Bài giải phần giải mạch P3

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Bài giải phần giải mạch P3

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Chapter 3, Solution 1. v1 8Ω 2Ω 40 Ω v2 6A 10 A At node 1, 6 = v1/(8) + (v1 - v2)/4 At node 2, v1 - v2/4 = v2/2 + 10 Solving (1) and (2), v1 = 9.143V, v2 = -10.286 V 2 v1 (9.143)2 P8Ω = = = 10.45 W 8 8 48 = 3v1 - 2v2 (1) 40 = v1 - 3v2 (2) P4Ω = (v 1 − v 2 )2 4 = 94.37 W v1 (= 10.286)2 = 52.9 W P2Ω = 2 = 2 2 Chapter 3, Solution 2 At node 1, v − v2 − v1 v1 − = 6+ 1 10 5 2 At node 2, v2 v −...

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  1. Chapter 3, Solution 1. v1 40 Ω v2 6A 8Ω 10 A 2Ω At node 1, 6 = v1/(8) + (v1 - v2)/4 48 = 3v1 - 2v2 (1) At node 2, v1 - v2/4 = v2/2 + 10 40 = v1 - 3v2 (2) Solving (1) and (2), v1 = 9.143V, v2 = -10.286 V v1 (9.143)2 2 P8Ω = = = 10.45 W 8 8 (v 1 − v 2 )2 P4Ω = = 94.37 W 4 v1 (= 10.286)2 P2Ω = 2 = = 52.9 W 2 2 Chapter 3, Solution 2 At node 1, − v1 v1 v − v2 − = 6+ 1 60 = - 8v1 + 5v2 (1) 10 5 2 At node 2, v2 v − v2 = 3+ 6+ 1 36 = - 2v1 + 3v2 (2) 4 2 Solving (1) and (2), v1 = 0 V, v2 = 12 V
  2. Chapter 3, Solution 3 Applying KCL to the upper node, v0 vo vo v 10 = + + +2+ 0 v0 = 40 V 10 20 30 60 v0 v v v i1 = = 4 A , i2 = 0 = 2 A, i3 = 0 = 1.33 A, i4 = 0 = 67 mA 10 20 30 60 Chapter 3, Solution 4 v1 2A v2 i1 i2 i3 i4 5Ω 10 Ω 10 Ω 5Ω 5A 4A At node 1, 4 + 2 = v1/(5) + v1/(10) v1 = 20 At node 2, 5 - 2 = v2/(10) + v2/(5) v2 = 10 i1 = v1/(5) = 4 A, i2 = v1/(10) = 2 A, i3 = v2/(10) = 1 A, i4 = v2/(5) = 2 A Chapter 3, Solution 5 Apply KCL to the top node. 30 − v 0 20 − v 0 v 0 + = v0 = 20 V 2k 6k 4k
  3. Chapter 3, Solution 6 v 2 − 12 v 0 v 0 − 10 i1 + i2 + i3 = 0 + + =0 4 6 2 or v0 = 8.727 V Chapter 3, Solution 7 At node a, 10 − Va Va Va − Vb = + → 10 = 6Va − 3Vb (1) 30 15 10 At node b, Va − Vb 12 − Vb − 9 − Vb + + =0 → 24 = 2Va − 7Vb (2) 10 20 5 Solving (1) and (2) leads to Va = -0.556 V, Vb = -3.444V Chapter 3, Solution 8 3Ω i1 v1 i3 5Ω i2 + 3V + V0 2Ω – + 4V0 – – 1Ω v1 v1 − 3 v1 − 4 v 0 i1 + i2 + i3 = 0 + + =0 5 1 5 2 8 But v 0 = v1 so that v1 + 5v1 - 15 + v1 - v1 = 0 5 5 or v1 = 15x5/(27) = 2.778 V, therefore vo = 2v1/5 = 1.1111 V
  4. Chapter 3, Solution 9 3Ω i1 v1 i3 6Ω + v0 – i2 + 12V + v1 8Ω + – 2v0 – – At the non-reference node, 12 − v1 v1 v1 − 2 v 0 = + (1) 3 8 6 But -12 + v0 + v1 = 0 v0 = 12 - v1 (2) Substituting (2) into (1), 12 − v1 v1 3v1 − 24 = + v0 = 3.652 V 3 8 6 Chapter 3, Solution 10 At node 1, v 2 − v1 v = 4+ 1 32 = -v1 + 8v2 - 8v0 (1) 1 8 1Ω 2i0 4A v0 v1 v2 8Ω 2Ω 4Ω i0
  5. At node 0, v0 v 4= + 2I 0 and I 0 = 1 16 = 2v0 + v1 (2) 2 8 At node 2, v 2 − v1 v 2 v 2I0 = + and I 0 = 1 v2 = v1 (3) 1 4 8 From (1), (2) and (3), v0 = 24 V, but from (2) we get v 4− o io = 2 = 2 − 24 = 2 − 6 = - 4 A 2 4 Chapter 3, Solution 11 4Ω i1 v i2 3Ω i3 10 V + – 5A 6Ω Note that i2 = -5A. At the non-reference node 10 − v v +5= v = 18 4 6 10 − v i1 = = -2 A, i2 = -5 A 4 Chapter 3, Solution 12 10 Ω v1 20 Ω v2 50 Ω i3 24 V + – 5A 40 Ω
  6. 24 − v 1 v − v 2 v1 − 0 At node 1, = 1 + 96 = 7v1 - 2v2 (1) 10 20 40 v1 − v 2 v 2 At node 2, 5 + = 500 = -5v1 + 7v2 (2) 20 50 Solving (1) and (2) gives, v1 = 42.87 V, v2 = 102.05 V v v i1 = 1 = 1.072 A, v2 = 2 = 2.041 A 40 50 Chapter 3, Solution 13 At node number 2, [(v2 + 2) – 0]/10 + v2/4 = 3 or v2 = 8 volts But, I = [(v2 + 2) – 0]/10 = (8 + 2)/10 = 1 amp and v1 = 8x1 = 8volts Chapter 3, Solution 14 5A 8Ω v0 v1 1Ω 2Ω 20 V – 4Ω + 40 V + – v1 − v 0 40 − v 0 At node 1, +5= v1 + v0 = 70 (1) 2 1 v1 − v 0 v v + 20 At node 0, +5= 0 + 0 4v1 - 7v0 = -20 (2) 2 4 8 Solving (1) and (2), v0 = 20 V
  7. Chapter 3, Solution 15 5A 8Ω v0 v1 1Ω 2Ω 20 V – 4Ω + 40 V + – Nodes 1 and 2 form a supernode so that v1 = v2 + 10 (1) At the supernode, 2 + 6v1 + 5v2 = 3 (v3 - v2) 2 + 6v1 + 8v2 = 3v3 (2) At node 3, 2 + 4 = 3 (v3 - v2) v3 = v2 + 2 (3) Substituting (1) and (3) into (2), − 56 2 + 6v2 + 60 + 8v2 = 3v2 + 6 v2 = 11 54 v1 = v2 + 10 = 11 i0 = 6vi = 29.45 A 2 2 v1  54  P65 = = v1 G =   6 = 144.6 W 2 R  11  2  − 56  P55 = v G =  2 2  5 = 129.6 W  11  P35 = (v L − v 3 ) G = (2) 2 3 = 12 W 2
  8. Chapter 3, Solution 16 2S v1 v2 8S v3 i0 + 13 V + 2A 1S v0 4S – – At the supernode, 2 = v1 + 2 (v1 - v3) + 8(v2 – v3) + 4v2, which leads to 2 = 3v1 + 12v2 - 10v3 (1) But v1 = v2 + 2v0 and v0 = v2. Hence v1 = 3v2 (2) v3 = 13V (3) Substituting (2) and (3) with (1) gives, v1 = 18.858 V, v2 = 6.286 V, v3 = 13 V Chapter 3, Solution 17 i0 4Ω 2Ω 10 Ω 60 V 8Ω 60 V + 3i0 –
  9. 60 − v1 v1 v1 − v 2 At node 1, = + 120 = 7v1 - 4v2 (1) 4 8 2 60 − v 2 v1 − v 2 At node 2, 3i0 + + =0 10 2 60 − v1 But i0 = . 4 Hence 3(60 − v1 ) 60 − v 2 v1 − v 2 + + =0 1020 = 5v1 - 12v2 (2) 4 10 2 60 − v1 Solving (1) and (2) gives v1 = 53.08 V. Hence i0 = = 1.73 A 4 Chapter 3, Solution 18 –+ v2 v1 v3 10 V 2Ω 2Ω + + 5A v1 v3 4Ω 8Ω – – (a) (b) v 2 − v1 v 2 − v3 At node 2, in Fig. (a), 5 = + 10 = - v1 + 2v2 - v3 (1) 2 2 v 2 − v1 v 2 − v 3 v1 v 3 At the supernode, + = + 40 = 2v1 + v3 (2) 2 2 4 8 From Fig. (b), - v1 - 10 + v3 = 0 v3 = v1 + 10 (3) Solving (1) to (3), we obtain v1 = 10 V, v2 = 20 V = v3
  10. Chapter 3, Solution 19 At node 1, V1 − V3 V1 − V2 V1 5 = 3+ + +  → 16 = 7V1 − V2 − 4V3 (1) 2 8 4 At node 2, V1 − V2 V2 V2 − V3 = +  → 0 = −V1 + 7V2 − 2V3 (2) 8 2 4 At node 3, 12 − V3 V1 − V3 V2 − V3 3+ + + =0  → − 36 = 4V1 + 2V2 − 7V3 (3) 8 2 4 From (1) to (3),  7 − 1 − 4  V1   16        − 1 7 − 2 V2  =  0   → AV = B 4 2 − 7  V3   − 36       Using MATLAB,  10  V = A −1 B =  4.933     → V1 = 10 V, V2 = 4.933 V, V3 = 12.267 V 12.267    Chapter 3, Solution 20 Nodes 1 and 2 form a supernode; so do nodes 1 and 3. Hence V1 V2 V3 + + =0 → V1 + 4V2 + V3 = 0 (1) 4 1 4 . V1 . V2 2Ω V3 4Ω 1Ω 4Ω
  11. Between nodes 1 and 3, − V1 + 12 + V3 = 0  → V3 = V1 − 12 (2) Similarly, between nodes 1 and 2, V1 = V2 + 2i (3) But i = V3 / 4 . Combining this with (2) and (3) gives . V2 = 6 + V1 / 2 (4) Solving (1), (2), and (4) leads to V1 = −3V, V2 = 4.5V, V3 = −15V Chapter 3, Solution 21 4 kΩ 2 kΩ 3v0 v3 3v0 v1 v2 + + + + + v0 v3 v2 3 mA 1 kΩ – – – (b) (a) Let v3 be the voltage between the 2kΩ resistor and the voltage-controlled voltage source. At node 1, v − v 2 v1 − v 3 3x10 −3 = 1 + 12 = 3v1 - v2 - 2v3 (1) 4000 2000 At node 2, v1 − v 2 v1 − v 3 v 2 + = 3v1 - 5v2 - 2v3 = 0 (2) 4 2 1 Note that v0 = v2. We now apply KVL in Fig. (b) - v3 - 3v2 + v2 = 0 v3 = - 2v2 (3) From (1) to (3), v1 = 1 V, v2 = 3 V
  12. Chapter 3, Solution 22 12 − v 0 v1 v − v0 At node 1, = +3+ 1 24 = 7v1 - v2 (1) 2 4 8 v1 − v 2 v 2 + 5v 2 At node 2, 3 + = 8 1 But, v1 = 12 - v1 Hence, 24 + v1 - v2 = 8 (v2 + 60 + 5v1) = 4 V 456 = 41v1 - 9v2 (2) Solving (1) and (2), v1 = - 10.91 V, v2 = - 100.36 V Chapter 3, Solution 23 v1 v 2 At the supernode, 5 + 2 = + 70 = v1 + 2v2 (1) 10 5 Considering Fig. (b), - v1 - 8 + v2 = 0 v2 = v1 + 8 (2) Solving (1) and (2), v1 = 18 V, v2 = 26 V v1 v2 –+ 5A 2A 8V + + 10 Ω 5Ω v1 v2 – – (a) (b)
  13. Chapter 3, Solution 24 6mA 1 kΩ 2 kΩ 3 kΩ V1 V2 + io - 30V 15V - 4 kΩ 5 kΩ + At node 1, 30 −V 1 V V − V2 =6+ 1 + 1 → 96 = 7V1 − 2V2 (1) 1 4 2 At node 2, (−15 −V 2) V2 V2 − V1 6+ = + → 30 = −15V1 + 31V2 (2) 3 5 2 Solving (1) and (2) gives V1=16.24. Hence io = V1/4 = 4.06 mA Chapter 3, Solution 25 i0 v0 20V + 2Ω 10V + – – 4Ω 40V + 1Ω – 2Ω Using nodal analysis, 20 − v 0 40 − v 0 10 − v 0 v −0 + + = 0 v0 = 20V 1 2 2 4 20 − v 0 i0 = = 0A 1
  14. Chapter 3, Solution 26 At node 1, 15 − V1 V − V3 V1 − V2 = 3+ 1 + → − 45 = 7V1 − 4V2 − 2V3 (1) 20 10 5 At node 2, V1 − V2 4 I o − V2 V2 − V3 + = (2) 5 5 5 V − V3 But I o = 1 . Hence, (2) becomes 10 0 = 7V1 − 15V2 + 3V3 (3) At node 3, V − V3 − 10 − V3 V2 − V3 3+ 1 + + =0 → − 10 = V1 + 2V2 − 5V3 (4) 10 5 5 Putting (1), (3), and (4) in matrix form produces  7 − 4 − 2  V1   − 45        7 − 15 3 V2  =  0   → AV = B 1 2 − 5  V3   − 10       Using MATLAB leads to  − 9.835  −1   V = A B =  − 4.982   − 1.96    Thus, V1 = −9.835 V, V2 = −4.982 V, V3 = −1.95 V Chapter 3, Solution 27 At node 1, 2 = 2v1 + v1 – v2 + (v1 – v3)4 + 3i0, i0 = 4v2. Hence, 2 = 7v1 + 11v2 – 4v3 (1) At node 2, v1 – v2 = 4v2 + v2 – v3 0 = – v1 + 6v2 – v3 (2) At node 3, 2v3 = 4 + v2 – v3 + 12v2 + 4(v1 – v3)
  15. or – 4 = 4v1 + 13v2 – 7v3 (3) In matrix form, 7 11 − 4  v 1   2  1 − 6 1   v  =  0    2    4 13 − 7  v 3   − 4      7 11 −4 2 11 −4 ∆ = 1 −6 1 = 176, ∆ 1 = 0 −6 1 = 110 4 13 −7 −4 13 −7 7 2 −4 7 11 2 ∆2 = 1 0 1 = 66, ∆ 3 = 1 − 6 0 = 286 4 −4 −7 4 13 − 4 ∆ 1 110 ∆ 66 v1 = = = 0.625V, v2 = 2 = = 0.375V ∆ 176 ∆ 176 ∆3 286 v3 = = = 1.625V. ∆ 176 v1 = 625 mV, v2 = 375 mV, v3 = 1.625 V. Chapter 3, Solution 28 At node c, Vd − Vc Vc − Vb Vc = + → 0 = −5Vb + 11Vc − 2Vd (1) 10 4 5 At node b, Va + 45 − Vb Vc − Vb Vb + = → − 45 = Va − 4Vb + 2Vc (2) 8 4 8 At node a, Va − 30 − Vd Va Va + 45 − Vb + + =0  → 30 = 7Va − 2Vb − 4Vd (3) 4 16 8 At node d, Va − 30 − Vd Vd Vd − Vc = +  → 150 = 5Va + 2Vc − 7Vd (4) 4 20 10 In matrix form, (1) to (4) become
  16.  0 − 5 11 − 2  Va   0        1 − 4 2 0  Vb   − 45   7 − 2 0 − 4  V  =  30  → AV = B     c   5 0 2 − 7 V   150    d    We use MATLAB to invert A and obtain  − 10.14    −1  7.847  V = A B= − 1.736     − 29.17    Thus, Va = −10.14 V, Vb = 7.847 V, Vc = −1.736 V, Vd = −29.17 V Chapter 3, Solution 29 At node 1, 5 + V1 − V4 + 2V1 + V1 − V2 = 0  → − 5 = 4V1 − V2 − V4 (1) At node 2, V1 − V2 = 2V2 + 4(V2 − V3 ) = 0  → 0 = −V1 + 7V2 − 4V3 (2) At node 3, 6 + 4(V2 − V3 ) = V3 − V4 → 6 = −4V2 + 5V3 − V4 (3) At node 4, 2 + V3 − V4 + V1 − V4 = 3V4  → 2 = −V1 − V3 + 5V4 (4) In matrix form, (1) to (4) become  4 − 1 0 − 1 V1   − 5        − 1 7 − 4 0 V2   0   0 − 4 5 − 1 V  =  6  → AV = B      3  − 1 0 − 1 5 V   2    4    Using MATLAB,  − 0.7708    −1  1.209  V = A B= 2.309     0.7076    i.e. V1 = −0.7708 V, V2 = 1.209 V, V3 = 2.309 V, V4 = 0.7076 V
  17. Chapter 3, Solution 30 v2 –+ 40 Ω I0 120 V v1 20 Ω v0 10 Ω 1 2 + 100 V – 4v0 + 2I0 – 80 Ω At node 1, v 1 − v 2 100 − v 1 4 v o − v 1 = + (1) 40 10 20 But, vo = 120 + v2 v2 = vo – 120. Hence (1) becomes 7v1 – 9vo = 280 (2) At node 2, vo − 0 Io + 2Io = 80  v + 120 − v o  v o 3 1 =  40  80 or 6v1 – 7vo = -720 (3) 7 − 9  v 1   280  from (2) and (3), 6 − 7  v  =  − 720   o    7 −9 ∆= = −49 + 54 = 5 6 −7 280 − 9 7 280 ∆1 = = −8440 , ∆2 = = −6720 − 720 − 7 6 − 720
  18. ∆1 − 8440 ∆ − 6720 v1 = = = −1688, vo = 2 = − 1344 V ∆ 5 ∆ 5 Io = -5.6 A Chapter 3, Solution 31 1Ω + v0 – v1 v2 2v0 v3 2Ω i0 1A 10 V + 4Ω 1Ω 4Ω – At the supernode, v1 v 2 v1 − v 3 1 + 2v0 = + + (1) 4 1 1 But vo = v1 – v3. Hence (1) becomes, 4 = -3v1 + 4v2 +4v3 (2) At node 3, v2 10 − v 3 2vo + = v1 − v 3 + 4 2 or 20 = 4v1 + v2 – 2v3 (3) v3 At the supernode, v2 = v1 + 4io. But io = . Hence, 4 v2 = v1 + v3 (4) Solving (2) to (4) leads to, v1 = 4 V, v2 = 4 V, v3 = 0 V.
  19. Chapter 3, Solution 32 5 kΩ v3 10 V 20 V –+ +– v1 v2 + + loop 1 v3 v1 loop 2 10 kΩ 12 V + – – – 4 mA (b) (a) We have a supernode as shown in figure (a). It is evident that v2 = 12 V, Applying KVL to loops 1and 2 in figure (b), we obtain, -v1 – 10 + 12 = 0 or v1 = 2 and -12 + 20 + v3 = 0 or v3 = -8 V Thus, v1 = 2 V, v2 = 12 V, v3 = -8V. Chapter 3, Solution 33 (a) This is a non-planar circuit because there is no way of redrawing the circuit with no crossing branches. (b) This is a planar circuit. It can be redrawn as shown below. 4Ω 3Ω 5Ω 12 V + 2Ω – 1Ω
  20. Chapter 3, Solution 34 (a) This is a planar circuit because it can be redrawn as shown below, 7Ω 2Ω 1Ω 3Ω 6Ω 10 V + 5Ω – 4Ω (b) This is a non-planar circuit. Chapter 3, Solution 35 30 V + 20 V + – – + i1 i2 v0 4 kΩ 2 kΩ 5 kΩ – Assume that i1 and i2 are in mA. We apply mesh analysis. For mesh 1, -30 + 20 + 7i1 – 5i2 = 0 or 7i1 – 5i2 = 10 (1) For mesh 2, -20 + 9i2 – 5i1 = 0 or -5i1 + 9i2 = 20 (2) Solving (1) and (2), we obtain, i2 = 5. v0 = 4i2 = 20 volts.
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