Problems With Assistance Module 2 – Problem 6

Problems With AssistanceModule 2 – Problem 6 Filename: PWA_Mod02_Prob06.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, vS. 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, vS. 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, vS. • 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, vS. 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, vS. 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 current through the voltage source, we can get this by converting the circuit connected to the source to a single resistor. As it turns out, there is a pair of resistors in this circuit that are in series or in parallel. Let’s find that pair. Look carefully at this circuit, to determine where the series or parallel combinations can be made.

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, vS. This is not 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 current through the voltage source, we can get this by converting the circuit connected to the source to a single resistor. While there is at least one delta combination, and at least one wye combination, there is a pair of resistors in this circuit that are in series or in parallel. Let’s find that pair. Look carefully at this circuit, to determine where the series or parallel combinations can be made.

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, vS. 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.

Series or Parallel Combinations? Find an expression for the power delivered by the source in this circuit as a function of the value of the source, vS. • We are looking for a series or parallel combination of resistors. Which of the following are true? • Resistors R1 and R2 are in parallel. • Resistors RX and R3 are in series. • Resistors Rm and RX are in parallel. • Resistors Rm and RX are in series.

You Said That Resistors R1 and R2 Are in Parallel Find an expression for the power delivered by the source in this circuit as a function of the value of the source, vS. You said that resistors R1 and R2 are in parallel. This is not correct. If they were in parallel, they would have the same voltage across them. Because of the possible voltage across Rm, this is not true. Perhaps even more clear, it should be noted that two resistors are in parallel when the two ends of each are connected together. That is not the case here. Go back, and try again.

You Said That Resistors RX and R3 Are in Series Find an expression for the power delivered by the source in this circuit as a function of the value of the source, vS. You said that resistors RX and R3 are in series. This is not correct. If they were in series, they would have the same current through them. Because of the possible current through R1 or Rm, this is not true. Go back, and try again.

You Said That Resistors Rm and RX Are in Parallel Find an expression for the power delivered by the source in this circuit as a function of the value of the source, vS. You said that resistors Rm and RX are in parallel. This is correct. They have the same voltage across them. Also, the two ends of each are connected together. Do not be misled by the wire between the left ends of these two resistors. This wire is marked in red here. Remember that two nodes separated by a wire are really one node. Let’s combine these parallel resistors, and replace them with their equivalent.

You Said That Resistors Rm and RX are in series Find an expression for the power delivered by the source in this circuit as a function of the value of the source, vS. You said that resistors Rm and RX are in series. This is not correct. If they were in series, they would have the same current through them. Because of the possible current through R2, this is not true. Go back, and try again.

Next Slide Replacing the Parallel Resistors Find an expression for the power delivered by the source in this circuit as a function of the value of the source, vS. We have decided to replace Rm and RX with their parallel equivalent resistor. We have made this replacement in the circuit shown here. The equivalent resistor is called R4. The next step is to recognize that resistors R2 and R4 are in series, and replace them with their equivalent resistor. In this case, we have placed the equivalent resistor in the place of RX, but this choice was arbitrary. Because the left hand nodes, shown in red, are connected by a wire, they are really the same node. The two positions are equivalent.

Next Slide Replacing the Series Resistors Find an expression for the power delivered by the source in this circuit as a function of the value of the source, vS. We have decided to replace R2 and R4 with their series equivalent resistor. We have made this replacement in the circuit shown here. The equivalent resistor is called R5. The next step is to recognize that resistors R1 and R5 are in parallel, and replace them with their equivalent resistor. In this case, we have placed the equivalent resistor in the place of R2, but this choice was arbitrary. Either position would be fine.

Next Slide Replacing the Parallel Resistors Again Find an expression for the power delivered by the source in this circuit as a function of the value of the source, vS. We have decided to replace R1 and R5 with their parallel equivalent resistor. We have made this replacement in the circuit shown here. The equivalent resistor is called R6. The next step is to recognize that resistors R3 and R6 are in series, and replace them with their equivalent resistor.

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, vS. We are now going to replace the series combination of R6 and R3 with their equivalent resistor which we will call R7. This has been done here. 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 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 R7, so we can just find the power absorbed by R7. 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. • If you choose to use the delta-to-wye transformations, you will also get this same answer. However, generally, this takes longer. Our rule of thumb is that we avoid these transformations when we can. Go back to Overviewslide.

Problems With Assistance Module 2 – Problem 6