Filename PWAMod 04Prob 01 ppt Use source transformations

Filename: PWA_Mod 04_Prob 01. ppt Use source transformations to solve for the current i. X. Problems With Assistance Module 4 – Problem 1 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 • Source 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 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

Coverage in this Module The material for this problem is covered in this module in the following presentation: • DPKC_Mod 04_Part 01 Next slide

Problem Statement Use source transformations to solve for the current i. X. Next slide

Solution – First Step – Where to Start? Use source transformations to solve for the current i. X. How should we start this problem? What is the first step? Next slide

Problem Solution – First Step Use source transformations to solve for the current i. X. How should we start this problem? What is the first step? a) Use the voltage-divider rule to find the voltage across R 5. b) Replace v. S 1 and R 1 with a current source in parallel with a resistance. c) Replace v. S 2 and R 2 with a current source in parallel with a resistance. d) Replace i. S 1 and R 4 with a voltage source in series with a resistance. e) Replace i. S 1 and R 3 with a voltage source in series with a resistor.

Your choice for First Step – Use the voltage-divider rule to find the voltage across R 5 Use source transformations to solve for the current i. X. This is not a good choice for the first step. One problem is that we were asked to use Source Transformations to solve this problem, and this would not be using them. However, there is a much bigger problem; resistors R 1 and R 5 are not in series, and the voltage across them is not known. Also, resistors R 3 and R 5 are not in series, and the voltage across them is not known. We can’t use the voltage-divider rule in this case. Go back and try again.

Your choice for First Step – Replace v. S 1 and R 1 with a current source in parallel with a resistance Use source transformations to solve for the current i. X. This is a good choice. This would be a reasonable first step, since once we had done this, the new resistance would then be in parallel with R 3, and we could simplify further. In fact, we will take this step later in the problem. However, simply by choice, we will pick another, equally good, first step. So, even though you made a good choice, please go back and try again.

Your choice for First Step – Replace v. S 2 and R 2 with a current source in parallel with a resistance Use source transformations to solve for the current i. X. This is a good choice for the first step, and the one that we will choose here. The voltage source v. S 2 and the resistor R 2 are in series, and can be replaced by a current source in parallel with a resistance. Once we do that, the resulting current source will be in parallel with i. S 1, and the resulting resistance will be in parallel with R 4. Let’s go ahead and make this replacement.

Your choice for First Step was – Replace i. S 1 and R 4 with a voltage source in series with a resistance Use source transformations to solve for the current i. X. This is possible, but is not a good choice for the first step. This is possible because i. S 1 and R 4 are indeed in parallel, and therefore, we can replace them with a voltage source in series with a resistance. However, if we did this, there is no advantage in terms of further simplification. It just doesn’t help us. Not every replacement is an improvement. Therefore, we recommend that you go back and try again.

Your choice for First Step was – Replace i. S 1 and R 3 with a voltage source in series with a resistor This is not a good choice. Use source transformations to solve for the current i. X. The i. S 1 current source and the R 3 resistor are not in parallel, nor are they in series. Therefore, we can not make any replacements of them. Please go back and try again.

Replacing v. S 2 and R 2 with a Current Source in Parallel with a Resistance Use source transformations to solve for the current i. X. We are going to replace the v. S 2 voltage source and the R 2 resistor with a current source in parallel with a resistance. Note that we need to be careful about polarities and signs. Note that the voltage source is defined at the bottom with respect to the top. So, we need to use a current source with the polarity arrow pointing down. Let’s make the replacement. Next slide

First Equivalent Circuit Replacement Use source transformations to solve for the current i. X. We have replaced the voltage source in series with resistor R 2, with a current source, i. S 2, in parallel with the same resistor R 2. Now, it should be clear that R 4 is in parallel with R 2, and i. S 2 is in parallel with i. S 1. Combining these, we get the new circuit in the next slide. Next slide

Parallel Equivalents Inserted Use source transformations to solve for the current i. X. We have replaced the parallel resistors and parallel current sources with their equivalents. It is now going to be useful to replace the parallel combination of i. S 3 and R 6 with a voltage source in series with a resistor. Let’s consider this in the next slide. Next slide

Which Equivalent is Correct? Use source transformations to solve for the current i. X. Equivalent #1 We have replaced the current source and the resistor in parallel with it (R 6) with a voltage source (v. S 3) in series with that resistor. Two possible ways of doing this are shown here. Which way is correct? Click on one to choose your answer. Equivalent #2

You Chose Equivalent #1 Next slide You chose the incorrect way to insert the equivalent circuit. Use source transformations to solve for the current i. X. Original Circuit Look at the original circuit on the left, and the replacement you chose on the right, in the circuits below. The two nodes of the source transformation equivalent are marked with dashed red lines. Can you find these two nodes in the equivalent on the right? They are not there. The other point is that R 6 and v. S 3 are supposed to be in series, but in this circuit they are not. Equivalent #1 Next slide

You Chose Equivalent #2 Use source transformations to solve for the current i. X. You have chosen the correct equivalent circuit. Note that the two terminals of the equivalent, marked in both circuits with dashed red lines, remain in place in both versions of the circuit. Next, we will replace v. S 1 and R 1 with a current source in parallel with a resistor, in the next slide. Original Circuit Equivalent #2 Next slide

What Polarity for the Current Source? Use source transformations to solve for the current i. X. Polarity #1 The voltage source v. S 1 and resistor R 1 have been replaced with a current source in parallel with a resistance. The key question here is which polarity should be used for the current source. Choose one of the polarities by clicking on it. Polarity #2

You Chose Polarity #1 Use source transformations to solve for the current i. X. Polarity #1 You made the correct choice, Polarity #1. Do not be confused by the change from a vertical alignment to a horizontal one. The relationship between the polarities with respect to the terminals is all that matters. The two terminals are marked here. Compare the polarities of the sources here with those in the equivalent circuits given in the definition. This polarity is correct. Now combine the parallel resistors.

You Chose Polarity #2 Use source transformations to solve for the current i. X. Polarity #2 You did not choose correctly. Do not be confused by the change from a vertical alignment to a horizontal one. The relationship between the polarities with respect to the terminals is all that matters. The two terminals are marked here. Compare the polarities of the sources here with those in the equivalent circuits given in the definition. These polarities are different from the definition. The correct choice was Polarity #1.

Combine Parallel Resistors Use source transformations to solve for the current i. X. Here we have combined the parallel resistors and replaced them with a resistance R 7. Now, can we replace the two series resistors, R 6 and R 7 with their series equivalent? Click on your choice. Yes, we can replace them. No, we cannot replace them.

You Chose: Yes, We Can Replace Them Use source transformations to solve for the current i. X. This choice was not correct. No, we cannot replace R 6 and R 7 with their series equivalent, because these two resistors are not in series. It is tempting to think that we can make this equivalent, but the alignment of the resistors does not make them in series. They are in series if they have the same current through them. Because of the current source, they do not have the same current through them. Go to the next slide.

You Chose: No, We Cannot Replace Them Use source transformations to solve for the current i. X. This choice was correct. No, we cannot replace R 6 and R 7 with their series equivalent, because these two resistors are not in series. It is tempting to think that we can make this equivalent, but the alignment of the resistors does not make them in series. They are in series if they have the same current through them. Because of the current source, they do not have the same current through them. Instead, let’s replace i. S 4 and R 7 with a voltage source in series with a resistance.

Replacing Current Source and Resistor Use source transformations to solve for the current i. X. We have replaced i. S 4 and R 7 with a voltage source in series with a resistance. Now we have resistors R 6 and R 7 in series, and also have two voltage sources in series. Note that the polarities of the voltage sources are such that they will subtract. We make the replacement in the next slide.

Series Resistors and Voltage Sources Use source transformations to solve for the current i. X. We have replaced the resistors R 6 and R 7 in series, and the two voltage sources in series. This circuit is simple enough that we can solve it directly. The current i. X is Go to Comments slide. Go back to Overview slide.

Was This the Easiest Way to Solve? • There are other ways to solve this problem. For example, we could use the node voltage method, and get only two equations. From that solution, we could find i. X. The prime benefit of Source Transformations is that we never have more than one simultaneous equation. This can be of help not only in the solution, but in understanding how this component value or that source affects the solution. This is particularly helpful in designing circuits. • Please note, though, that no two resistors, and no two sources, are in series or in parallel in the original circuit. If we are to use equivalent circuits, we must use Source Transformations. Go back to Overview slide.