivers 200 mW [c] 200% z 200(—-0.0052) = ~—1.04 V pdep source m 2002'A2'1 = (—1.04)(0.0028) = ~2.912 mW The dependent source delivers 2.912 mW. P 4.50 [a] 200 2—— 85i1 — 2525 — 5023 0 : —-75i1 + 352'2 + 1502'3 (supermesh) 63 ~— i2 = 4.3(i1 — i2) Solving, 2'] = 4.6 A; 732 2: 5.7 A; 2'3 = 0.97 A ia=i2=5.7 A; 25:23 =4.6A ic=i3=0.97A; id=i1—i2=—1.1A z'ezil—i32363A [b] 1025 + to + 25(2‘2 — 2'1) x 0 v0 2 ~57 ~ 27.5 =3 #845 V p4_3id = ~v0(4.3'id) = -(—-84.5)(4.3)(——1.1) = —399.685 W(dev) pgogv = ——200(4.6) = —920 W(dev) Zpdev = 1319.685W ZPdis z (5.7)210+(1.1)2(25)+(0.97)2100+(4.6)2(10)+ (363)2(50) = 1319.685W ZPdeV = ZPdis = 1319.685 W P 4.56 [a] The node voltage method requires summing the currents at two supernodes in terms of four node voltages and using two constraint equations to reduce the system of equations to two unknowns. If the connection at the bottom of the circuit is used as the reference node, then the voltages controlling the dependent sources are node voltages. This makes it easy to formulate the constraint equations. The current in the 10 V source is obtained by summing the currents at either terminal of the source. The mesh current method requires summing the voltages around the two meshes not containing current sources in terms of four mesh currents. In addition the voltages controlling the dependent sources must be expressed in terms of the mesh currents. Thus the constraint equations are more complicated, and the reduction to two equations and two unknowns involves more algebraic manipulation. The current in the 10 V source is found by subtracting two mesh currents. Because the constraint equations are easier to formulate in the node voltage method, it is the preferred approach. ’02 : (Um; _’21: : inn "4 _ v3 2: 274m Solving, v1 = 50 V; '02 2 40 V, '03 = -20 V, - ’Ul Um = M .fl _ :1 _ A 2" 25 2 8 plov = —-10i0 = W Thus, the 10 V source absorbs 180 W. ...
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This note was uploaded on 02/17/2009 for the course EEL 3111C taught by Professor Srivastava during the Spring '08 term at University of Florida.

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HW4 - ’Ul ’01—’U2.4 ——— 2 P49 2 125 25...

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