# ECE2262 Electric Circuit

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1 ECE2262 Electric Circuit Chapter 7: FIRST AND SECOND-ORDER RL AND RC CIRCUITS Response to First-Order RL and RC Circuits Response to Second-Order RL and RC Circuits 1

2 2

3 7.1. Introduction 3

4 4

5 In dc steady state, a capacitor looks like an open circuit and an inductor looks like a short circuit. The voltage across a capacitor must be a continuous function of time The current flowing through an inductor must be a continuous function of time 5

6 We wish to examine the voltages and currents, instantaneous power, and energy stored on a capacitor or inductor when the capacitor or inductor is connected with resistors, sources and switches. Natural Response The respond of the circuit to the initial energy stored in the capacitor or inductor until all the energy is spent. Step Response The response of a circuit to a dc input signal (step input) 6

7 7.2. Response to First-Order RL and RC Circuits to Constant Sources 7

8 Natural Response : t = 0! The currents and voltages that arise when stored energy in an inductor or capacitor is suddenly released to a resistive network. This happens when the inductor or capacitor is abruptly disconnected from its DC source. 8

9 Step Response : t = 0 + The currents and voltages that arise when energy is being acquired by an inductor or capacitor due to the sudden application of a DC voltage or current source (initial values are zero!) 9

10 Total Response = Natural Response + Step Response Natural Response = Zero Input Response! v( 0!) Step Response = Zero-State Response (Forced Response)! v 0! ( ) = 0 10

11 The First Order Differential Equation - Constant Source dx( t) dt x t + ax( t) = A with x( t 0 ) = x 0 ( ) = x p ( t)! +! x t h particular solution ( ) natural response dx p ( t) dt + ax p ( t) = A, dx h ( t) dt + ax h ( t) = 0 11

12 Particular Solution Let x p ( t) = K 1! ak 1 = A! K 1 = A a The particular solution of dx( t) dt + ax( t) = A! dx( t) dt + ax( t) = A x p ( t ) = A a The complete solution is x( t) = A a + x ( t) h where x h ( t) is the solution of the homogenous equation 12

13 Solution of the homogenous equation: dx h dt Choose (?) the solution in the form: x h ( t) ( t) = K 2 e!t + ax h ( t) = 0! K 2!e!t + ak 2 e!t = 0! dx h ( t) dt + ax h ( t) = 0!! + a = 0!! = "a! x h ( t ) = K 2 e!at. 13

14 Total Response - the general solution of the 1 st order differential equation dx( t) dt + ax( t) = A x( t) = A a + K 2e! at! If x( 0!) is given then K 2 can be uniquely determined! x (!) = A a assuming that a > 0 14

15 x( t) a =!2 a =!1 a = 1 a = 4 t 15

16 Let x( 0!) = x 0! x( t) = A a + K 2 e!at! K 2 = x 0! A a! ( ) = A a x t ( 1! ) e!at +! x 0 e!at!# "# \$ forced response natural response x (!) = A a! a > 0 x( t) = A a ( 1! ) e!at if x 0 = 0! a > 0 16

17 A. Natural Response of RL and RC Circuits The currents and voltages that arise when stored energy in an inductor or capacitor is suddenly released to a resistive network. This happens when the inductor or capacitor is abruptly disconnected from its DC source. 17

18 A1. Natural Response of RL Circuits For t < 0 the inductor appears as a short circuit and therefore all the source current I s appears in the inductive branch, i L 0! We wish to find i t ( ) = I s, i R ( 0!) = 0 ( ) and v( t) after the switch has been opened, i.e., when the source has been disconnected and the inductor begins releasing energy. 18

19 The circuit for t! 0 KVL: L di( t) dt + Ri t ( ) = 0! di( t) dt + R L i( t) = 0! i( t) = K 2 e! R L t! i( t) = i( 0!)e! R L t! i 0 + ( ) = i 0! ( ) 19

20 Natural Response of an RL circuit i( t) = I 0 e!t/" " = L R I 0 e!t/" 20

21 The voltage across resistor v( t) = i( t)r = RI 0 e!t/" v( t) = RI 0 e!t/" v( 0!) = 0, v( 0 + ) = RI 0 21

22 The energy delivered to the resistor during any interval of time after the switch has been opened t ( ) = p z w t t! ( )dz =! Ri 2 ( z)dz 0 0 = RI 0 2 # t 0 e!2z/" dz = I 0 2 L 2 { 1! e!2t/" } As t! " the energy dissipated in the resistor approaches the initial energy 2 L stored in the inductor = I

23 Time Constant! = L R! = 1! = 0.3 i( t) = I 0 e!t/" 23

24 Graphical Interpretation of the Time Constant 24

25 Calculation of the natural response of an RL circuit 1. Find the initial current I 0 through the inductor! t < 0 2. Find the time constant of the circuit:! = L R! t > 0 3. Use i( t) = I 0 e!t/" to generate i( t) from I 0 and! 25

26 Example In the following circuit the switch has been closed for a long time before it is opened at t = 0 1. At t = 0! the voltage across inductor is zero (short circuit)! Note: i 0 ( 0!) = 0 i L ( 0!) = i L ( 0 + ) = 20A 26

27 2. t = 0 + i L The time constant! = L R eq = 2 10 = 0.2sec R eq = 2 + ( 40 10) = 10! i L ( t) = I 0 e!t/" = 20e!5t, t! 0, i L ( 0!) = 20A 27

28 3. The current in the 40! resistor i L i 0 ( t) = ! "i t L ( ( ) ) =!4e!5t, t! 0 +, i 0 ( 0!) = 0 v 0 ( t) = 40i 0 ( t) =!160e!5t, t!

29 4. Power and Energy dissipated in the 10! resistor i L ( ) 2 t p 10! ( t) = v 0 10 = 2560e!10t W, t! 0 + " w 10! = # p 10! ( z)dz = 256 J 0 The initial energy stored in the 2 H inductor: w L ( 0) = 1 2 Li 2 L ( 0 + ) = 1 2 2! 202 = 400 J.!100% = 64 % - the percentage of energy dissipated in the 10! resistor 29

30 Example In the following circuit the initial currents in inductors L 1 and L 2 have been established by sources not shown. i i 4 1. Find i 1,i 2,i 3 for t! 0! need v( t) and i( t) Equivalent Circuit: L : L 1 L 2 = 5! = 4H R eq : {( 10 15) + 4} 40 = 8! 30

31 Equivalent Circuit: I 0 The time constant! = L R eq = 4 / 8 = 0.5 sec.! i( t) = I 0 e!t/" = 12e!2t, t! 0, i( 0!) = 12! v( t) = 8i( t) = 96e!2t, t! 0 +, v( 0!) = 0 31

32 i i 4 CD: i 3 = i( t) = 12e!2t, t! 0, v( t) = 96e!2t, t > 0 + i 4 = i = i ( ) i = 4 5 i! i = i i 3 = 12 25!12e"2t = 5.76e!2t t!

33 i i 4 i( t) = 12e!2t, t! 0, v( t) = 96e!2t, t! 0 + i 1 ( t) = 1 t v( z) L 1! dz +!8 0 ( ) = 1.6! 9.6e!2t, t! 0, i 1 ( 0!) =!8 i 2 ( t) = 1 t v( z) L 2! dz +!4 0 ( ) =!1.6! 2.4e!2t, t! 0, i 2 ( 0!) =!4 33

34 2. The initial energy stored in the inductors i i 4 w = 1 2 5! ! 42 = 320 J 3. As t! " then i 1! 1.6, i 2! "1.6 Hence, a long time after the switch has been opened, the energy trapped in the two inductors is w = 1 2 5! ! ( "1.6 )2 = 32 J 34

35 4. The total energy delivered to the resistive network I 0 ( )! 8 i( t) = 12e!2t w = # i 2 z dz = 1152# e!4 z dz = 288 J! = " This is the difference between the initially stored energy (320 J) and the energy trapped in the parallel inductors ( 32 J) 0 " 35

36 i( t) i 1 ( t ) i 2 ( t) i( 0!) = 12, i 1 ( 0!) =!8, i 2 ( 0!) =!4 36

37 37

38 A2. Natural Response of RC Circuits t = 0!! the capacitor behaves as an open circuit! v( 0!) = V g t =

39 Natural Response C dv( t) dt + v( t) R = 0 with v( 0!) = v( 0 + ) = V = V g 0 v( t) = V 0 e!t/", t! 0 Time Constant:! = RC 39

40 v( t) = V 0 e! t/" " = RC 40

41 Calculation of the natural response of an RC circuit 1. Find the initial voltage V 0 across the capacitor! t < 0 2. Find the time constant of the circuit:! = RC! t > 0 3. Use v( t) = V 0 e!t/" to generate v( t) from V 0 and! 41

42 Example ( ) = 100V, t = 0 + (position y)! v C ( 0 + ) = 100V and = 80k! 1. t = 0! (position x)! v C 0! Time Constant = R eq C = ( 80!10 3 )( 0.5!10 "6 ) = 40 ms. v C ( t) = V 0 e!t/" = 100e!25t V, t! 0 42

43 2. v 0 ( t) VD: v 0 ( t) = = 48k! v C t ( ) = 60e!25t V, t! i 0 t ( ) = v ( t) 0 60k = e!25t ma, t! 0+ 43

44 4. The power and energy dissipated in the 60 k! resistor p 60 ( t) = i 2 0 ( t)! 60k = 60e!50t mw, t! 0+ ( ) w 60 = " p 60 z dz = 1.2 mj! 48 % of the initial energy The initial energy stored in the capacitor: 0! w C = 1 2 Cv 2 C = !10"6!100 2 = 2.5 mj 44

45 45

46 B. Step Response for RC and RL Circuits The currents and voltages that arise when energy is being acquired by an inductor or capacitor due to the sudden application of a DC voltage or current source 46

47 B1. Step Response of RC Circuits t = 0 R t = t 0 V s We wish to determine v t ( ) for t > 0, given the initial value v( 0!) and the values of the DC voltage source V s and the resistance R. v( 0!) - the capacitor voltage just before the switch is closed. v( 0!) = v( 0 + ) due to the continuity of the capacitor voltage 47

48 For t > 0 by KVL R V s V s = Ri( t) + v( t) and i( t) = C dv( t) V s = RC dv( t) dt + v( t) dt dv( t) dt + 1 RC v t ( ) = 1 RC V s 48

49 dv( t) dt + 1! v( t) = 1! V s,! = RC - Time Constant ( ) with respect ( ) (natural response) The first order differential equation yielding the response v t to the forcing input V s and the initial value v 0! 49

50 RC - circuit (step) response to constant source V s! page 16 v( 0) = v( 0!) ( ) = v( 0!)! V s v t "# \$ % e!t/& + V s for t! 0 v (!) = V s - steady state value Under DC conditions the capacitor becomes an open circuit, taking the full value of the DC voltage source V s 50

51 The solution ( ) = v( 0!)! V s v t "# \$ % e!t/& + V s can be rewritten ( ) = V s 1! e!t/" v t #\$ %!#" # \$ & + v!# 0! "# \$ zero-state response ( )e!t/" zero-input response! see page 16! zero-input response = natural response! zero-state response = forced response 51

52 If the switching time is not t = 0 but t = t 0 then the step response is ( ) = v( t 0!)!V s v t "# \$ % e! ( t!t 0 )/& + V s for t! t 0 Note: Important for circuits with multiple switching 52

53 Example (a) v( 0!) = 4V, V s = 2V, R = 2!, C = 0.5F! " = 1s ( ) = "# v( 0!)! V s \$ % e!t/& + V s! v t v t ( ) = 2e!t + 2, t! 0!2 t!1 (b) v( 1! ) = 4V, V s = 6V,! = 0.5s! v( t) =!2e ( ) + 6, t! 1 53

54 Example The switch is moved from position a to b at t = 0. The switch in position a has been for a long time. i 1 2! b a 2! V s i2 2! + ( ) " v t 1F 1! 2A 1. t < 0 - the capacitor is in the DC state and behaves like an open circuit, i.e., the source 2 A flows trough the 1! resistor. Hence, v( 0!) = 2 "1 = 2V 54

55 2. t > 0 i 1 2! b a 2! V s i2 2! + ( ) " v t 1F 1! 2A Thevenin equivalence: V oc = ! V s = 1 2 V s - steady state value R Th = 2 2 =1! 55

56 1! b a 2! V s s 2 + ( ) " v t 1F 1! 2A Time Constant:! = R Th " C = 1 s, v (!) = V s 2, v( 0 + ) = 2V ( ) = 2! V s v t " # \$ 2 % & ' e!t + V s 2 = 2e!t + V s 2 1! e!t "# \$ % ( ) = v( 0!)! V s v t "# \$ % e!t/& + V s 56

57 3. i 1 ( t) and i 2 ( t) for t > 0 i 1 2! b a 2! V s i2 2! + ( ) " v t 1F 1! 2A i 1 ( ) ( t) = V! v t s 2 ( ) and i 2 ( t) = v t 2 i 1 ( t) =!e!t + V s 4 1+ e!t "# \$ %, i 1 ( 0 + ) = V s 2!1 i 2 ( t) = e!t + V s 4 1! e!t "# \$ %, i 2 ( 0 + ) = 1 57

58 4. i 1 ( t) and i 2 ( t) for t < 0 i 1 2! b a 2! V s i2 2! + ( ) " v t 1F 1! 2A i 1 t ( ) = i 2 t ( ) = V s 4 58

59 5. Plot v( t), i 1 ( t), i 2 ( t) if V s = 8V 59

60 60

61 B2. Step Response of RL Circuits I s R We wish to determine i t ( ) for t > 0, given the initial value i 0! values of the DC current source I s and the resistance R. i( 0!) - the initial current just before the switch is closed. i( 0!) = i( 0 + ) due to the continuity of the inductor current ( ), the 61

62 For t > 0 by KCL I s R di( t) dt I s = v( t) R + i( t) and v( t) = L di( t) dt + 1! i( t) = 1! I s,! = L / R - Time Constant The first order differential equation yielding the response i t the forcing input I s and the initial value i 0! 62 ( ) with respect to ( ) (natural response)

63 RL - circuit (step) response to constant source V s :! page 16 i( 0) = i( 0!) ( ) = i 0! i t "# ( )! I s \$ % e!t/& + I s for t! 0 i (!) = I s - steady state value Under DC conditions the conductor becomes a short circuit, taking the full value of the DC current source I s 63

64 The solution ( ) = i( 0!)! I s i t "# \$ % e!t/& + I s can be rewritten ( ) = I s 1! e!t/" i t #\$ %!#" # \$ & + i! 0 #! "# \$ zero-state response ( )e!t/" zero-input response! see page 16! zero-input response = natural response! zero-state response = forced response 64

65 If the switching time is not t = 0 but t = t 0 then the step response is ( ) = i t 0! i t "# ( )! I s \$ % e! ( t!t 0 )/& + I s for t! t 0 Note: Important for circuits with multiple switching 65

66 66

67 B3. A General Solution for Step and Natural Responses In the RC and RL circuits the unknown variable (signal) can be determined from the general formula ( ) = x f + " x( t 0 )! x f x t # ( \$ % e! t!t 0 )/& x f - the final value (steady state) - the value of x (!) x( t 0 ) - initial value at either t 0! or t 0 + t 0 -switching time, commonly t 0 = 0! - time constant:! = RC,! = L / R 67

68 1. Identify the variable of interest for the circuit. For RC circuits, it is most convenient to choose the capacitive voltage; for RL circuits, it is best to choose the inductive current. 2. Determine the initial value of the variable, which is its value at t 0. Note that if you choose capacitive voltage or inductive current as your variable of interest, it is not necessary to distinguish between t 0! and t 0 +. This is because they both are continuous variables. If you choose another variable, you need to remember that its initial value is defined at t Calculate the final value of the variable, which is its value as t! " 4. Calculate the time constant for the circuit. 68

69 Example The switch in position a has been for a long time. 1. t = 0!! the capacitor looks like an open circuit! v C ( 0!) = "!40 ( ) =!30 V 69

70 2. t! " After the switch has been in position b for a long time the the capacitor looks like an open circuit! v C! ( ) = 90V 3. Time constant:! = RC = 400k! 0.5µF = 200!10 "3 = 0.2 s. 4. v C ( t) = 90 + [!30! 90]e!t/0.2 = 90!120e!5t, t! 0. x( t) = x (!) + x( 0 + ) " x (!) #\$ % & e "t/' 70

71 5. i( t), t! 0 + x( t) = x (!) + x( 0 + ) " x (!) #\$ % & e "t/' i 0 + ( ) = 90! v c ( 0 + ) i (!) = 0, 400k time constant:! = RC = 0.2 s = 90! (!30 ) 400k = 300 µa i( t) = 0 + [ 300! 0]e!5t = 300e!5t µa, t! 0 + Note: i t ( ) = C dv ( t) C dt gives the same result 71

72 Note: v C ( t ) = 0 if 90!120e!5t = 0! t = log! 4 \$ " # 3 % & = ms.! power p C ( t) = v C ( t)! i( t) = # % \$ &% < 0 for 0 " t " t 0 > 0 for t > t 0 72

73 Example No initial charge on the capacitor. Determine i( t) and v( t) 1. v C ( 0!) = 0! v C ( 0 + ) = 0 - no initial charge i( 0!) = 0 x( t) = x (!) + x( 0 + ) " x (!) #\$ % & e "t/' 73

74 2. Since v C ( 0 + ) = 0! i( 0 + ) = ! 7.5m = 3 ma. 3. i (!) = 0 (the capacitor is an open circuit) x( t) = x (!) + x( 0 + ) " x (!) #\$ % & e "t/' 74

75 4. Time Constant:! = R Th C! R Th = = 50k! R Th! = 50 "10 3 " 0.1"10 #6 = 5 ms 5. i( t) = 0 + [ 3! 0]e!t/5"10!3 = 3e!200t ma, t > 0 +, i( 0!) = 0 75

76 6. v( t): v( t) = v C ( t) + 30k! i( t), i( t) = 3e!200t ma v C ( 0 + ) = 0 v C ( t) = v C (!) + v C ( 0 + ) " v C (!) #\$ % & e "t/' v C (!) (the capacitor is an open circuit) = 7.5m! 20k = 150V v C ( t) = [ 0!150]e!200t = 150!150e!200t V 76

77 v( t) = v C ( t) + 30k! i( t) = 150! 60e!200t V v( 0!) = 20k " 7.5m = 150V v( 0 + ) = 20k! i 20 = 20! 30 50! 7.5m = 90V 77

78 Example 7.3. Find i( t). The circuit has been in the steady-state before the switch is closed. Note: for t! 0 + we have i( t) = v C ( t) / 6k! v C ( t) = v C (!) + v C ( 0 + ) " v C (!) #\$ % & e "t/'! 78

79 1. t = 0! v C ( 0!)! k " i( 0!) = 0 and i( 0!) = 36!12 12k = 2mA v C ( 0!) = 36! 2k " 2m = 36-4 = 32 V 79

80 2. v C (!) + v C! " ( ) v C (!) = " 36 = 27V 80

81 3. Time Constant:! = R Th C = 3 2!103!100!10 "6 = 0.15 s R Th = 2 6 = 3 2 k! 81

82 4. v C ( t) = v C (!) + v C ( 0 + ) " v C (!) #\$ % & e "t/', v C (!) = 27V, v C ( 0!) = v C ( 0 + ) = 32V,! = 0.15 s v C ( t) = e!t/0.15 V, t! 0 i( t) = v C ( t) / 6k, t! 0 + and i( 0!) = 2mA i( t) = e!t/0.15 ma, t!

83 Example Find v t ( ).The circuit has been in the steady-state before the switch closure at t = 0. 83

84 1. t = 0! 6A 4! i L 0! ( ) = " 6A = 8 3 A 84

85 2. t = 0 +! i L ( 0!) = 8 3 A = i ( 0 + ) L KCL at v 1 ( 0 + ): v ( )! 24 + v ( 0 + ) v ( 0 + ) 1 12 = 0! v 1 ( 0 + ) = 20 3 V!24 + v( 0 + ) + v 1 ( 0 + ) = 0! v( 0 + ) = 24! v 1 ( 0 + )! v( 0 + ) = 52 3 V! 85

86 3. The steady-state: v (!) = 24V 86

87 4. The Thevenin equivalent resistance and the time constant R Th = { 4 6} 12 = 2!! = L / R Th = 4 / 2 = 2s 87

88 5. x( t) = x (!) + x( 0 + ) " x (!) v t Note: v( 0!) v (!) = 24V, v( 0 + ) = 52 3 V,! = 2s " # \$ ( ) = ! 24 % & ' e!t/2 = 24! 20 3 e!t/2 V, t! 0 + #\$ % & e "t/' i 4! ( 0 ") = = 24 6 = 4A! v 0! ( ) = 4" # 4A = 16V 88

89 v( t) = 24! 20 3 e!t/2, V, t! 0 +, v( 0!) = 16V, v( 0 + ) = 52 3 V = V 89

90 90

91 91

92 92

93 93

94 Example Find i L ( t)for t! 0. The general form of the solution: i L ( t) = i L (!) + i L ( 0 + ) " i L (!) #\$ % & e "t/! time constant:! = L / R Th 94

95 1. t = 0! v 1 ( 0!) i L ( 0!) 5v 1 ( 0!) KVL:!v 1 ( 0!) +12i L ( 0!)! 5v 1 ( 0!) = 0! i L ( 0!) = 1 2 v ( 0!) (A) 1 Since v 1 ( 0!) = 10V! i L ( 0!) = 5A 2. t = 0 +! i L ( 0!) = i L ( 0 + ) = 5A! 95

96 3. t =! 4A v 1 (!) i L (!) 5v 1 (!) By source transformation we obtain the following equivalent circuit 8V 4A 2# + ( ) v 1! i L (!) " Note: Mesh Equations 96 5v 1 (!)

97 8V 4A 2# + ( ) v 1! " i L (!) 5v 1 (!) Loop 1:!v 1 (") +12i L (")! 5v 1 (") = 0! v 1 (!) = 2i L (!) Loop 2:!8 + ( 2 +12)i L (")! 5v 1 (") = 0! i L (!) = 2A! 97

98 4. R Th and! = L / R Th. We wish to find R Th using the test source approach. 4A v 1 +! I T V T R Th = V T I T 5v 1 98

99 4A v 1 +! I T V T R Th = V T I T 5v 1!v 1!12I T + V T! 5v 1 = 0 and v 1 = 2I T! V T = 24I T! V T I T = 24!! R Th = 24!!! = L / R Th = 2 24 = 1 12 s! 99

100 5. i L ( t) = i L (!) + i L ( 0 + ) " i L (!) #\$ % & e "t/! i L (!) = 2A i L ( 0 + ) = 5A! = 1 12 s i L ( t) = 2 + [ 5! 2]e!12t = 2 + 3e!12t (A), t > 0 i L ( 0!) = 5 100

101 i L ( t) = 2 + 3e!12t 101

102 102

103 Problem Switch 1 is opened at t = 0 and Switch 2 is closed at t = 0.5ms. Find the voltage v t ( ). 10V V 1 SW1 SW1 SW 2 R 1 6k! 4k! R 2 v t + ( ) C " R 3 6k! 0.25µF 20V V 2 v( t) = \$ & % '& 4e!1000t, 0 " t " 0.5 #10!3! ( t!0.5#10!3) 8! e, t > 0.5 #10!3 103

104 104

105 Sequential Switching 105

106 Unbounded Response 106

107 Example 107

108 v 0 ( t) = 10e 40t V - it grows without limit 108

109 109

110 110

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