Problems With Assistance Module 6 – Problem 5 Filename: PWA_Mod06_Prob05.ppt For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. Problems With Assistance Module 6 – Problem 5 Go straight to the First Step You can see a brief introduction starting on the next slide, or go right to the problem. Go straight to the Problem Statement Next slide

Overview of this Problem In this problem, we will use the following concepts: RC Natural Response Equivalent Circuits Energy Stored in Capacitors Go straight to the First Step Go straight to the Problem Statement Next slide

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

Coverage in this Module The material for this problem is covered in this module in the following presentations: DPKC_Mod06_Part01 DPKC_Mod06_Part02 This is the material in this computer module that you might consult for more explanation. These are presentations of key concepts that you should find in this problem. Next slide

Problem Statement For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. This is the basic problem. We will take it step by step. Next slide

Solution – First Step – Where to Start? For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. Try to decide on the first step before going to the next slide. How should we start this problem? What is the first step? Next slide

Problem Solution – First Step For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. Click on the step that you think should be next. How should we start this problem? What is the first step? Redraw the circuit for t < 0. Define the capacitive voltage across the 10[mF] capacitor. Define the capacitive voltage across the 22[mF] capacitor. Redraw the circuit for t > 0.

Your Choice for First Step – Redraw the circuit for t < 0 For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. This is not the best choice. It is true that we will be doing this very shortly. However, it should not be done yet. The problem here is that the current iX may change instantaneously when a switch is thrown. Thus, solving for it directly can cause problems. Ask yourself, what should I solve for first, and then go back and try again.

Your Choice for First Step – Define the capacitive voltage across the 10[mF] capacitor For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. This is the best choice for the first step. Generally speaking, in switching problems with capacitors, it is wise to solve for capacitive voltages first. Here, we have more than one capacitor, and should define all of the capacitive voltages before moving forward. Since the voltage across the 22[mF] capacitor is already defined (vX), at this point we simply need to define the voltage across the 10[mF] capacitor. Let’s define this voltage.

Your Choice for First Step – Define the capacitive voltage across the 22[mF] capacitor For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. This is a generally good idea, but in this case is not a good choice for the first step. Generally, we wish to define capacitive voltages in our first step. However, this voltage is already defined, and does not need to be redefined. Please go back, and find something that needs to be done, and try again.

Your Choice for First Step – Redraw the circuit for t > 0 For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. This is not the best choice. It is true that we will be doing this eventually. It is premature to begin doing these things now, since we need initial conditions before the solution can move forward. Ask yourself, what should I solve for first, and then go back and try again.

Defining the Capacitive Voltage For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. The polarity that we choose for this does not matter. However, choosing a polarity does matter. Click on the step that you think should be next. Both capacitive voltages have been defined. What should the second step be? Redraw the circuit for t < 0, with the capacitors as short circuits. Redraw the circuit for t < 0, with the capacitors as open circuits. Redraw the circuit for t > 0 , with the capacitors as short circuits. Redraw the circuit for t > 0 , with the capacitors as open circuits. Find the initial values of the capacitor voltages from the voltage sources.

Your Choice for Second Step – Redraw the circuit for t < 0, capacitor as short circuits For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. This is not a good choice for the second step. The capacitors will not act as a short circuit for this time period. Please go back and try again.

Your Choice for Second Step – Redraw the circuit for t < 0, capacitors as open circuits For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. This is not the best choice for the second step. While we have been told that the circuit had been in this condition for a long time, since the voltage source is across the capacitor, we know the voltage across the capacitor at t = 0-, so we don’t need to perform this step. Please go back and try again.

Your Choice for Second Step – Redraw the circuit for t > 0, capacitors as short circuits For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. This is not a good choice for the second step. For t > 0, the capacitors will not be in steady state, at least, not until time has passed. Thus, we cannot use steady-state equivalents. In addition, the steady-state equivalent for a capacitor would be an open circuit, not a short circuit. Please go back and try again.

Your Choice for Second Step – Redraw the circuit for t > 0, capacitors as open circuits For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. This is not a good choice for the second step. For t > 0, the capacitors will not be in steady state, at least, not until time has passed. Thus, we cannot use steady-state equivalents. This steady-state equivalent, capacitors as open circuits, will be valid after many time constants. Please go back and try again.

Your Choice for Second Step – Find the initial values of the capacitor voltages from the voltage sources For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. This is the best choice for the second step. For t < 0, the capacitors have a voltage source across them. This sets the voltage across the capacitor. Thus, we need to do nothing except note that the voltage across the capacitors for t = 0- must equal the value for t = 0+. Let’s set these values.

Finding the Initial Values of the Capacitor Voltages For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. Since the voltage sources were across the capacitors for t < 0, we can say that Now, we want to draw the circuit for t > 0, which we will do in the next slide.

Redrawing the Circuit for t > 0 For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. We have removed the sources, since they are no longer a part of the circuit that we are interested in. What should be the third step?

What is the Third Step? For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. We have redrawn for t > 0. What should be the third step? Find the time constant for the 10[mF] capacitor. Find the time constant for the 22[mF] capacitor. Replace the series capacitors with their equivalent capacitor. Find the final value for vX(t). Click on the step that you think should be next.

Your Choice for Third Step – Find the time constant for the 10[mF] capacitor For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. This is not a good choice for the third step. Finding the time constant for a capacitor means finding the equivalent resistance seen by that capacitor. However, the 10[mF] capacitor also sees the other capacitor. This is not a condition that we can handle with our solution technique. We need to have one of the six circuits. Can we simplify to one of these? Please go back and try again.

Your Choice for Third Step – Find the time constant for the 22[mF] capacitor For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. This is not a good choice for the third step. Finding the time constant for a capacitor means finding the equivalent resistance seen by that capacitor. However, the 22[mF] capacitor also sees the other capacitor. This is not a condition that we can handle with our solution technique. We need to have one of the six circuits. Can we simplify to one of these? Please go back and try again.

Your Choice for Third Step – Replace the series capacitors with their equivalent For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. This is the best choice for the third step. By doing this, we will simplify the circuit, and have one of the six circuits that we know the solution for. At that point, we will be able to solve for the solution. Let’s make this equivalent. Note that we need to be careful about the initial condition for this equivalent circuit.

Your Choice for Third Step – Find the final value for vX(t) For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. This is not a good choice for the third step. This is, indeed, something that we are going to do eventually. However, we do not know how to get this directly at this point. It might look as though the final value will be zero, but this is not correct. With one capacitor and a resistor, the final value is zero, since it is the natural response. But, that’s not what we have here. Please go back and try again.

Replacing Series Capacitors with Equivalent For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. Next Slide We have drawn the equivalent for the series capacitors. Note that the voltage across this equivalent must be the same as was across the two capacitors, or vX - v10. Now, we have a natural response circuit, and can solve.

Finding Initial Condition and Time Constant For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. The initial condition has been solved, above, and is –38[V]. The time constant will be CEQ(2.2[kW]) = 15[ms]. Thus, we can write Note that as a natural response, the final value for vEQ is zero. Next Slide

Finding iX(t) Next Slide For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. Now, we have the voltage across the 2.2[kW] resistor, so we can write Next Slide

What is the Fourth Step? For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. Click on the step that you think should be next. We have found the solution for part a). Now, we have part b). What should be the fourth step? Use the voltage divider rule to find vX. Use vEQ to find vX. Use iX to find vX.

Your Choice for Fourth Step – Use the voltage divider rule to find vX For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. This is not a good choice for the fourth step. The problem here is that the voltage divider rule, applied to capacitors, is more complicated than for resistors, because of the initial conditions. We have not derived it in these modules. We would be better off by using things we know and understand. Let’s go back and try again.

Your Choice for Fourth Step – Use vEQ to find vX For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. This is not a good choice for the fourth step. It is not clear how to use vEQ to get vX. We do not have a voltage divider rule for capacitors, or at least we have not derived it in these modules. We would be better off by using things we know and understand. Let’s go back and try again.

Your Choice for Fourth Step – Use iX to find vX For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. This is the best choice for the fourth step. We have the current through the capacitor, which is iX(t). We can integrate that, using the initial condition, to get an expression for the voltage across the capacitor, vX(t). Let’s perform the integration.

Integrating iX to find vX For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. We integrate iX(t) to get an expression for the voltage across the capacitor, vX(t). Noting that iX and vX are in the active sign convention for the capacitor, we have Next Slide

Performing the Integration For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. Now, we have to perform the integration. It is important to take note of the units in this expression. We get Next Slide

Solution for Part b) Next Slide For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. We have the expression for vX(t). Now, we want the value after many time constants after t = 0. Note, after many time constants, the exponential expression will be negligible, and we are left with Next Slide

Solution for Part b) – Note 1 For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. This result may surprise you. Even though we had a natural response problem, with no independent sources for t > 0, we have a nonzero final value. There is a reason for this. We have two capacitors in series. With two capacitors in series, it is possible to have trapped energy. This trapped energy cannot be removed from the series combination unless we break the series combination. This is what leads to the nonzero final value. Next Slide

Solution for Part b) – Note 2 For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. In these trapped energy problems, we will find the the sum of the initial energies for the two series capacitors will be more than the initial energy for the equivalent capacitor that we found using the series equivalent. The difference between the sum for the two series capacitors, and that for the equivalent capacitor, will be the trapped energy. This is covered in several circuit analysis texts. Next Slide

Solution for Part b) – Note 3 For the circuit shown, the switches had been closed for a long time. At t = 0 both switches open. For the time interval t > 0, find the equation for the current iX(t). Find the steady-state value of the voltage vX, that is, the value many time constants after t = 0. It is also possible to have trapped energy when we have inductors in parallel. Take care in both cases. Go to the Comments Slide

What is the deal with trapped energy? Trapped energy results from the defining equations for capacitors and inductors. Most students seem to find it easy to see that some charge can get stuck between the two series capacitors, and remain there until the series combination is broken. This causes the trapped energy. It is less clear to many students what is happening when inductors are in parallel. Energy will be trapped in this case, but the concept of trapped flux linkages is harder to picture. Most people are well served by simply remembering that with series capacitors, and with parallel inductors, energy can be trapped, and the final value may not be zero. Go back to Overview slide.