use source transformations to solve for the current i x. problems with assistance module 4 –...

Dave ShattuckUniversity of Houston

© Brooks/Cole Publishing Co.

Use source transformations to solve for the current iX.

iS1=0.5[A] R4=

11[W]

R3=39[W]

R5=22[W]+

-vS2=

12[V]

R1=27[W]

+ -

vS1=5[V]

R2=56[W]

iX

Problems With AssistanceModule 4 – Problem 1

Filename: PWA_Mod04_Prob01.ppt

Next slide

Go straight to the Problem Statement

Go straight to the First Step



Overview of this Problem

In this problem, we will use the following concepts:

• Equivalent Circuits

• Source Transformations

Next slide

Go straight to the Problem Statement

Go straight to the First Step



Textbook Coverage

The material for this problem is covered in your textbook in the following sections:

• Circuits by Carlson: Sections #.#• Electric Circuits 6th Ed. by Nilsson and Riedel: Sections

#.#• Basic Engineering Circuit Analysis 6th Ed. by Irwin and

Wu: Section #.#• Fundamentals of Electric Circuits by Alexander and

Sadiku: Sections #.#• Introduction to Electric Circuits 2nd 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_Mod04_Part01

Next slide


© Brooks/Cole Publishing Co. Problem Statement

Next slide


iS1=0.5[A] R4=

11[W]

R3=39[W]

R5=22[W]+

-vS2=

12[V]

R1=27[W]

+ -

vS1=5[V]

R2=56[W]

iX


© Brooks/Cole Publishing Co. Solution – First Step – Where to Start?How should we start this problem? What is the first step?

Next slide


iS1=0.5[A] R4=

11[W]

R3=39[W]

R5=22[W]+

-vS2=

12[V]

R1=27[W]

+ -

vS1=5[V]

R2=56[W]

iX



Problem Solution – First StepHow should we start this

problem? What is the first step?

a) Use the voltage-divider rule to find the voltage across R5.

b) Replace vS1 and R1

with a current source in parallel with a resi

stance

.

c) Replace vS2 and R2

with a current source in parallel with a resi

stance

.

d) Replace iS1 and R4

with a voltage source in series with a resist

ance

.

e) Replace iS1 and R3

with a voltage source in series with a resist

or

.


iS1=0.5[A] R4=

11[W]

R3=39[W]

R5=22[W]+

-vS2=

12[V]

R1=27[W]

+ -

vS1=5[V]

R2=56[W]

iX


© Brooks/Cole Publishing Co. Your choice for First Step – Use the voltage-divider rule to find the voltage across R5

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 R1 and R5 are not in series, and the voltage across them is not known. Also, resistors R3 and R5 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.


iS1=0.5[A] R4=

11[W]

R3=39[W]

R5=22[W]+

-vS2=

12[V]

R1=27[W]

+ -

vS1=5[V]

R2=56[W]

iX


© Brooks/Cole Publishing Co. Your choice for First Step – Replace vS1 and R1 with a current source in parallel with a

resistanceThis 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 R3, 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.


iS1=0.5[A] R4=

11[W]

R3=39[W]

R5=22[W]+

-vS2=

12[V]

R1=27[W]

+ -

vS1=5[V]

R2=56[W]

iX


© Brooks/Cole Publishing Co. Your choice for First Step – Replace vS2 and R2 with a current source in parallel with a resistance

This is a good choice for the first step, and the one that we will choose here.

The voltage source vS2 and the resistor R2 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 iS1, and the resulting resistance will be in parallel with R4. Let’s go ahead and make this replacement.


iS1=0.5[A] R4=

11[W]

R3=39[W]

R5=22[W]+

-vS2=

12[V]

R1=27[W]

+ -

vS1=5[V]

R2=56[W]

iX


© Brooks/Cole Publishing Co. Your choice for First Step was – Replace iS1 and R4 with a voltage source in series with a resistance

This is possible, but is not a good choice for the first step.

This is possible because iS1 and R4 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.


iS1=0.5[A] R4=

11[W]

R3=39[W]

R5=22[W]+

-vS2=

12[V]

R1=27[W]

+ -

vS1=5[V]

R2=56[W]

iX


© Brooks/Cole Publishing Co. Your choice for First Step was – Replace iS1 and R3 with a voltage source in series with a resistor

This is not a good choice.

The iS1 current source and the R3 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.


iS1=0.5[A] R4=

11[W]

R3=39[W]

R5=22[W]+

-vS2=

12[V]

R1=27[W]

+ -

vS1=5[V]

R2=56[W]

iX



Replacing vS2 and R2 with a Current Source in Parallel with a Resistance

We are going to replace the vS2 voltage source and the R2 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.


iS1=0.5[A] R4=

11[W]

R3=39[W]

R5=22[W]+

-vS2=

12[V]

R1=27[W]

+ -

vS1=5[V]

R2=56[W]

iX

Next slide


© Brooks/Cole Publishing Co. First Equivalent Circuit ReplacementWe have replaced the voltage source in

series with resistor R2, with a current source, iS2, in parallel with the same resistor R2. Now, it should be clear that R4 is in parallel with R2, and iS2 is in parallel with iS1. Combining these, we get the new circuit in the next slide.

iS1=0.5[A] R4=

11[W]

R3=39[W]

R5=22[W]

R1=27[W]

+ -

vS1=5[V]

iXR2=

56[W]

iS2=12[V]/56[W]=

0.21[A]


Next slide


© Brooks/Cole Publishing Co. Parallel Equivalents InsertedWe 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 iS3 and R6 with a voltage source in series with a resistor. Let’s consider this in the next slide.


iS3=0.29[A] R6=

9.2[W]

R3=39[W]

R5=22[W]

R1=27[W]

+ -vS1=5[V]

iX

Next slide


© Brooks/Cole Publishing Co. Which Equivalent is Correct?We have replaced the current source

and the resistor in parallel with it (R6) with a voltage source (vS3) 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.


R3=39[W]

R5=22[W]

R1=27[W]

+ -

vS1=5[V]

iX

R6=9.2[W]

+

-

vS3=2.7[V]

R3=39[W]

R5=22[W]

R1=27[W]

+ -

vS1=5[V]

iX

+

-

vS3=2.7[V]

R6=9.2[W]

Equivalent #1Equivalent #2



iS3=0.29[A] R6=

9.2[W]

R3=39[W]

R5=22[W]

R1=27[W]

+ -

vS1=5[V]

iX

You Chose Equivalent #1You chose the incorrect way to insert the equivalent 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 R6 and vS3 are supposed to be in series, but in this circuit they are not.


R3=39[W]

R5=22[W]

R1=27[W]

+ -

vS1=5[V]

iX

R6=9.2[W]

+

-

vS3=2.7[V]

Original Circuit Equivalent #1

Next slide

Next slide


© Brooks/Cole Publishing Co. You Chose Equivalent #2You 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 vS1 and R1 with a current source in parallel with a resistor, in the next slide.


Equivalent #2

iS3=0.29[A] R6=

9.2[W]

R3=39[W]

R5=22[W]

R1=27[W]

+ -

vS1=5[V]

iX

Original Circuit

R3=39[W]

R5=22[W]

R1=27[W]

+ -

vS1=5[V]

iX

+

-

vS3=2.7[V]

R6=9.2[W]

Next slide


© Brooks/Cole Publishing Co.What Polarity for the Current Source?

The voltage source vS1 and resistor R1 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.


R3=39[W]

R5=22[W]

R1=27[W]

iX

+

-

vS3=2.7[V]

R6=9.2[W]

iS4=5[V]/27[W]=0.185[A]

R3=39[W]

R5=22[W]

R1=27[W]

iX

+

-

vS3=2.7[V]

R6=9.2[W]

iS4=5[V]/27[W]=0.185[A]

Polarity #1 Polarity #2


© Brooks/Cole Publishing Co.You Chose 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.


Polarity #1

R3=39[W]

R5=22[W]

R1=27[W]

+ -

vS1=5[V]

iX

+

-

vS3=2.7[V]

R6=9.2[W]R3=

39[W]

R5=22[W]

R1=27[W]

iX

+

-

vS3=2.7[V]

R6=9.2[W]

iS4=5[V]/27[W]=0.185[A]


© Brooks/Cole Publishing Co.You Chose 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.


Polarity #2

R3=39[W]

R5=22[W]

R1=27[W]

+ -

vS1=5[V]

iX

+

-

vS3=2.7[V]

R6=9.2[W]R3=

39[W]

R5=22[W]

R1=27[W]

iX

+

-

vS3=2.7[V]

R6=9.2[W]

iS4=5[V]/27[W]=0.185[A]


© Brooks/Cole Publishing Co.Combine Parallel Resistors

Here we have combined the parallel resistors and replaced them with a resistance R7.

Now, can we replace the two series resistors, R6 and R7 with their series equivalent? Click on your choice.

Yes, we can replace them.

No, we cannot replace them.


R7=16[W]

R5=22[W]

iX

+

-

vS3=2.7[V]

R6=9.2[W]

iS4= 0.185[A]


© Brooks/Cole Publishing Co.You Chose: Yes, We Can Replace Them

This choice was not correct.

No, we cannot replace R6 and R7 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.


R7=16[W]

R5=22[W]

iX

+

-

vS3=2.7[V]

R6=9.2[W]

iS4= 0.185[A]


© Brooks/Cole Publishing Co.You Chose: No, We Cannot Replace Them

This choice was correct.

No, we cannot replace R6 and R7 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 iS4 and R7 with a voltage source in series with a resistance.


R7=16[W]

R5=22[W]

iX

+

-

vS3=2.7[V]

R6=9.2[W]

iS4= 0.185[A]


© Brooks/Cole Publishing Co.Replacing Current Source and Resistor

We have replaced iS4 and R7 with a voltage source in series with a resistance.

Now we have resistors R6 and R7 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.


R7=16[W]

R5=22[W]

iX

+

-

vS3=2.7[V]

R6=9.2[W]

+ -

vS4=(0.185[A])(16[W])=

3.0[V]


© Brooks/Cole Publishing Co.Series Resistors and Voltage Sources

We have replaced the resistors R6 and R7 in series, and the two voltage sources in series.

This circuit is simple enough that we can solve it directly. The current iX is


R8=25.2[W]

R5=22[W]iX

+

-

vS4=-0.3[V]

4

8 5

0.3[V]

25.2[ ] 22[ ]

6.4[mA].

SX

X

vi

R R

i

W W

Go back to Overview

slide.

Go to Comments

slide.


© Brooks/Cole Publishing Co. 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 iX. 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.

iS1=0.5[A] R4=

11[W]

R3=39[W]

R5=22[W]+

-vS2=

12[V]

R1=27[W]

+ -

vS1=5[V]

R2=56[W]

iX

use source transformations to solve for the current i x. problems with assistance module 4 –...

Documents