1. Find the Norton Equivalent as seen by resistor R 7. Problems With Assistance Module 4 – Problem 3 Filename: PWA_Mod04_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: Equivalent Circuits Norton’s Theorem 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 presentation: DPKC_Mod04_Part03 Next slide

5. Problem Statement Next slide Find the Norton Equivalent as seen by resistor R 7.

6. Solution – First Step – Where to Start? How should we start this problem? What is the first step? Next slide Find the Norton Equivalent as seen by resistor R 7.

7. Problem Solution – First Step How should we start this problem? What is the first step? a)Define the open- circuit voltage.Define the open- circuit voltage b)Label the terminals of resistor R 7 and remove it.Label the terminals of resistor R 7 and remove it c)Define the short- circuit current.Define the short- circuit current d)Combine resistors R 2 and R 3 in parallel.Combine resistors R 2 and R 3 in parallel e)Combine resistors R 6 and R 7 in parallel.Combine resistors R 6 and R 7 in parallel Find the Norton Equivalent as seen by resistor R 7.

8. Your choice for First Step – Define the open-circuit voltage This is not a good choice for the first step. The key here is that we are finding an equivalent seen by something; when we do this, the first step should always be to get rid of that something. Here, we need to get rid of R 7 before doing anything else. Remember that in circuits, a component or device does not “see” itself. We suggest that you go back and try again.try again Find the Norton Equivalent as seen by resistor R 7.

9. Your choice for First Step – Label the terminals of resistor R 7 and remove it This is the best choice for this problem. The key in these kinds of problems where we have been asked for an equivalent “as seen by” something, is to remove the something as the first step. The something is assumed to not “see itself”, and thus it needs to be removed. Its terminals become the place where the equivalent is found. Failure to do this can lead to errors. Let’s go ahead and remove R 7.remove R 7 Find the Norton Equivalent as seen by resistor R 7.

10. Your choice for First Step – Define the short-circuit current This is could be the first step, since R 7 will not affect the short circuit current. However, it is a dangerous choice. The key is get rid of R 7 right away, and then we don’t have to worry about whether it has an effect or not. So, let’s go back and try again. try again Find the Norton Equivalent as seen by resistor R 7.

11. Your choice for First Step was – Combine resistors R 2 and R 3 in parallel This is a valid step, but is not a good choice for the first step. The key here is that we are finding the equivalent seen by R 7. Thus, we need to handle R 7 right at the beginning. Therefore, although R 2 and R 3 are in parallel, we recommend that you go back and try again.try again Find the Norton Equivalent as seen by resistor R 7.

12. Your choice for First Step was – Combine resistors R 6 and R 7 in parallel This is not a good choice. The problem here is that we are finding the equivalent as seen by R 7. If we combine R 6 and R 7 in parallel, then R 7 will be gone, and we will not be able to complete the problem correctly. Please go back and try again.try again Find the Norton Equivalent as seen by resistor R 7.

13. Label the terminals of resistor R 7 and remove it We have named the terminals of R 7, and then removed resistor R 7 from the circuit. We named the terminals A and B. Let’s go to the next slide and consider what to solve for first.next slide Next slide Find the Norton Equivalent as seen by resistor R 7.

14. What Should We Solve for First? We need to consider what to solve for first. Click on your choice from the choices below. The open-circuit voltage.open-circuit voltage The short-circuit current.short-circuit current The equivalent resistance.equivalent resistance The total power dissipated.total power dissipated Find the Norton Equivalent as seen by resistor R 7.

15. You Chose: The open-circuit voltage You said that the first thing to solve for would be the open-circuit voltage. This solution is not the easiest available to us. For example, this would have five essential nodes, and would require four simultaneous equations. There are better choices. Please go back and try again.try again Find the Norton Equivalent as seen by resistor R 7.

16. You Chose: The short-circuit current You said that the first thing to solve for would be the short-circuit current. This is a good choice. Once we apply the short circuit, the R 6 resistor could be neglected, we would have four essential nodes, and three equations, one of which would be very simple. Let’s find the short-circuit current.find the short-circuit current Find the Norton Equivalent as seen by resistor R 7.

17. You Chose: The equivalent resistance You said that the first thing to solve for would be the equivalent resistance. This is a good choice. However, while this will be the next thing we will choose, there is another thing that must be found as well; it will be either the open-circuit voltage, or the short- circuit current. Go back, and determine which you think would be better.which you think would be better Find the Norton Equivalent as seen by resistor R 7.

18. You Chose: The total power dissipated You said that the first thing to solve for would be the total power dissipated. This is not a good choice. The total power dissipated is not useful to us at this point. Go back and try again. try again Find the Norton Equivalent as seen by resistor R 7.

19. Finding the Short-Circuit Current Let’s find the short-circuit current. The first step is to define the polarity. That is done in the circuit shown here. We have also defined some voltages to help us get this value. Find the Norton Equivalent as seen by resistor R 7. Next slide

20. Writing the KCL’s (The Node-Voltage Method) We wish to find v C and v D first, and then use those values to get i SC. We write the equations here: Find the Norton Equivalent as seen by resistor R 7. Solving for v C and v D we get This simplifies to yield Next slide

21. Finding i SC Using the values for v C and v D that are shown here, we can find i SC. We write a KCL for the red dashed closed surface: Find the Norton Equivalent as seen by resistor R 7. Solving for i SC we get Next slide

22. Next Step We have found one of the three possible quantities. The best choice for the next quantity to find is probably the equivalent resistance. To find this, we set the independent sources equal to zero, which will simplify the circuit a great deal. Let’s do this in the next slide.next slide Find the Norton Equivalent as seen by resistor R 7. Next slide

23. Finding the Equivalent Resistance, Step 1 To find the equivalent resistance, we set the independent sources equal to zero. With this we have the circuit below. We note that R 4 and R 5 are in series, and R 2 and R 3 are in parallel. Let’s combine these and redraw.combine these and redraw Find the Norton Equivalent as seen by resistor R 7. next slide Next slide

24. Finding the Equivalent Resistance, Step 2 Having combined these resistors, it should be clear that the series combination of R 1 and R 9 is in parallel with R 8, which is in parallel with R 6. Note that each is connected between terminals A and B. Thus we have the equivalent resistance, as seen by A and B, as Find the Norton Equivalent as seen by resistor R 7. Next slide

25. The Norton Equivalent The Norton equivalent is a current source equal to the short-circuit current, in parallel with the equivalent resistance. Thus, we have as the answer the circuit drawn below. Find the Norton Equivalent as seen by resistor R 7. Check this solution Go back to Problem Statement

26. Checking the Solution Let’s check our answer to see if it works. Let’s find the current through the R 7 resistor for the Norton equivalent. We will call this current i X. We have Next slide

27. Checking the Solution, Original Circuit Let’s check our answer to see if it works. Let’s find the current through the R 7 resistor for the original circuit. Again, we will call this current i X. We will also define node voltages, to allow us to solve. We have the circuit below. Next slide

28. Checking the Solution, Original Circuit Equations The node-voltage equations are: The solutions are: Using this, we can find i X, which is This is the same answer as before. Go to Comments Slide

29. Was This Worth It? This is a good question. Again, the best answer is, “It depends.” We have gone through a fair amount of work, but by doing so we have a simpler circuit. Whether it was worth the work depends on what we were going to use the circuit for. For example, if we were to connect the circuit to 12 different resistors, or to 12 different current sources, it would be much easier to solve the simpler circuit each time, and in the end it would be worth it. For one resistor, it was probably not a good use of our time. Note, though, that Norton’s Theorem also has benefits as a way of thinking about a circuit. This will pay off in many areas, among them when we are designing circuits. Go back to Overview slide. Overview