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Unformatted text preview: ’Ul ’01—’U2
. .4 ——— 2
P49 2 +125+ 25 0
(“201 +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 = 684736 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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 Spring '08
 SRIVASTAVA
 Mesh Analysis, constraint equations, Zpabs

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