Problems With Assistance Module 1 – Problem 6

1. Problems With AssistanceModule 1 – Problem 6 Filename: PWA_Mod01_Prob06.ppt This problem is adapted from: Exam #1, Problem 2 – ECE 2300 – February 26, 1994 Department of Electrical and Computer Engineering University of Houston Houston, TX, 77204-4793 Go straight to the First Step Go straight to the Problem Statement Next slide

2. Overview of this Problem In this problem, we will use the following concepts: • Kirchhoff’s Voltage Law • Kirchhoff’s Current Law • Ohm’s Law • Power and Sign Conventions Go straight to the First Step Go straight to the Problem Statement Next slide

3. Textbook Coverage The material for this problem is covered in your textbook in the following sections: • Circuits by Carlson: Chapter 1 • Electric Circuits 6th Ed. by Nilsson and Riedel: Chapters 1 & 2 • Basic Engineering Circuit Analysis 6th Ed. by Irwin and Wu: Chapters 1 & 2 • Fundamentals of Electric Circuits by Alexander and Sadiku: Chapters 1 & 2 • Introduction to Electric Circuits 2nd Ed. by Dorf: Chapters 1 - 3 Next slide

4. Coverage in this Module The material for this problem is covered in this module in the following presentations: • DPKC_Mod01_Part02, DPKC_Mod01_Part03, and DPKC_Mod01_Part04. Next slide

5. Problem Statement • Find the power absorbed by the following elements in the circuit given here. • The dependent current source. • The independent current source. • The 200[W] resistor. Next slide

6. Solution – First Step – Where to Start? • Find the power absorbed by the following elements in the circuit given here. • The dependent current source. • The independent current source. • The 200[W] resistor. How should we start this problem? What is the first step? Next slide

7. Problem Solution – First Step • Find the power absorbed by the following elements in the circuit given here. • The dependent current source. • The independent current source. • The 200[W] resistor. • How should we start this problem? What is the first step? • Write a KCL equation for every node • Write a KVL equation for every loop • Define variable names for voltages and currents needed • Redraw the circuit to show power relationships • Convert all voltages and currents to the passive convention

8. Your choice for First Step –Write a KCL equation for every node • Find the power absorbed by the following elements in the circuit given here. • The dependent current source. • The independent current source. • The 200[W] resistor. This is not a good choice. We will certainly be writing KCL equations for this problem, but there are six nodes, and we do not need to write KCL for each of these nodes. One important habit to begin developing is to approach problems efficiently. In these kinds of problems that means solving only for the things we need for the solution requested. Go back and try again.

9. Your choice for First Step –Write a KVL equation for every loop • Find the power absorbed by the following elements in the circuit given here. • The dependent current source. • The independent current source. • The 200[W] resistor. This is not a good choice. We will certainly be writing KVL equations for this problem, but there are six loops, just following elements. We do not need to write KVL for each of these loops. One important habit to begin developing is to approach problems efficiently. In these kinds of problems that means solving only for the things we need for the solution requested. Go back and try again.

10. Your choice for First Step –Define variable names for voltages and currents needed • Find the power absorbed by the following elements in the circuit given here. • The dependent current source. • The independent current source. • The 200[W] resistor. This is the best choice given here. We want to find the power absorbed by each of the current sources. To get power for a current source, we need to find the current and voltage for that source. The current is set by the value of the current source, so we need to define two voltages, one for each current source. The polarity for each is arbitrary. For the 200[W] resistor, we need either the current through it, or the voltage across it. Conveniently, the current through it is already defined. Let’s define the voltages across the current sources.

11. Your choice for First Step –Redraw the circuit to show power relationships • Find the power absorbed by the following elements in the circuit given here. • The dependent current source. • The independent current source. • The 200[W] resistor. This is not a good choice. It is not even clear what this proposed step means. One might try to redraw the circuit to place the components that absorb positive power together, and those that deliver positive power together. However, while we know that the resistors all absorb positive power, we don’t know which sources deliver positive power, and which absorb positive power. Any of the four could do either. Even if we knew that, it is not clear how to redraw the circuit to place components on this basis. Go back and try again.

12. Your choice for First Step –Convert all voltages and currents to the passive convention • Find the power absorbed by the following elements in the circuit given here. • The dependent current source. • The independent current source. • The 200[W] resistor. This is not a good choice. While it is true that the problem requests power absorbed, we do not need to have our variables in the passive convention to get this. So, while we could define new variables to achieve this goal, it is just not needed, and it takes time. We would like to be efficient in our approach. Go back and try again.

13. Define the Voltages Across the Current Sources • Find the power absorbed by the following elements in the circuit given here. • The dependent current source. • The independent current source. • The 200[W] resistor. • We have chosen to name these two voltages vX and vY. These choices were arbitrary, as were the polarities for each. Now, to get the solution, we need to find values for these two voltages, and for the current iX. Let’s consider this next. It would be nice to pick a simple place to start. Are there any of these three variables that we can find in one step? • vX • vY • iX

14. Can We Find vX in One Step? • Find the power absorbed by the following elements in the circuit given here. • The dependent current source. • The independent current source. • The 200[W] resistor. Can we find vX in one step? If you look at the circuit, we can write a KVL around the loop marked in red. However, the value of the dependent voltage source depends on iY, which is unknown. So this will take more than one step. Go back and try again.

15. Can We Find vY in One Step? • Find the power absorbed by the following elements in the circuit given here. • The dependent current source. • The independent current source. • The 200[W] resistor. Can we find vY in one step? If you look at the circuit, we can write a KVL around the loop marked in red. However, the value of the dependent current source, which dictates the voltage across the resistor, depends on iX, which is unknown. So this will take more than one step. Go back and try again.

16. Can We Find iX in One Step? • Find the power absorbed by the following elements in the circuit given here. • The dependent current source. • The independent current source. • The 200[W] resistor. Can we find iX in one step? Yes, we can. If you look at the circuit, we can write a KVL around the loop marked in red. The equation that results depends only on iX, which means we can solve for it directly. Let’s write the equation.

17. Writing KVL for the Loop in Red to Get iX Let’s write a KVL equation, using the loop marked in red, going clockwise starting at the lower left of the loop. The equation that results depends only on iX, which means we can solve for it directly. Note that the term iXR3is the voltage across the R3 resistor. The polarity of this voltage is shown in the diagram for clarity. The polarity of this voltage is determined by the the reference polarity for the current through this resistor, and by the definition of Ohm’s Law. Let’s solve.

18. Solving for iX Let’s solve for iX. Next, we can use this to find vY. As we noted in an earlier slide, we can find vY if we have iX. Let’s write that KVL next.

19. Writing KVL to get vY Let’s write a KVL equation, using the loop marked in red, going clockwise starting at the lower left of the loop. The equation that results depends on iX and vY, which means we can solve for it. Note that iS2is equal to 2.5 times iX which we solved for already (it was –40[mA]). Let’s solve this for vY now.

20. Solving for vY Solving for vY, we get The last variable that we wanted to solve for was vX. Let’s write an equation for that.

21. Let’s write a KVL equation, using the loop marked in red, going clockwise starting at the lower left of the loop. The equation that results depends on vX and iY, which means we will need another equation as well. We write a KCL equation for the node labeled A to get that needed equation. Writing KVL to get vX Note that the voltage across the R4resistor is determined by the current through it, which in this case the value of the iS1current source. Be careful of the polarity for this voltage, as always. Let’s solve for vX.

22. Let’s solve for vX , Solving for vX Now we have everything that we said that we needed to find the powers that we were asked for. Let’s solve for the requested powers.

23. The Power Absorbed by the Dependent Current Source • Find the power absorbed by the following elements in the circuit given here. • The dependent current source. • The independent current source. • The 200[W] resistor. • The power absorbed by the dependent current source, labeled iS2, will be the current times the voltage. Which expression is correct for this? • pabs,iS2= -vY iS2 • pabs,iS2= vY iS2 • pdel,iS2= vY iS2 • pdel,iS2= -vY iS2

24. The Power Absorbed by the Dependent Current SourceOption 1) • Find the power absorbed by the following elements in the circuit given here. • The dependent current source. • The independent current source. • The 200[W] resistor. Option 1) was correct. The power absorbed by the dependent current source, labeled iS2, will be pabs,iS2= -vY iS2. The voltage and current reference polarities are in the active convention, and this follows. Thus, the answer for part a), the power absorbed by the dependent current source is +700[mW]. Go to part b).

25. The Power Absorbed by the Dependent Current SourceOption 2) • Find the power absorbed by the following elements in the circuit given here. • The dependent current source. • The independent current source. • The 200[W] resistor. Option 2), pabs,iS2= vY iS2, was not correct. The power absorbed by the dependent current source, will not be vY iS2. The voltage and current reference polarities are in the active convention. Go back and try again.

26. The Power Absorbed by the Dependent Current SourceOption 3) • Find the power absorbed by the following elements in the circuit given here. • The dependent current source. • The independent current source. • The 200[W] resistor. Option 3), pdel,iS2= vY iS2, is a valid expression for the power delivered by this source, since the voltage and current reference polarities are in the active convention. However, we were asked for the power absorbed. It is correct to find this, and then take the negative of this result to get the answer. However, this adds another step, so we suggest that you go back and try again.

27. The Power Absorbed by the Dependent Current SourceOption 4) • Find the power absorbed by the following elements in the circuit given here. • The dependent current source. • The independent current source. • The 200[W] resistor. Option 4), pdel,iS2= -vY iS2, was not correct. The power delivered by the dependent current source, will not be -vY iS2. The voltage and current reference polarities are in the active convention. Go back and try again.

28. The Power Absorbed by the Independent Current Source • Find the power absorbed by the following elements in the circuit given here. • The dependent current source. • The independent current source. • The 200[W] resistor. The power absorbed by the independent current source, labeled iS1, will be pabs,iS1= -vX iS1. The voltage and current reference polarities are in the active convention, and this follows. Thus, the answer for part b), the power absorbed by the independent current source is –86.32[W]. Go to part c).

29. The Power Absorbed by the 200[W] Resistor • Find the power absorbed by the following elements in the circuit given here. • The dependent current source. • The independent current source. • The 200[W] resistor. The power absorbed by the 200[W] resistor will be pabs,R2= iX2R2. Thus, the answer for part c), the power absorbed by the 200[W] resistor is 320[mW]. Go to Notes

30. Why do we have to worry about all these signs and directions? • A major issue here is the sign of the answer, and then signs of the individual terms in the equations along the way. Indeed, this is one of the challenging issues in this course. • The idea in all these problems is that the sign matters. The goal is not only to get the answer right, but to have the sign associated with the magnitude also be correct. To do this, we need a system that will always work for us. Once possible approach to getting the signs right has been given in this module. Go back to Overviewslide. Go to Next Notes Slide

31. Isn’t this problem pretty hard? • When this problem was given as an exam problem early in the semester, some students felt that it was a fairly difficult problem. Many of these students would have had no difficulty later in the course. However, the beginning students often have to work on this kind of problem. • One reason for this is that the first step to take is not immediately obvious. However, once we recognize that we can get the current iX pretty easily, and then get the current iY from that, the problem is much easier to complete. We need to learn how to look at such problems and see how to start them. Go back to Overviewslide.