Problems With Assistance Module 2 – Problem 5

Problems With AssistanceModule 2 – Problem 5 Filename: PWA_Mod02_Prob05.ppt Go straight to the First Step Go straight to the Problem Statement Next slide

Overview of this Problem In this problem, we will use the following concepts: • Equivalent Circuits • Series and Parallel Combinations of Resistors • Delta-to-Wye Transformations 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 chapters: • Circuits by Carlson: Chapters 2, 4 • Electric Circuits 6th Ed. by Nilsson and Riedel: Chapter 3 • Basic Engineering Circuit Analysis 6th Ed. by Irwin and Wu: Chapters 2, 10 • Fundamentals of Electric Circuits by Alexander and Sadiku: Chapter 2 • Introduction to Electric Circuits 2nd Ed. by Dorf: Chapter 3 Next slide

Coverage in this Module The material for this problem is covered in this module in the following presentations: • DPKC_Mod02_Part01. Next slide

Problem Statement Find an expression for the power delivered by the source in this circuit as a function of the value of the source, iS. Next slide

Solution – First Step – Where to Start? Find an expression for the power delivered by the source in this circuit as a function of the value of the source, iS. How should we start this problem? What is the first step? Next slide

Problem Solution – First Step Find an expression for the power delivered by the source in this circuit as a function of the value of the source, iS. • How should we start this problem? What is the best first step? • Write a series of KVL or KCL equations. • Combine resistors in series and in parallel to simplify the circuit. • Use delta-to-wye transformations to simplify this circuit. • Define currents and voltages for each of the elements in the circuit.

Your Choice for First Step –Write a Series of KVL or KCL Equations Find an expression for the power delivered by the source in this circuit as a function of the value of the source, iS. This is not the best choice for the first step. We could indeed write a set of KVL or KCL equations, once we had defined currents for the resistor elements. This would result in a set of simultaneous equations that could be solved for the voltage across the source. However, there are better approaches. We advocate an approach that allows us to avoid solving simultaneous equations. Go back and try again.

Your Choice for First Step Was –Combine Resistors in Series and in Parallel to Simplify the Circuit Find an expression for the power delivered by the source in this circuit as a function of the value of the source, iS. This is not a possible choice for the first step. The goal is to simplify the circuit, to make the solution easier and faster. Since all we really need in this problem is the voltage across the current source, we can get this by converting the circuit connected to the source to a single resistor. However, there are no series resistors in this circuit, and no parallel resistors. Go back and try again. Verify for yourself that there are no series resistors in this circuit, and that there are no parallel resistors in this circuit. Notice as well that this is not given to us as being a “balanced bridge”, as in a Wheatstone Bridge. Thus, there is current through Rm. The resistors R2 and R4 are not in series. The resistors R3 and RX are not in series.

Your Choice for First Step Was –Use Delta-to-wye Transformations to Simplify This Circuit Find an expression for the power delivered by the source in this circuit as a function of the value of the source, iS. This is the best possible choice for the first step. The goal is to simplify the circuit, to make the solution easier and faster. Since all we really need in this problem is the voltage across the current source, we can get this by converting the circuit connected to the source to a single resistor. We can use the delta-to-wye transformations, or wye-to-delta transformations, as the first step. Let’s go ahead and do this. There are at least two different delta configurations in this circuit, and at least three different wye configurations. Of these, any transformation would make the circuit subject to further simplification by replacing series combinations or parallel combinations of resistors with their equivalents.

Your Choice for First Step Was –Define Currents and Voltages for Each of the Elements in the Circuit Find an expression for the power delivered by the source in this circuit as a function of the value of the source, iS. This is not the best choice for the first step. In general, we do like to define currents and voltages. However, if it is clear that we are not going to be using the variables we define, then this is not a good use of our time. In this problem, there is a better approach. At some point we may need to define variables, but it is best to wait until you have a good idea of which ones you need. Go back and try again.

Using Delta-to-Wye Transformations Find an expression for the power delivered by the source in this circuit as a function of the value of the source, iS. We have decided to simplify this circuit by performing delta-to-wye transformations. In this problem, it really does not matter which one we choose. Therefore, we will simply pick one, and make the transformation. We will transform the delta configuration that is marked in red in this figure. Let’s prepare to make the transformation. Generally, we might want to redraw the equivalent after the transformation, for example with a quick sketch, just to be sure that this transformation will be of benefit. This could be done quickly, and help us choose the right approach. You might even be able to do this in your mind.

Next Slide Preparing to Make a Delta-to-Wye Transformations Find an expression for the power delivered by the source in this circuit as a function of the value of the source, iS. We have decided to simplify this circuit by performing delta-to-wye transformations. To prepare to make the transformation, we want to name the terminals. The point is this: we are going to be taking resistors out and putting others in their place, and there will be three different values. We need to be careful to get them all in the proper places. Using appropriate names for the terminals and resistors will help us with this. In these modules, we have labeled the terminals A, B, and C when we were defining the equations for the delta-to-wye transformation. Thus, we follow that rule. If possible, it is a good idea to have slides 36 and 37 from DPKC Part 1 Module 2 available, or other versions available, when working problems.

Next Slide Renaming Resistors Find an expression for the power delivered by the source in this circuit as a function of the value of the source, iS. We also choose to rename the resistors. We do this so that we can use the formulas we have derived, and never substitute the wrong value in. We leave the existing names in place as well. Since there were no resistors with these same names before, this works satisfactorily. We have used the rule that the resistor opposite to node A is called resistor RA, the resistor opposite to node B is called resistor RB, the resistor opposite to node C is called resistor RC. This is the same rule we used when we derived the equations for the delta-to-wye transformation.

The Formulas Find an expression for the power delivered by the source in this circuit as a function of the value of the source, iS. Now, we use the formulas below to make the transformation. The transformed circuit is shown. Notice that we have used primes (') in the names, R1', R2', and R3', since there will be confusion with the names of resistors that were already in the circuit. Other names would be possible, of course, but these names make keeping things straight a little easier, since they are close to the ones in our formulas. Next Slide

The Values for the Transformed Circuit Find an expression for the power delivered by the source in this circuit as a function of the value of the source, iS. Substituting values into these equations, we get Now, we are ready to make series and parallel combinations. At this point, it is probably clear that R1 and R1' are in series, R2' and R4 are in series, and R3' and RX are in series. Let's combine these next. Next Slide

Combining Series Resistors Find an expression for the power delivered by the source in this circuit as a function of the value of the source, iS. We replace the series combination R1 and R1' with their equivalent, which we will call R5. We replace the series combination R2' and R4 with their equivalent, which we will call R6. We replace the series combination R3' and RX with their equivalent, which we will call R7. The next step is to replace the parallel combination of R6 and R7 with their equivalent resistor which we will call R8. Next Slide

Combining Parallel Resistors Find an expression for the power delivered by the source in this circuit as a function of the value of the source, iS. We are now going to replace the parallel combination of R6 and R7 with their equivalent resistor which we will call R8. It is clear now that R5 and R8 are in series, and this combination can be replaced with an equivalent resistor, which we will call R9. Next Slide

Combining Series Resistors Again Find an expression for the power delivered by the source in this circuit as a function of the value of the source, iS. We are now going to replace the series combination of R5 and R8 with an equivalent resistor, which we will call R9. We are in position now to complete the solution. The power delivered by the source must be the power absorbed by the rest of the circuit it is connected to. The rest of the circuit is equivalent to resistor R9, so we can just find the power absorbed by R9. Go to Comments Slide

What if I chose another method? Is that a big deal? • If you picked another method, such as writing a set of equations using KVL and KCL, it does not make that much difference. One advantage of the approach taken here is that we do not have to solve simultaneous equations. However, if we have access to a good calculator or a computer, the solution can be done easily taking many other approaches. • While this is true, we still recommend that you learn this approach to solving circuits. There are many ways in which equivalent circuits can help us, and they are crucial tools. They are worth the time it take to understand them. Go back to Overviewslide.

Problems With Assistance Module 2 – Problem 5