1. Chapter 7 In chapter 6, we noted that an important attribute of inductors and capacitors is their ability to store energy In this chapter, we are going to determine the currents and voltages that arise when energy is either released or acquired by an inductor or capacitor in response to an abrupt change in a dc voltage or current Energy acquired by a capacitor Energy released by a capacitor In this chapter, we will focuses on circuits that consist only of sources, resistors, and either (but not both) inductors or capacitors Such configurations are called RL (resistor-inductor) and RC (resistor-capacitor) circuits

2. Our analysis of the RL and RC circuits will be divided into three phases: First Phase we consider 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 and allowed to discharge through a resistor The currents and voltages that arise in this configuration are referred to as the natural response of the circuit to emphasize that the nature of the circuit itself, not external sources excitation determine its behavior you can think of this as when a tank of water is opened suddenly, will the water in the tank disappear instantaneously in zero second or will it takes some time no matter how small to empty the tank

3. Second Phase we consider the currents and voltages that arise when energy is being acquired by an inductor or capacitor due to the sudden application to a dc voltage or current source you can think of this as when a tank of water is being filled suddenly, will the water in the tank rise instantaneously in zero second or will it takes some time no matter how small for the water to rise in the tank This response is referred to as the step response

4. Third Phase The process for finding both the natural response ( First Phase ) and step response ( Second Phase ) is the same thus in the third phase we will develop a general method that can be used to find the response of RL and RC circuit to any abrupt change in a dc voltage or current

5. 7.1 The Natural Response of an RL Circuits Let the circuit shown which contain an inductor is shown Suppose the switch has been in a closed position for a long time we will define long time later For the time being long time All currents and voltages have reached a constant value Only constant or DC currents can exist in the circuit just prior to the switch’s being opened The inductor appears as a short circuit

6. Because the voltage across the inductive branch is zero There can be no current in either R 0 or R All the source current I S appears in the inductive branch Finding the natural response requires finding the voltages and currents at any branch in the circuit after the switch has been opened.

7. If we let t = 0 denote the instant when the switch is opened The problem become one of finding v (t) and i (t) ( or i and v ) for t ≥ 0 For t ≥ 0 the circuit become

8. Deriving the Expression for the current KVL around the loop This is a first order differential equation because it contains terms involving the ordinary derivative of the unknown di/dt. The highest order derivative appearing in the equation is 1.Hence the first order We can still describe the equation further. Since the coefficients in the equation R and L are constant that is not functions of either the dependent variable i or the independent variable t Thus the equation can also be described as an ordinary differential equation with constant coefficients

9. To solve the differential equation we processed as follows: Integrating both side to obtain explicit expression for i as a function of t Here t 0 = 0 Based on the definition of the natural logarithm

10. Since the current through an indicator can not change abruptly or instanteously Were I 0 is the initial current on the inductor just before the switch opened ( or in some cases closed) In the circuit above I 0 = I s Therefore Which shows that the current start from initial value I 0 and decreases exponentially toward zero as t increases

11. Note the voltage v (t) here is defined for t > 0 because the voltage across the resistor R is zero for t < 0 ( all the current I s was going through L and zero current through R) We derive the voltage across the resistor form direct application of Ohm’s law Note the voltage v (t) across the resistor and across the indicator can change instanousley or have a jump

12. We derive the power dissipated in the resistor or Whichever form is used, the resulting expression can be reduce to Note the current i (t) through the resistor is zero for t < 0

13. We derive the energy delivered to the resistor The energy deliver to the resistor is after the switch is opened because before that there was no current passing through the resistor and the voltage across it was zero

14. Note that just before the switch is opened the current on the indicator is I S = I 0 The initial energy stored in the indictor

15. The significant of the time constant Example Seconds The coefficient at the exponential term namely R/L determine the rate at witch the exponential term in the current approaches zero The reciprocal of this ratio (L/R ) is the time constant which has the units of seconds the time constant for the R,L circuit is It is convenient to think of the time elapsed after switching in terms of integral multiple of  After one time constant, the maximum of the current I 0 has dropped to 37% of its value

16. Table showing the value of exponential e  t  for integral multiples of  Note that when the time elapsed after switching exceed five time constants The current is less than 1% of its initial value I 0 Thus sometime we say that five time constant after switching has occurred, the currents and voltages have for most practical purposes reached their final values Thus with 1% accuracy along time Five or more time consent

17. The existence of the current in the RL circuit is momentary event and is referred to as transient response of the circuit The response that exist a long time after the switching has taken place is called steady-state response The steady-state response in the RL circuit is zero, that were the i (t) will go to as t goes to infinity

18. Determining the time constant The time constant can be determined as follows (1) If the RL circuit can be put as Were L is an equivalent inductor and R is an equivalent resistor Seen by the equivalent indictor Example

19. (2) If we know the differential equation  is the inverse of this constant Example Suppose the differential equation is given as

20. (2) If a trace or graph of i (t) is given Differentiating i (t) we have

26. We need to Find v (t) Since the current through an inductor is given as

27. The circuit reduce to Finding v (t)   

30. =

31. The voltage across the capacitor =

32. For t >= 0 KCL on the upper junction

36. Solving for t

37. For the circuit shown above, the switch was in position a for a long time, then at time t=0 the switch move to position b. Find the voltage v(t) for t > 0 ? Solution

42. We need to Find i(t) Similar to the previous example 7.2, the voltage across a capacitor is given as Question: how to find i(t) ?

43. We need to Find i(t) Question: how to find i(t) ? Do we find i(t) through the capacitor voltage The answer is no of course because we do not know However we can find i(t) through the resistor 250 k  as we need to find v(t) as will be shown next

44. The circuit reduce to The solution of the voltage v(t) is given as were

45. The circuit reduce to Note the direction of the current i (t) and the initial voltages and the polarity of

46. The circuit reduce to

48. d) Show that the total energy delivered to the 250 kW resistor is the difference between the results in (b) and (c) (a) (b) (c) Comparing the results obtained in (b) and in (c)

50. 7.3 Step Response of RL and RC Circuits We are going to find the currents and voltages in 1 st order RL and RC circuits when a DC voltage or current is suddenly applied. The Step Response of an RL Circuit The switch is closed at t = 0, the task is to find the expressions for the current in the circuit and for the voltage across the inductor after the switch has been closed

51. After the switch has being closed, we have We can solve the differential equation similar to what we did previously with the natural response by separating the variables i and t and integrating as follows

52. Were I 0 is the current at t = 0 and i(t) is the current at any t > 0 Performing the Integrating and the substitution of the limits When no initial current on the inductor I 0 = 0 Integrating both side to obtain explicit expression for i as a function of t We know separate the variables i and t

53. The equation for the no initial current (I 0 = 0) indicate that the after the switch is closed the current will increase exponentially to its final value of V s / R When no initial current on the inductor I 0 = 0 When initial current on the inductor I 0 ≠ 0

54. At one time constant t = L / R the current will be The current will reached 63% of Its final value When no initial current on the inductor I 0 = 0

55. The rate of change of i(t) or di(t)/dt current will be The rate of change of i(t) or di(t)/dt at t = 0 ( The tangent at t=0 ) will be Which when drawn on the plot of i(t) will be as Here  = L/R therefore the slop of the tangent at t =0 is

56. The voltage across the inductor will be Before the switch close the voltage across the inductor is zero Just after the switch close the voltage across the inductor will jump to When initial current on the inductor I 0 ≠ 0 If the initial current on the inductor I 0 = 0, the current will be and the voltage across the inductor will jump to Note the inductor voltage can jump however the current is not allowed to jump because initial current on the inductor I 0. This will make the voltage drop across the resistor after the switch was closed at t = 0 + to be RI 0 → voltage drop across the inductor is (V s -RI 0 )

57. When no initial current on the inductor I 0 = 0 When initial current on the inductor I 0 ≠ 0 When no initial current on the inductor I 0 = 0

59. Solving the differential equation when the initial current on the inductor I 0 ≠ 0, we have What we want to do next is to find the indictor current i(t) without finding the differential equation or solving it Let us look at the the indictor current i(t) The constant part which is the steady state value of the current or the value of the current at t = ∞ ( i(∞) ) initial current on the inductor i(0) reciprocal of the time constant 

60. You find it after you move the switch and the inductor in the DC state or is short You find it before you move the switch and the inductor in the DC state or is short The time constant  You find it after you move the switch and it is

61. Example 7.5The switch in the circuit shown has been in position a for a long time At t = 0, the switch moves from position a to position b. (a) Find the expression for i(t) for t >0. Inductor is short For t < 0.

62. Inductor is short

63. We need  ( the time constant) We look at the circuit t > 0 We deactivate independent sources ( leave dependent) We now find the equivalent resistant seen by the indictor

64. The time constant

65. The Step Response of an RC Circuit The switch is closed at t = 0, the task is to find the expressions for the voltage across And the current through the capacitor after the switch has been closed

66. After the switch has being closed, we have We can solve the differential equation similar to what we did previously with the step response for RL circuit by separating the variables v and t and integrating we obtain Were V 0 is the initial voltage at the capacitor

67. Let us look at the the capacitor voltage current v(t) The steady state value of the voltage v C (∞) Initial voltage on the capacitor v(0) The time constant 

69. The current through the capacitor will be When initial voltage on the capacitor V 0 ≠ 0 Before the switch close the current through the capacitor is zero ( open circuit ) Just after the switch close the current through the capacitor will jump to because initial voltage on the capacitor V 0. This will make the current through the resistor after the switch was closed at t = 0 + to be V 0 /R → current through the capacitor is (I s  V 0 /R) Note the capacitor current can jump however the voltage is not allowed to jump

70. Summary of Step Response of RL and RC Circuits because initial current on the inductor I 0. This will make the voltage drop across the resistor after the switch was closed at t = 0 + to be RI 0 → voltage drop across the inductor is (V s -RI 0 ) because initial voltage on the capacitor V 0. This will make the current through the resistor after the switch was closed at t = 0 + to be V 0 /R → current through the capacitor is (I s -V 0 /R)

71. Example 7.11

72. For t < 0 both switches are closed causing the 150 mH inductor to short the 18  resistor Using source transformations, we find that i L (0  ) = 6 A For 0 ≤ t ≤ 35 ms For t ≤ 0 Switch 1 is open ( switch 2 is closed) The 60 V voltage source and 4  and 12  are disconnected from the circuit i L (0  ) = 6 A

73. For 0 ≤ t ≤ 35 ms Switch 1 is open ( switch 2 is closed) The 60 V voltage source and 4  and 12  are disconnected from the circuit The indictor is no longer behaving as a short circuit because the DC source is no longer in The circuit The 18  resistor is no longer short-circuited For t ≤ 0 For 0 ≤ t ≤ 35 ms i L (0  ) = 6 A

74. For 0 ≤ t ≤ 35 ms t ≥ 35 ms i L (0  ) = 6 A For t ≤ 0 We find i L (0.035  ) from the circuit before For 0 ≤ t ≤ 35 ms

75. t ≥ 35 ms We find i L (0.035  ) from the circuit before For 0 ≤ t ≤ 35 ms

76. i L (0  ) = 6 A For 0 ≤ t ≤ 35 ms t ≥ 35 ms

78. Example 7.11 Unbounded Response

79. The Integrating Amplifier