Assignment3_Solutions

# Assignment3_Solutions - P2.48 P2.49 P253 i Vl—VZ zoJr 10...

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Unformatted text preview: P2.48* P2.49* P253 i Vl—VZ zoJr 10 At node 2 we have: V—3+ V3 _ V1 5 10 In standard form, the equations become 0.15V1—0.1V2 =1 =1 At node 1 we have: — 0.1V1+ 0.3V2 = 2 Solving, we find V1 : 14.29 Vand v2 : 11.43 V. V1 _ V2 20.285714. Then we have ti 2 Writing a KVL equation, we have v1 — V2 =10. V1 V2 At the reference node, we write a KCL equation: g +5 =1. Solving, we find V1 2 6.667 and v2 = —3.333. V2 _ V1 V1 _ Then, writing KCL at node 1, we have 1;. = 5 Writing KCL equations at nodes 1, 2, and 3, we have V1 V1 _ V2 —+ +£2.20 Rs 1’94 V2_V1+V2_V3+ﬁ:0 424 Pa 4"35 V3 V3_V2 =I. 121+ + R.) 1 In standard form, we have: 0.151/1— 0.101»2 = —5 — 0.101/1 + 0.47513 — 0.251!3 = 0 — 0.251/Z + 0.301/3 = 5 Solving using Matlab, we have 6 = [0.15 -O.10 O; -O.10 0.475 -0.25; O -0.25 0.30] I = [-5: 0: 5] V = 6\I v1 = —30.56 v v2 = 4.167 v v3 = 20.14 v V1‘ V2 5 Then, wr'iTing KCL equa’rions of nodes 1 and 2, we have: 3+6, =1 and i+0.5.ﬂ,—{, =0 P2.56* FirsT, we can write: 1;, = Subs’ri‘ru’ring for r; and simplifying, we have 0.3v1— 0.2V2 =1 — 0.1V1+ 0.15V2 = 0 Solving, we have v1 = 60nd V2 = 4. Then, we have i, 2 V1— V2 2 0.4 A. P2.57* VX = V2 7 V1 Wr‘i’ring KCL at nodes 1 and 2: ﬂ Vl—va Vl—V2:1 5 + 15 + 10 V_2 V272VX Vail/1:2 5 + 10 + 10 Subsfi‘rufing and simplifying, we have 15v1— 7vZ = 30 and v1+ 21/2 2 20. Solving, we find v1 = 5.405 and v2 = 7.297. P259 Fir‘sT, we can wire: I. _ 5;; — v2 " 10 Simplifying, we find i, = —0.2V2. Then wr'i’re KCL of nodes 1 and 2: V1 — 5i, v2 5 10 X Subsfi‘rufing for' r; and simplifying, we have V1 — V2 =15 and 0.3V2 = —3 which yield V1: 25 V and V2 = —10V. P2150 The circuit with a l-A current source connected is: V1 L’1 _ V2 — + 20 10 V —v v 2 1+—2—O.1vx=0 10 5 Using the first equation to substitute for' VX and simplifying, we have 0.15 v1 — 0.1V2 =1 — 0.2v1+ 0.4V2 = 0 Solving we find v1 =10. However, the equivalent resistance is equal in value to 14 so we have Reg = 10 Q. P2.63 We write equations in which voltages are in voltsl resistances are in k9, and currents are in mA. KCL node 2: (V3 _ V1) + (V2 _ V3) : V2 _ 4 3 2 KCL node 3: (V3 _ V3) + (V3 _ V1) + 0’3 _ V4) : 2 3 1 2 . 2 v_4 2 KCL ref node. 2 + 5 2 V1 — [/4 = Using Matlab: >> 6 = [-1/4 (1/2 +1/3 + 1/4) -1/3 0; —1—1/3 (1+1/2 +1/3) —1/2; 0 1/2 01/5: 10 O —1] >> I = [0; 2: 2: 10] >> V : G\I V : 9.3659 4.2537 6.8000 -0.6341 ...
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Assignment3_Solutions - P2.48 P2.49 P253 i Vl—VZ zoJr 10...

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