The_solutions_to_CHT3s_homework

# The_solutions_to_CHT3s_homework - Chapter 3 Solution 1 Let...

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Unformatted text preview: Chapter 3, Solution 1. Let V}; be the voltage at the node between l-kQ and 4-14! resistors. 9—vx +6—Vx _ vk > ‘vﬂ< : 6 1k 4K 2K ' V I; : —" : 3 mA 2k Chapter 3, Solution 2. At node 1, 41—1126+VIM‘F‘2 —> 60=-8vl+5v: (1) 10 5 2 At node 2, :3 23+6+V1;V2 —> 36=-2vl+3Vg (2) Solving (1) and (2), v1: v3 = 12 v Chapter 3, Solution 5. Applyr KC L to the top node. 30—* 20—: r —"0+—‘”“ =10 —p v0=2ov 2k 5k 4k Chapter .3, Solution 13. At node number 21 [(v; + 2) — 0}“10 + V354 = 3 01' V3 = 8 volts But, I = [(v: + 2) — 0]..-"'10 = (8 + 2):"l{} = 1 amp and v1 = 8x1 = Svolts Chapter 3, Solution 11. At the top node, KVL gives V0 —36 + V0 —0 + V0 —(—12) Z 1 2 4 0 1.?51’0 =33 01‘ VD: 18.85TV P1Q = (36—1385. 7)3_.--'1 = 293.9 w Pm = (VD)3..-='2 = (13.357f7‘2 = 177.79 w P49 = (18.85?+12)27’4 = 233 w. Chapter 3, Solution 18. "1 I "'3 i 10 x 5 i + + i 4 Q I — — l i t E (a) (b) At node 2, in Fig. (3), 5 = Vg gvl V3 gv3 —p 10 = - vl + 2V2 - V3 (1) At the supernode V3 _ VI + V2 _ V3 : V—1—|—V—3 —> 40 = 2V1 + V3 (2) 2 2 4 8 From Fig. (b), - V1 - 10 + V3 = 0—? V3 = vl + 10 (3) Solving (1) to (3), we obtain v1 = 1|] V, V2 = 20 V = V3 Chapter .3, Selutinn 39. For mesh L —10—2II +1|3If1 —6I3 : 0 But IA. : II —I2. Hence, 10:—ZI]+2I2+1011—612 —> 52411—213 (1) For mesh 2, 12+832—61120 a. 6:31’1—41’2 (2) Solving (1) and (2) leads to I1 : (ELSA, I2 2-0.9A Chapter 3, Solution 46. 1301' loop 1, —12+11f1—8f2:[} —> 11f1—8f2212 (1) 1301‘ loop 2, —8.1F1+142‘2 +2120 : 0 But 120 : 33‘1, —8.1F1+142‘2 +6f1: 0 —> £1: Ti: (2) Substituting (2) into (1)? TTE2 —8:"2 : 12 —> £220.1739A and i1: Ti; 21.217 A Chapter 3, Solution 56. For loop 1, 12 = 4i1 — 2i: — 2i3 which leads to 6 = 2i1 — i2 — i3 (1) For 100]) 2, 0 = 61': —2i1 — 2 i3 which leads to 0 = -i1 + 3l2 — i3 (2) For 100]) 3, 0 = 6i3 — 2i1 — 2i: which leads to 0 = -i1 — i2 + 3i3 (3) 111 matrix form (1), (2), and (3) become, 2 —1 —1 i1 6 —1 3 —1 i2 : 0 _—1 —1 3 i3 0 2 —1 —1 2 6 —l A=—l 3 —1:8, \$2=—1 3 —l:24 —l —1 3 —1 0 3 2 —1 6 A3: —1 3 0 :24,the1'eforei3=i3=24..-"'8=3A, —1 —1 0 V] = 2i: = 6 volts, v = 2i3 = 6 volts ...
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## This note was uploaded on 02/16/2010 for the course ELECE 5501 taught by Professor Ahuja during the Spring '10 term at Texas Tech.

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The_solutions_to_CHT3s_homework - Chapter 3 Solution 1 Let...

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