Bài giải mạch P4

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Bài giải mạch P4

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  1. Chapter 4, Solution 1. 1Ω i 5Ω io + 1V 8Ω 3Ω − 1 1 8 (5 + 3) = 4Ω , i = = 1+ 4 5 1 1 io = i= = 0.1A 2 10 Chapter 4, Solution 2. 1 6 (4 + 2) = 3Ω, i1 = i 2 = A 2 1 1 io = i1 = , v o = 2i o = 0.5V 2 4 5Ω i1 4Ω io i2 1A 8Ω 6Ω 2Ω If is = 1µA, then vo = 0.5µV Chapter 4, Solution 3. R 3R io 3R 3R + + 3R 1.5R + R 1V Vs vo − − − (b) (a)
  2. (a) We transform the Y sub-circuit to the equivalent ∆ . 3R 2 3 3 3 3 R 3R = = R, R + R = R 4R 4 4 4 2 vs vo = independent of R 2 io = vo/(R) When vs = 1V, vo = 0.5V, io = 0.5A (b) When vs = 10V, vo = 5V, io = 5A (c) When vs = 10V and R = 10Ω, vo = 5V, io = 10/(10) = 500mA Chapter 4, Solution 4. If Io = 1, the voltage across the 6Ω resistor is 6V so that the current through the 3Ω resistor is 2A. 2A 2Ω 2Ω 1A 3A 3A i1 + 3Ω 6Ω 4Ω Is 2Ω 4Ω Is v1 − (a) (b) vo 3 6 = 2Ω , vo = 3(4) = 12V, i1 = = 3A. 4 Hence Is = 3 + 3 = 6A If Is = 6A Io = 1 Is = 9A Io = 6/(9) = 0.6667A
  3. Chapter 4, Solution 5. 2Ω v1 3Ω vo + Vs 6Ω 6Ω 6Ω − 1 If vo = 1V, V1 =   + 1 = 2V 3 2 10 Vs = 2  + v1 = 3 3 10 If vs = vo = 1 3 3 Then vs = 15 vo = x15 = 4.5V 10 Chapter 4, Solution 6 R2 R3 RT Let RT = R2 // R3 = , then Vo = Vs R2 + R3 RT +R1 R2 R3 V RT R2 + R3 R2 R3 k= o = = = Vs RT + R1 R2 R3 R1 R2 + R2 R3 + R3 R1 + R1 R2 + R3
  4. Chapter 4, Solution 7 We find the Thevenin equivalent across the 10-ohm resistor. To find VTh, consider the circuit below. 3Vx 5Ω 5Ω + + 4V 15 Ω VTh - 6Ω - + Vx - From the figure, 15 V x = 0, VTh = (4) = 3V 15 + 5 To find RTh, consider the circuit below: 3Vx 5Ω 5Ω V1 V2 + 4V 15 Ω 1A - 6Ω + Vx - At node 1, 4 − V1 V V − V2 = 3V x + 1 + 1 , V x = 6 x1 = 6  → 258 = 3V2 − 7V1 (1) 5 15 5 At node 2,
  5. V1 − V2 1 + 3V x + =0 → V1 = V2 − 95 (2) 5 Solving (1) and (2) leads to V2 = 101.75 V 2 V V 9 RTh = 2 = 101.75Ω, p max = Th = = 22.11 mW 1 4 RTh 4 x101.75 Chapter 4, Solution 8. Let i = i1 + i2, where i1 and iL are due to current and voltage sources respectively. 6Ω i2 i1 6Ω 4Ω 5A + 4Ω 20V − (a) (b) 6 20 i1 = (5) = 3A, i 2 = = 2A 6+4 6+4 Thus i = i1 + i2 = 3 + 2 = 5A Chapter 4, Solution 9. Let i x = i x1 + i x 2 where i x1 is due to 15V source and i x 2 is due to 4A source, 12 Ω i ix1 ix2 -4A + 10 Ω 40Ω 12Ω 10Ω 40Ω 15V − (a) (b)
  6. For ix1, consider Fig. (a). 10||40 = 400/50 = 8 ohms, i = 15/(12 + 8) = 0.75 ix1 = [40/(40 + 10)]i = (4/5)0.75 = 0.6 For ix2, consider Fig. (b). 12||40 = 480/52 = 120/13 ix2 = [(120/13)/((120/13) + 10)](-4) = -1.92 ix = 0.6 – 1.92 = -1.32 A p = vix = ix2R = (-1.32)210 = 17.43 watts Chapter 4, Solution 10. Let vab = vab1 + vab2 where vab1 and vab2 are due to the 4-V and the 2-A sources respectively. 3vab1 3vab2 10 Ω 10 Ω +− +− + + + 4V vab1 2A vab2 − − − (a) (b) For vab1, consider Fig. (a). Applying KVL gives, - vab1 – 3 vab1 + 10x0 + 4 = 0, which leads to vab1 = 1 V For vab2, consider Fig. (b). Applying KVL gives, - vab2 – 3vab2 + 10x2 = 0, which leads to vab2 = 5 vab = 1 + 5 = 6 V
  7. Chapter 4, Solution 11. Let i = i1 + i2, where i1 is due to the 12-V source and i2 is due to the 4-A source. 6Ω io i1 + 2Ω 3Ω 12V − (a) i2 4A ix2 4A 6Ω 2Ω 3Ω 2Ω 2Ω (b) For i1, consider Fig. (a). 2||3 = 2x3/5 = 6/5, io = 12/(6 + 6/5) = 10/6 i1 = [3/(2 + 3)]io = (3/5)x(10/6) = 1 A For i2, consider Fig. (b), 6||3 = 2 ohm, i2 = 4/2 = 2 A i = 1+2 = 3A Chapter 4, Solution 12. Let vo = vo1 + vo2 + vo3, where vo1, vo2, and vo3 are due to the 2-A, 12-V, and 19-V sources respectively. For vo1, consider the circuit below. 2A 2A 5Ω 4Ω io 5 Ω + vo1 − + vo1 − 6Ω 3Ω 12 Ω 5Ω
  8. 6||3 = 2 ohms, 4||12 = 3 ohms. Hence, io = 2/2 = 1, vo1 = 5io = 5 V For vo2, consider the circuit below. 6Ω 5Ω 4Ω 6Ω 5Ω + vo2 − + + vo2 − + 3Ω 12 Ω + 3Ω 3Ω 12V 12V v1 − − − 3||8 = 24/11, v1 = [(24/11)/(6 + 24/11)]12 = 16/5 vo2 = (5/8)v1 = (5/8)(16/5) = 2 V For vo3, consider the circuit shown below. 5Ω 4Ω 5Ω 4Ω + vo3 − + vo3 − + + 2Ω + 6Ω 12 Ω 19V 12 Ω v2 3Ω − − 19V − 7||12 = (84/19) ohms, v2 = [(84/19)/(4 + 84/19)]19 = 9.975 v = (-5/7)v2 = -7.125 vo = 5 + 2 – 7.125 = -125 mV Chapter 4, Solution 13 Let io = i1 + i2 + i3 , where i1, i2, and i3 are the contributions to io due to 30-V, 15-V, and 6-mA sources respectively. For i1, consider the circuit below.
  9. 1 kΩ 2 kΩ 3 kΩ + i1 30V - 4 kΩ 5 kΩ 3//5 = 15/8 = 1.875 kohm, 2 + 3//5 = 3.875 kohm, 1//3.875 = 3.875/4.875 = 0.7949 kohm. After combining the resistors except the 4-kohm resistor and transforming the voltage source, we obtain the circuit below. i1 30 mA 4 kΩ 0.7949 k Ω Using current division, 0.7949 i1 = (30mA) = 4.973 mA 4.7949 For i2, consider the circuit below. 1 kΩ 2 kΩ 3 kΩ i2 - 15V 4 kΩ 5 kΩ + After successive source transformation and resistance combinations, we obtain the circuit below: 2.42mA i2 4 kΩ 0.7949 k Ω Using current division, 0.7949 i2 = − (2.42mA) = −0.4012 mA 4.7949
  10. For i3, consider the circuit below. 6mA 1 kΩ 2 kΩ 3 kΩ i3 4 kΩ 5 kΩ After successive source transformation and resistance combinations, we obtain the circuit below: 3.097mA i3 4 kΩ 0.7949 k Ω 0.7949 i3 = − (3.097mA) = −0.5134 mA 4.7949 Thus, io = i1 + i2 + i3 = 4.058 mA Chapter 4, Solution 14. Let vo = vo1 + vo2 + vo3, where vo1, vo2 , and vo3, are due to the 20-V, 1-A, and 2-A sources respectively. For vo1, consider the circuit below. 6Ω 4Ω 2Ω + + vo1 3Ω − 20V − 6||(4 + 2) = 3 ohms, vo1 = (½)20 = 10 V
  11. For vo2, consider the circuit below. 6Ω 6Ω 4V 4Ω 2Ω 4Ω 2Ω −+ + + 1A vo2 3Ω vo2 3Ω − − 3||6 = 2 ohms, vo2 = [2/(4 + 2 + 2)]4 = 1 V For vo3, consider the circuit below. 6Ω 2A 2A 4Ω 2Ω 3Ω + vo3 3Ω 3Ω − − vo3 + 6||(4 + 2) = 3, vo3 = (-1)3 = -3 vo = 10 + 1 – 3 = 8 V Chapter 4, Solution 15. Let i = i1 + i2 + i3, where i1 , i2 , and i3 are due to the 20-V, 2-A, and 16-V sources. For i1, consider the circuit below. io + 20V 1Ω − i1 4Ω 2Ω 3Ω
  12. 4||(3 + 1) = 2 ohms, Then io = [20/(2 + 2)] = 5 A, i1 = io/2 = 2.5 A For i3, consider the circuit below. + 1Ω 4Ω 2Ω i3 vo ’ − 16V 3Ω + − 2||(1 + 3) = 4/3, vo’ = [(4/3)/((4/3) + 4)](-16) = -4 i3 = vo’/4 = -1 For i2, consider the circuit below. 1Ω 2A 1Ω 2A 2Ω (4/3)Ω i2 4Ω i2 3Ω 3Ω 2||4 = 4/3, 3 + 4/3 = 13/3 Using the current division principle. i2 = [1/(1 + 13/2)]2 = 3/8 = 0.375 i = 2.5 + 0.375 - 1 = 1.875 A p = i2R = (1.875)23 = 10.55 watts
  13. Chapter 4, Solution 16. Let io = io1 + io2 + io3, where io1, io2, and io3 are due to the 12-V, 4-A, and 2-A sources. For io1, consider the circuit below. io1 4Ω 3Ω 2Ω + 10 Ω 5Ω 12V − 10||(3 + 2 + 5) = 5 ohms, io1 = 12/(5 + 4) = (12/9) A For io2, consider the circuit below. 4A io2 3Ω 2Ω 4Ω 10Ω 5Ω i1 2 + 5 + 4||10 = 7 + 40/14 = 69/7 i1 = [3/(3 + 69/7)]4 = 84/90, io2 =[-10/(4 + 10)]i1 = -6/9 For io3, consider the circuit below. io3 3Ω 2Ω i2 4Ω 10 Ω 5Ω 2A 3 + 2 + 4||10 = 5 + 20/7 = 55/7 i2 = [5/(5 + 55/7)]2 = 7/9, io3 = [-10/(10 + 4)]i2 = -5/9 io = (12/9) – (6/9) – (5/9) = 1/9 = 111.11 mA
  14. Chapter 4, Solution 17. Let vx = vx1 + vx2 + vx3, where vx1,vx2, and vx3 are due to the 90-V, 6-A, and 40-V sources. For vx1, consider the circuit below. 30 Ω 10 Ω 20 Ω + − vx1 + 60 Ω 30 Ω 90V − io 10 Ω + − vx1 3A 20 Ω 12 Ω 20||30 = 12 ohms, 60||30 = 20 ohms By using current division, io = [20/(22 + 20)]3 = 60/42, vx1 = 10io = 600/42 = 14.286 V For vx2, consider the circuit below. 10 Ω i ’ 10 Ω i ’ o o + vx2 − + vx2 − 30 Ω 60 Ω 6A 30 Ω 20 Ω 20 Ω 6A 12 Ω io’ = [12/(12 + 30)]6 = 72/42, vx2 = -10io’ = -17.143 V For vx3, consider the circuit below. 10 Ω 10 Ω 10 Ω io” + − + − vx3 vx3 30 Ω 60 Ω 30 Ω + 20 Ω 7.5Ω 40V 4A − io” = [12/(12 + 30)]2 = 24/42, vx3 = -10io” = -5.714 vx = 14.286 – 17.143 – 5.714 = -8.571 V
  15. Chapter 4, Solution 18. Let ix = i1 + i2, where i1 and i2 are due to the 10-V and 2-A sources respectively. To obtain i1, consider the circuit below. 2Ω 10i1 i1 1Ω i1 1Ω 2Ω +− + 5i1 4Ω + 4Ω 10V 10V − − -10 + 10i1 + 7i1 = 0, therefore i1 = (10/17) A For i2, consider the circuit below. 10i2 i2 1Ω 2Ω i io 1Ω 2Ω o +− +− 10i2 4Ω + 4Ω 2A 2V − -2 + 10i2 + 7io = 0, but i2 + 2 = io. Hence, -2 + 10i2 +7i2 + 14 = 0, or i2 = (-12/17) A vx = 1xix = 1(i1 + i2) = (10/17) – (12/17) = -2/17 = -117.6 mA Chapter 4, Solution 19. Let vx = v1 + v2, where v1 and v2 are due to the 4-A and 6-A sources respectively. ix v1 ix v2 + + 2Ω 4A 8Ω v1 2Ω 6A 8Ω v2 −+ − −+ − 4ix 4ix (a) (b)
  16. To find v1, consider the circuit in Fig. (a). v1/8 = 4 + (-4ix – v1)/2 But, -ix = (-4ix – v1)/2 and we have -2ix = v1. Thus, v1/8 = 4 + (2v1 – v1)/8, which leads to v1 = -32/3 To find v2, consider the circuit shown in Fig. (b). v2/2 = 6 + (4ix – v2)/8 But ix = v2/2 and 2ix = v2. Therefore, v2/2 = 6 + (2v2 – v2)/8 which leads to v2 = -16 Hence, vx = –(32/3) – 16 = -26.67 V Chapter 4, Solution 20. Transform the voltage sources and obtain the circuit in Fig. (a). Combining the 6-ohm and 3-ohm resistors produces a 2-ohm resistor (6||3 = 2). Combining the 2-A and 4-A sources gives a 6-A source. This leads to the circuit shown in Fig. (b). i i 2A 6Ω 2Ω 3Ω 4A 2Ω 2Ω 6A (a) (b) From Fig. (b), i = 6/2 = 3 A Chapter 4, Solution 21. To get io, transform the current sources as shown in Fig. (a). io 6Ω 3Ω i + + + 6Ω 3Ω vo 2 A − 12V − 6V 2 A − (a) (b)
  17. From Fig. (a), -12 + 9io + 6 = 0, therefore io = 666.7 mA To get vo, transform the voltage sources as shown in Fig. (b). i = [6/(3 + 6)](2 + 2) = 8/3 vo = 3i = 8 V Chapter 4, Solution 22. We transform the two sources to get the circuit shown in Fig. (a). 5Ω 5Ω − 4Ω 10Ω + 10V 2A (a) i 1A 10Ω 4Ω 10Ω 2A (b) We now transform only the voltage source to obtain the circuit in Fig. (b). 10||10 = 5 ohms, i = [5/(5 + 4)](2 – 1) = 5/9 = 555.5 mA
  18. Chapter 4, Solution 23 If we transform the voltage source, we obtain the circuit below. 8Ω 10 Ω 6Ω 3Ω 5A 3A 3//6 = 2-ohm. Convert the current sources to voltages sources as shown below. 10 Ω 8Ω 2Ω + + 10V 30V - - Applying KVL to the loop gives − 30 + 10 + I (10 + 8 + 2) = 0  → I = 1A p = VI = I 2 R = 8 W
  19. Chapter 4, Solution 24 Convert the current source to voltage source. 16 Ω 1Ω 4Ω + 5Ω + 48 V 10 Ω Vo - + - 12 V - Combine the 16-ohm and 4-ohm resistors and convert both voltages sources to current Sources. We obtain the circuit below. 1Ω 2.4A 20 Ω 5Ω 2.4A 10 Ω Combine the resistors and current sources. 20//5 = (20x5)/25 = 4 Ω , 2.4 + 2.4 = 4.8 A Convert the current source to voltage source. We obtain the circuit below. 4Ω 1Ω + + 19.2V Vo 10 Ω - - Using voltage division, 10 Vo = (19.2) = 12.8 V 10 + 4 + 1
  20. Chapter 4, Solution 25. Transforming only the current source gives the circuit below. 9Ω 18 V −+ + 5Ω 12V − i − 4Ω 30 V vo + + − +− 2Ω 30 V Applying KVL to the loop gives, (4 + 9 + 5 + 2)i – 12 – 18 – 30 – 30 = 0 20i = 90 which leads to i = 4.5 vo = 2i = 9 V Chapter 4, Solution 26. Transform the voltage sources to current sources. The result is shown in Fig. (a), 30||60 = 20 ohms, 30||20 = 12 ohms 10 Ω + vx − 3A 30Ω 60Ω 6A 30Ω 20Ω 2A (a) 20 Ω 10 Ω 12 Ω + vx − + + − 60V i − 96V (b)
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