1. The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). Problems With Assistance Module 6 – Problem 3 Filename: PWA_Mod06_Prob03.ppt Next slide Go straight to the Problem Statement Go straight to the First Step

2. Overview of this Problem In this problem, we will use the following concepts: Defining Equations for Capacitors RC Natural Response Next slide Go straight to the Problem Statement Go straight to the First Step

3. Textbook Coverage The material for this problem is covered in your textbook in the following sections: Circuits by Carlson: Sections #.# Electric Circuits 6 th Ed. by Nilsson and Riedel: Sections #.# Basic Engineering Circuit Analysis 6 th Ed. by Irwin and Wu: Section #.# Fundamentals of Electric Circuits by Alexander and Sadiku: Sections #.# Introduction to Electric Circuits 2 nd Ed. by Dorf: Sections #-# Next slide

4. Coverage in this Module The material for this problem is covered in this module in the following presentations: DPKC_Mod06_Part01 DPKC_Mod06_Part02 DPKC_Mod06_Part03 DPKC_Mod06_Part04 Next slide

5. Problem Statement Next slide The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]).

6. Solution – First Step – Where to Start? How should we start this problem? What is the first step? Next slide The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]).

7. Problem Solution – First Step How should we start this problem? What is the first step? a)Find the Thevenin equivalent seen by the capacitor.Find the Thevenin equivalent seen by the capacitor b)Find the Norton equivalent seen by the capacitor.Find the Norton equivalent seen by the capacitor c)Define the capacitive voltage.Define the capacitive voltage d)Redraw the circuit for t < 0.Redraw the circuit for t < 0 e)Redraw the circuit for t > 0.Redraw the circuit for t > 0 The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]).

8. Your Choice for First Step – Find the Thevenin equivalent seen by the capacitor The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). This is not a good choice. It is not even clear what this means. There are two switching events, and as a result there may well be a different equivalent for each of the three time periods, t 10[ms]. If we choose to find an equivalent, it will only be when we have identified a time period when no switching taking place. Please go back and try again.try again

9. Your Choice for First Step – Find the Norton equivalent seen by the capacitor The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). This is not a good choice. It is not even clear what this means. There are two switching events, and as a result there may well be a different equivalent for each of the three time periods, t 10[ms]. If we choose to find an equivalent, it will only be when we have identified a time period when no switching taking place. Please go back and try again.try again

10. Your Choice for First Step – Define the capacitive voltage This is the best choice for the first step. In almost all of these RC and RL circuits with switching, it is wise to solve first for the capacitive voltage or inductive current. This prevents problems finding initial conditions after switching take place, since these quantities cannot change instantaneously. Let’s define the capacitive voltage.define the capacitive voltage The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]).

11. Your Choice for First Step – Redraw the circuit for t < 0 This is not the best choice for the first step. We will probably be doing this very shortly. However, in almost all of these RC and RL circuits with switching, it is wise to solve first for the capacitive voltage or inductive current. To find the capacitive voltage, we need to define the polarity for the voltage first. Please go back and try again.try again The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]).

12. Your Choice for First Step – Redraw the circuit for t > 0 This is not the best choice for the first step. We will probably be doing this later in the problem. However, in almost all of these RC and RL circuits with switching, it is wise to solve first for the capacitive voltage or inductive current. To find the capacitive voltage, we need to define the polarity for the voltage first. Please go back and try again.try again The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]).

13. Defining the Capacitive Voltage The polarity that we choose for this does not matter. However, choosing a polarity does matter. The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). We have defined the capacitive voltage. What should the second step be? a)Solve for the current i X (t) in terms of the capacitive voltage.Solve for the current i X (t) in terms of the capacitive voltage b)Redraw the circuit for t < 0.Redraw the circuit for t < 0 c)Apply source transformations.Apply source transformations d)Redraw the circuit for t > 0.Redraw the circuit for t > 0

14. Your Choice for Second Step – Solve for the current i X (t) in terms of the capacitive voltage The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). This is not a good choice for the second step. It is not even clear what this would mean, since the current i X is not even attached to the capacitive voltage until after switch SW2 closes at t = 10[ms]. The key here is to draw the circuit for a given time period, where the switches are all in known positions, and then solve. Please go back and try again.try again

15. Your Choice for Second Step – Redraw the circuit for t < 0 The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). This is good choice for the second step, and the one that we will take. We want to draw the circuit for a given time period, where the switches are all in known positions, and then solve for quantities of interest. The quantity of interest here would be the initial value of the capacitive voltage, at t = 0. Redrawing for t < 0 helps us find that, since that voltage will not change when switch SW1 opens, since it is a capacitive voltage. Let’s go ahead and redraw the circuit.redraw the circuit

16. Your Choice for Second Step – Apply source transformations The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). This is not a good choice for the second step. We could do this, but we would be getting ahead of ourselves. The key here is to draw the circuit for a given time period, where the switches are all in known positions, and then solve. Please go back and try again.try again

17. Your Choice for Second Step – Redraw the circuit for t > 0 The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). This is good choice for the second step, but is not the one that we will take. We want to draw the circuit for a given time period, where the switches are all in known positions, and then solve for quantities of interest. One quantity of interest here would be the initial value of the capacitive voltage, at t = 0. Redrawing for t 0. Let’s go ahead and redraw the circuit.redraw the circuit

18. Redrawing the Circuit for t < 0 The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). We have redrawn the circuit for t < 0 in the circuit at right, above. Note that switch SW1 was closed during this time, and switch SW2 was open. Since switch SW2 was open, the circuit to the right of that switch has no effect, and has been omitted here. Note also that the capacitive voltage v C has been labeled as v C (0), since it must be the same value with the switching at t = 0. Let’s find v C (0).find v C (0) The capacitor has been redrawn as an open circuit, since the circuit was in this condition for a long time, and had the opportunity to reach a steady-state condition.

19. Solving for v C (0) The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). To solve for v C (0), we recognize that the current through the 3.3[k  ] resistor is 2[mA], and write Next Slide

20. What is the Third Step? The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). We have found the initial condition we wanted. What should be the third step? a)Redraw for t > 0.Redraw for t > 0 b)Redraw for 0 < t < 10[ms].Redraw for 0 < t < 10[ms] c)Redraw for t > 10[ms].Redraw for t > 10[ms]

21. Your Choice for Third Step – Redraw for t > 0 The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). This is not a good choice for the third step. It is not even clear what this would mean, since the switch SW2 closes at t = 10[ms]. The key here is to draw the circuit for a given time period, where the switches are all in known positions, and then solve. The time period given here does not meet this criterion, since switch SW2 moves in this time period. Please go back and try again.try again

22. Your Choice for Third Step – Redraw for 0 < t < 10[ms] The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). This is the best choice for the third step. This is the time period for the next time of interest. During this time period, no switches close or open. Note that in this time period, both switches are open. Let’s redraw for this time period.redraw for this time period

23. Your Choice for Third Step – Redraw for t > 10[ms] The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). This is a good choice for the third step, but it is not the choice that we will choose. During this time period, no switches close or open. However, it just seems reasonable to work through the problem chronologically, and so the time period 0 < t < 10[ms] makes sense as the next period to consider. In addition, the initial condition for the last time period will be obtained from the time period 0 < t < 10[ms]. Let’s redraw for this time period.redraw for this time period

24. Redrawing the Circuit for 0 < t < 10[ms] The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). We have redrawn the circuit for 0 < t < 10[ms] in the circuit at right, above. Note that switch SW1 and switch SW2 are open during this time. Since switch SW2 was open, the circuit to the right of that switch has no effect, and has been omitted here. Since switch SW1 was open, the circuit to the left of that switch has no effect, and has been omitted here. Let’s find v C (t) for 0 < t < 10[ms].find v C (t) for 0 < t < 10[ms] The capacitor has been redrawn as a capacitor, since it the circuit has not had a chance to reach a steady-state condition. In fact, as we shall see, this kind of a circuit is a special case. It is not one of our six basic circuits, so does not have an exponential response.

25. Solution for 0 < t < 10[ms] The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). The key at this point is to recognize that there is no current through the capacitor for this time period. As a result, the voltage across the capacitor cannot change. Remember that Next Slide

26. Solution for 0 < t < 10[ms], Note 1 The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). The key at this point is to recognize that there is no current through the capacitor for this time period. As a result, the voltage across the capacitor cannot change. Remember that Note that we use the symbol  in this solution, since it is for a capacitive voltage. A capacitive voltage up to the time of switching, must be valid at the time of switching, too, since it cannot change instantaneously. Next Slide

27. Solution for 0 < t < 10[ms], Note 2 The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). Some students become so fixated on RL and RC exponential responses that they begin to assume that all circuits fit into this response. Note that when we derived the exponential response equation, we did so only for six circuits, and for circuits that reduce to one of those six circuits. This circuit does not reduce to any of those. In these cases, it is necessary to go back to the basic defining equations for inductors and capacitors, which is what we did to get the solution below. Next Slide

28. What is the Fourth Step? The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). We have found the initial condition we wanted. What should be the fourth step? a)Solve for i X (t) for 0 < t < 10[ms].Solve for i X (t) for 0 < t < 10[ms] b)Replace the capacitor with an open circuit for t > 10[ms].Replace the capacitor with an open circuit for t > 10[ms] c)Replace the capacitor with a voltage source for t > 10[ms].Replace the capacitor with a voltage source for t > 10[ms] d)Redraw for t > 10[ms].Redraw for t > 10[ms]

29. Your Choice for Fourth Step – Solve for i X (t) for 0 < t < 10[ms] The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). This is not a good choice for the fourth step. We could certainly solve for i X (t) for 0 10[ms], since this current can change instantaneously when switch SW2 closes at t = 10[ms]. Please go back and try again.try again

30. Your Choice for Fourth Step – Replace the capacitor with an open circuit for t > 10[ms] The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). This is not a good choice for the fourth step. This would not be a valid step. For t > 10[ms], we have no reason to believe that the circuit will be in a steady-state condition. In fact, it will not be. Only when nothing is changing can the capacitor be replaced by an open circuit. Please go back and try again.try again

31. Your Choice for Fourth Step – Replace the capacitor with a voltage source for t > 10[ms] The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). This is not a good choice for the fourth step. This would not be a valid step. For t > 10[ms], the capacitor does not behave like a voltage source. A voltage source holds its voltage, independent of the current through it. For a capacitor, the voltage is a function of the integral of the current. In some solution techniques, a capacitor is considered as a voltage source, but only under very specific conditions. We believe that this can be misleading if not understood completely, and recommend that this not be done at this point. Please go back and try again.try again

32. Your Choice for Fourth Step – Redraw the circuit for t > 10[ms] The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). This is the best choice for the fourth step. We need to have the circuit that is valid after the next set of switching. This will allow us to solve for the capacitive voltage for t > 10[ms], and in addition, for i X (t) for t > 10[ms]. Let’s redraw the circuit for that time period.redraw the circuit

33. Redrawing the Circuit for t > 10[ms] The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). Here we have the circuit for the third time period. We should note that if we were to take the Thevenin equivalent with respect to the capacitor, we would have a natural response circuit. Note that there is no independent source, so that Thevenin voltage, as seen by the capacitor, would be zero. We already have the initial condition for this time period, so all we need is the Thevenin resistance, which will give us the time constant. Let’s find this resistance.find this resistance

34. Finding the Thevenin Resistance The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). We want to find the Thevenin equivalent with respect to the capacitor. We have a dependent source present, so it is advisable to apply a test source, in place of the capacitor. Note that we have no independent sources present, so we do not need to set any independent sources equal to zero. Let’s apply a test source.apply a test source

35. Applying a Test Source The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). We have applied a test current source, with a value of 1[A]. To solve for v T, we can apply KCL to the top node, and get Next, we will use KVL to find v T.use KVL

36. Writing KVL Around Loop The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). Writing KVL around the loop, we get Next, we will use this resistance to find the time constant, .find the time constant

37. Finding the Time Constant The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). Now, with this, we have the time constant, as Finally, we have everything we need to find v C (t).find v C (t)

38. Finding v C (t) The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). We can substitute what we know into the equation for v C (t), Finally, we have everything we need to find i X (20[ms]).find i X (20[ms])

39. Finding i X (20[ms]) The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). We use KCL to solve for the current through the 2.2[k  ] resistor in terms of i X. Then, using KVL, we can write an expression for i X (20[ms]). We get our solution on the next slide.next slide

40. The Solution for i X (20[ms]) The switch SW2 was open for a long time before t = 0. The switch SW1 was closed for a long time, and then opened at t = 0. Then, switch SW2 closed at t = 10[ms]. Find i X (20[ms]). Solving, we have Go to the Comments Slide Comments Slide

41. How do I know whether a solution will be exponential or not? The key is to look at the circuit, and determine whether it can be reduced, through equivalent circuits, to one of the six circuits that we described as having single time constant, exponential solutions. However, in the special cases, we will get a clue if we try to use this solution technique when it doesn’t work. If we had solved for the R EQ during the time 0 < t < 10[ms], we would have seen that R EQ = infinity, which would give an infinite time constant. This is not possible, and is a clue that we have chosen the wrong approach. Go back to Overview slide. Overview