EECE 210 Homework, Spring 2008

Homework 3 Solution P4.6 1200) 52 7 250 @40ma ev P4.11 [a] Note that we have chosen the lower node as the reference node, and that the voltage at the upper node with respect to the reference node is vp. Write a KCL equation (node voltage equation)by summing the currents leaving the upper node: 2 Hohe Uh Oa ef 120+5 25 Solve by multiplying both sides of the KCL equation by 125 and collecting the terms involving v, on one side of the equation and the constants on the other side of the equation: Vo +25+5u.+5=0 He 6up = —30 so Uo = —30/6 = -5 V %—110) wy—ve Wy — v3 2 8 16 Vg—V, . U2 . Ug — U3 8 3 24 vg+110 ug—vg vg —Vv > + 4 * =0 so 11v, — 2ue — v3 = 880 =0 so —3u; + 12v2 — v3 =0 =0 so —3v, — 2v2 + 29v3 = —2640 Solving, vu) = 74.64V; v2=11.79V; v3 = —82.5V — 1o-y% V1 — V2 Thus, i) = 2 =1768A ig= AS 786A . U2 P U2 — U3 => =3.93A = = 3: n= 3 is 71 3.93 A . v3 +110 . UV, — U3 = =13.75A = =9. or) 5) ig 16 9.82 A [b] $>Paey = 1104, + 110i = 3457.14 W DA19 Pais = i3(2) + (3) + 23(2) + 3(8) + #3(24) + #2(16) = 3457.14 W P4.14 40v@, The three node voltage equations are: vy —-40 vy v—v2 a nobis = 0 4 * 40 2 vg—U, | Vg— U3 2S _og = 0 a U 4 3 3 42 = 28 = 0 27 4 ‘ Place these equations in standard form: 1 = . ( 1) , W2,1,1 vi n(G+a 2 M9 oC) + GH) + #3) wo sal) + w(bH P 4.33 Ay 50 a 3n g 4Q + “—— Ay 4 = “yy oo" 1 13 AM My Why 20 10 an, 20 ‘| \ vy 40Q v 4Q yD28A g 1 2 oan ee AW AM 19 2 135 + 3(t; — ig) + 20(t, —i3)4+ 2%, = 0 5ig + 4(t2 — ig) + 3(%2 — iy) = 59 10i, + liz + 20(i3 — 21) + 4(ig —ig) = 0 ‘The dependent source constraint equation is: tg =n -ty Place these equations in standard form: 41(3 + 20 + 2) + ig(—3) + is(—20) +7,(0) = 135 i1(—3) + 42(5 + 4 + 3) + i3(—4) + i,(0) = 0 ty (—20) + ie(—4) + i3(1 +20 + 4) + 4,(10) 44 (1) + i2(—1) + 43(0) +1, (1) = 0 Solving, 4,=64.8A, i2=39A; ig = 68.4 A; i, =—25.8A Calculate the power: P200 = 20(68.4 — 64.8)? = 259.2 W Thus the 20 2 resistor dissipates 259.2 W. iT} Oo P 4.34 1200 a l AP The mesh current equations: —182 + li + 3(i, — 3) + (i, — ig) = 0 —Tig + 2(i2 — 41) + 10(ég — ig) = 0 5t3 + 10(ig — tg) + 3(izg — 11) = 0 The dependent source constraint equation: ig = tg — ig Place these equations in standard form: i(1 +3 + 2) + i9(—2) + i3(-3) 4 i4(0) = 132 ty(—2) + ig(10 + 2) + i3(—10) + to(—7) = 0 i1(—3) + io(—10) + i3(5 + 10 +3) 4 i,(0) = 0 4,(0) + ig(—1) + 43(1) + ég(1) = 0 Solving, 4, =48 A; %2=36A; %3 = 284A; Solve for the power: Pdep source = —T(ig)ie = —7(8)(36) = —2016 W Thus, the dependent source is developing 2016 W. ig =8A