1. RL and RC circuits first- order response Electric circuits ENT 161/4

2. RL and RC circuit original response A first-order circuit is characterized by a first-order differential equation. This circuit contain resistor and capacitor or inductor in one close circuit.

3. The natural response of a circuit refers to the behaviour ( in terms of voltages and currents) of the circuit itself, with no external sources of excitation. RL circuit: circuit that have resistor and inductor. RC circuit: circuit that have resistor and capacitor.

4. Natural response RC circuit

5. Consider these three condition : 1. At initially, t=0 -, switch doesn’t change for some time 2. At initial, t=0 +, switch doesn’t change for some time 3. At final condition, t→∞, switch doesn’t change for some time

6. Known t ≤ 0, v(t) = V 0. voltage  Therefore t ≥ 0:

7. For t > 0,

8. Natural response RC circuit graph

9. This show that the voltage response of the RC circuit is an exponential decay of the initial voltage. constant, τ = RC

10. Constant τ define how fast voltage reach stable condition :

11. Natural response RL circuit

12. Consider these three condition : 1. At initially, t=0 -, switch doesn’t change for some time 2. At initial, t=0 +, switch doesn’t change for some time 3. At final condition, t→∞, switch doesn’t change for some time

13. Known at t ≤ 0, i(t) = I 0  Therefore t > 0, Current

14. For t > 0,

15. EXAMPLE Switch in circuit for some time before open at t=0. Calculate a) I L (t) at t ≥ 0 b) I 0 (t) at t ≥ 0+ c) V 0 (t) at t ≥ 0+ d) Total energy percentage that stored in inductor 2H that absorb by 10Ω resistor.

17. Answer a) Switch close for some time until t=0, known voltage at inductor should be zero at t = 0 -. Therefore, initial current at inductor was 20A at t = 0-. Thus i L (0 + ) also become 20A, because immidiate changes for current didn’t exist in inductor.

18. Equivalent resistance from inductor and constant time

19. Therefore, current i L (t)

20. b) Current at resistor 40Ω could be calculate by using current divider law,

21. This current was at t ≥ 0 + because i 0 = 0 at t = 0 -. Inductor will become close circuit when switch open immediately and produce changes immediately at current i 0. Therefore,

22. c) V 0 could be calculate by using Ohm’s Law,

23. d) Total power absorb by 10Ω resistor

24. Total energy absorb by 10Ω resistor

25. Initial energy stored at 2H inductor

26. Therefore, energy percentage that absorb by 10Ω resistor

27. Step response RC circuit The step response of a circuit is its behaviour when the excitation is the step function, which may be a voltage or a current source.

28. Consider these three condition : 1. At initially, t=0 -, switch doesn’t change for some time 2. At initial, t=0 +, switch doesn’t change for some time 3. At final condition, t→∞, switch doesn’t change for some time

29. Known at t ≤ 0, v(t)=V 0 For t > 0, voltan

30. Current for step response RC circuit

31. Then, for t >0 Where

32. V f = Force voltage or known as steady-state response V n = known as transient response is the circuit’s temporary response that will die out with time.

33. Step response RC circuit graph force Natural total

34. Step Response RL circuit

35. Consider these three condition : 1. At initially, t=0 -, switch doesn’t change for some time 2. At initial, t=0 +, switch doesn’t change for some time 3. At final condition, t→∞, switch doesn’t change for some time

36. known i(t)=I 0 at t ≤ 0.  For t > 0, Current

37. Finally,

38. Question Switch in those circuit was at x position for some time. At t=0, switch move to position y immediately. Calculate, (a) Vc(t) at t ≥ 0 (b) V 0 (t) at t ≥ 0+ (c) i 0 (t) at t ≥ 0+ (d) Total energy absorb by 60kΩ resistor.

40. Answer (a) Constant for circuit  V C (0)=100V  equivalent resistor = 80kΩ.

41. Then, V C (t) for t ≥ 0:

42. Answer (b) V 0 (t) could be calculate by using voltage divider law.

43. Answer (c) current i 0 (t) can be calculated by using ohm’s law

44. Answer (d) Power absorb by 60kΩ resistor

45. Total energy

46. Second-order RLC circuit RLC circuit : circuit that contain resistor, inductor and capacitor Second-order response : response from RLC circuit Type of RLC circuit: 1. RLC series circuit 2. RLC parallel circuit

47. Original response for parallel RLC circuit

48. Take total current flows out from node

49. differential of t,

50. Take

51. Characteristic equation known as zero :

52. The root of the characteristic equation are

53. Response for RLC parallel circuit

54. The root of the characteristic equation are where:

55. summarize ParameterTerminologyValue in natural response s 1, s 2 characteristic equation α frequency Neper resonant radian frequency

56. Roots solution s 1 and s 2 depend on α and Consider these cases saperately: 1. If < α, voltage response was overdamped 2. If > α, voltage response was underdamped 3. If = α, voltage response was critically damped

57. Overdamped voltage response Solution for overdamped voltage

58. constant A1 and A2 can be determined from the initial conditions v(0+) and Known,

59. Here v(0+) = V0 and initial value for dv/dt was

60. Solution for overdamped natural response, v(t) : 1. Calculate characteristic equation, s1 and s2, using R, L and C value. 2. Calculate v(0+) and using circuit analysis.

61. 3. Calulate A1 and A2 by solve those equation 4. Insert s1, s2, A1 and A2 value to calculate overdamped natural response for t ≥ 0.

62. Example for overdamped natural response for v(0) = 1V and i(0) = 0

63. Underdamped voltage response At > α2, root of the characteristic equation was complex number and those response called underdamped.

64. Therefore ωd : damped radian frequency

65. underdamped voltage response for RLC parallel circuit was

66. constant B1 and B2 was real number. Solve those two linear equation to calculate B1 and B2,

67. Example for underdamped voltage response for v(0) = 1V and i(0) = 0

68. Critically Damped Voltage Response Second-order circuit was critically damped when = α. When circuit was critically damped, two characterictic root equation was real and same,

69. Solution for voltage Linear equation to calculate D1 and D2 value

70. Example for critically damped voltage response at v(0) = 1V and i(0) = 0

71. Step response RLC parallel circuits

72. From Kirchhoff current law

73. Known Therefore

74. Have,

75. There are two solution to solve the equation, direct approach and indirect approach.

76. Indirect approach From Kirchhoff’s current law:

77. Differential

78. Depend on characteristic equation root :

79. Insert in Kirchhoff’s current law eq:

80. Direct approach It’s simple to calculate constant for the equation directly by using initial value response function.

81. Constant of the equation could be calculate from and

82. The solution for a second-order differential equation with a constant forcing function equals the forced response plus a response funtion identical in form to the natural response.

83. If and Vf represent the final value of the response function. The final value may be zero,

84. Natural response for RLC Series circuit The procedures for finding the natural or step responses of a series RLC circuit are the same as those used to find the natural or step responses of a parallel RLS circuit, because both circuits are described by differential equations that have the same form.

85. RLC series circuit

86. Summing the voltages around the closed path in the circuit,

87. differential

88. Characteristic equation for RLC series circuit

89. Characteristic equation root @

90. Neper frequency (α) for RLC series circuit and resonant radian frequency was,

91. Current response Overdamped Underdamped critically damped

92. Three kind of solution

93. Step response for RLC series circuit The procedures for finding the step responses of series RLC circuit are the same as those used to find the step response of a parallel RLC circuit.

94. RLC series circuit

95. Using Kirchhoff’s voltage law,

96. Current known as,

97. Differential for current

98. Insert in Voltage current law equation

99. Three solution that possibly for v C

100. Contoh 1 Tenaga awal yang disimpan oleh litar berikut adalah sifar. Pada t = 0, satu punca arus DC 24mA diberikan kepada litar. Nilai untuk perintang adalah 400Ω. 1. Apakah nilai awal untuk i L ? 2. Apakah nilai awal untuk ? 3. Apakah punca-punca persamaan ciri? 4. Apakah ungkapan numerik untuk iL(t) pada t ≥ 0?

102. Jawapan 1. Tiada tenaga yang disimpan dalam litar sebaik sahaja punca arus digunakan, maka arus awal bagi induktor adalah sifar. Induktor mencegah perubahan yang serta-merta pada arus induktor, oleh itu i L (0)=0 sebaik sahaja suis dibuka.

103. 2. Nilai awal voltan kapasitor adalah sifar sebelum suis dibuka, oleh itu ia akan sifar sebaik sahaja suis dibuka. Didapati: maka

104. 3. Dari elemen-elemen dalam litar, diperolehi

105. Oleh kerana, maka punca-punca persamaan ciri adalah nyata

106. 4. sambutan arus induktor adalah overdamped dan persamaan penyelesaian adalah

107. Dua persamaan serentak:

108. Penyelesaian numerik:

109. Contoh 2 Tiada tenaga disimpan dalam inductor 100mH atau kapasitor 0.4µF apabila suis di dalam litar berikut ditutup. Dapatkan v C (t) untuk t ≥ 0.

111. Jawapan Punca-punca persamaan ciri:

112. Punca-punca adalah kompleks, maka sambutan voltan adalah underdamped. Oleh itu, diperolehi voltan v C :

113. Pada awalnya, tiada tenaga tersimpan dalam litar, maka:

114. Selesaikan untuk dan

115. penyelesaian untuk v C (t)