Download Problems With Assistance, Module 4, Problem 1

Filename: PWA_Mod04_Prob01.ppt
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Use source transformations to solve for the current iX.
vS1=
5[V]
R 1=
27[W]
-
+
Problems With Assistance
Module
4
–
Problem
1
R=
R=
2
3
56[W]
39[W]
Go
straight to
the First
Step
vS2=
12[V]
+
iS1=
0.5[A]
R 4=
11[W]
iX
R 5=
Go
22[W]
straight to
the
Problem
Statement
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
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
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
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
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Coverage in this Module
The material for this problem is covered in
this module in the following presentation:
• DPKC_Mod04_Part01
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Problem Statement
Use source
transformations to
solve for the current iX.
vS1=
5[V]
R 1=
27[W]
-
+
R 2=
56[W]
R 3=
39[W]
vS2=
12[V]
+
iS1=
0.5[A]
R 4=
11[W]
iX
R 5=
22[W]
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Solution – First Step – Where to Start?
Use source
transformations to
solve for the current iX.
vS1=
5[V]
How should
we start this
problem?
What is the
first step?
R 1=
27[W]
-
+
R 2=
56[W]
R 3=
39[W]
vS2=
12[V]
+
iS1=
0.5[A]
R 4=
11[W]
iX
R 5=
22[W]
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
How should we start this
problem? What is the
first step?
Problem Solution – First Step
Use source transformations
to solve for the current iX.
vS1=
5[V]
Use the voltage-divider
rule to find the voltage
across R5.
b)
Replace vS1 and R1 with a
current source in parallel
with a resistance.
c)
Replace vS2 and R2 with a
current source in parallel
with a resistance.
d)
Replace iS1 and R4 with a
voltage source in series
with a resistance.
e)
Replace iS1 and R3 with a
voltage source in series
with a resistor.
R 1=
27[W]
-
+
a)
R 2=
56[W]
R 3=
39[W]
vS2=
12[V]
+
iS1=
0.5[A]
R 4=
11[W]
iX
R 5=
22[W]
Dave Shattuck
University of Houston
Your choice for First Step –
Use the voltage-divider rule to find the voltage across R5
© Brooks/Cole Publishing Co.
This is not a good choice for
the first step.
Use source transformations
to solve for the current iX.
vS1=
5[V]
R 1=
27[W]
-
+
R 2=
56[W]
R 3=
39[W]
vS2=
12[V]
+
iS1=
0.5[A]
R 4=
11[W]
iX
R 5=
22[W]
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.
Dave Shattuck
University of Houston
Your choice for First Step –
Replace vS1 and R1 with a current source in parallel with a
resistance
© Brooks/Cole Publishing Co.
This is a good choice.
Use source transformations
to solve for the current iX.
vS1=
5[V]
R 1=
27[W]
-
+
R 2=
56[W]
R 3=
39[W]
vS2=
12[V]
+
iS1=
0.5[A]
R 4=
11[W]
iX
R 5=
22[W]
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.
Dave Shattuck
University of Houston
Your choice for First Step –
Replace vS2 and R2 with a current source in parallel with a resistance
© Brooks/Cole Publishing Co.
This is a good choice for the
first step, and the one that
we will choose here.
Use source transformations
to solve for the current iX.
vS1=
5[V]
R 1=
27[W]
-
+
R 2=
56[W]
R 3=
39[W]
vS2=
12[V]
+
iS1=
0.5[A]
R 4=
11[W]
iX
R 5=
22[W]
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.
Dave Shattuck
University of Houston
Your choice for First Step was –
Replace iS1 and R4 with a voltage source in series with a resistance
© Brooks/Cole Publishing Co.
This is possible, but is not a
good choice for the first
step.
Use source transformations
to solve for the current iX.
vS1=
5[V]
R 1=
27[W]
-
+
R 2=
56[W]
R 3=
39[W]
vS2=
12[V]
+
iS1=
0.5[A]
R 4=
11[W]
iX
R 5=
22[W]
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.
Dave Shattuck
University of Houston
Your choice for First Step was –
Replace iS1 and R3 with a voltage source in series with a resistor
© Brooks/Cole Publishing Co.
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.
Use source transformations
to solve for the current iX.
vS1=
5[V]
R 1=
27[W]
-
+
Please go back and try again.
R 2=
56[W]
R 3=
39[W]
vS2=
12[V]
+
iS1=
0.5[A]
R 4=
11[W]
iX
R 5=
22[W]
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Replacing vS2 and R2 with a Current Source in Parallel
with a Resistance
Use source transformations
to solve for the current iX.
vS1=
5[V]
R 1=
27[W]
-
+
R 2=
56[W]
R 3=
39[W]
vS2=
12[V]
+
iS1=
0.5[A]
R 4=
11[W]
iX
R 5=
22[W]
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.
Next slide
Dave Shattuck
University of Houston
First Equivalent Circuit Replacement
© Brooks/Cole Publishing Co.
We 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.
Use source transformations
to solve for the current iX.
vS1=
5[V]
R1=
27[W]
-
+
R3=
39[W]
iS2=
12[V]/56[W]=
0.21[A]
R2=
56[W]
iS1=
0.5[A]
R 4=
11[W]
iX
R5=
22[W]
Next slide
Dave Shattuck
University of Houston
Parallel Equivalents Inserted
© Brooks/Cole Publishing Co.
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 iS3 and R6 with a
voltage source in series with a
resistor. Let’s consider this in the
next slide.
Use source transformations
to solve for the current iX.
vS1=
5[V]
R1=
27[W]
-
+
R3=
39[W]
iS3=
0.29[A]
R 6=
9.2[W]
iX
R5=
22[W]
Next slide
Dave Shattuck
University of Houston
Which Equivalent is Correct?
© Brooks/Cole Publishing Co.
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.
Use source transformations
to solve for the current iX.
Equivalent #1
vS1=
5[V]
Equivalent #2
R1=
27[W]
vS1=
5[V]
+
+
-
vS3=
2.7[V]
R3=
39[W]
R6=
9.2[W]
iX
R5=
22[W]
+
-
vS3=
2.7[V]
-
+
R6=
9.2[W]
R 1=
27[W]
R3=
39[W]
iX
R 5=
22[W]
You chose the incorrect way to insert the equivalent circuit.
© Brooks/Cole Publishing Co.
Use source transformations
to solve for the current iX.
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.
Original Circuit
Equivalent #1
vS1=
5[V]
R1=
27[W]
+
-
+
R6=
9.2[W]
R 3=
39[W]
iS3=
0.29[A]
R 6=
9.2[W]
iX
R5=
22[W]
+
-
vS3=
2.7[V]
R1=
27[W]
-
vS1=
5[V]
Next slide
You Chose Equivalent #1
Dave Shattuck
University of Houston
R3=
39[W]
iX
R5=
22[W]
Next slide
Dave Shattuck
University of Houston
You Chose Equivalent #2
© Brooks/Cole Publishing Co.
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.
Use source transformations
to solve for the current iX.
Next, we will replace vS1 and R1 with a current source in
parallel with a resistor, in the next slide.
Original Circuit
vS1=
5[V]
R1=
27[W]
Equivalent #2
-
+
R 3=
39[W]
R6=
9.2[W]
iS3=
0.29[A]
R 6=
9.2[W]
iX
R5=
22[W]
+
-
vS3=
2.7[V]
R 1=
27[W]
-
+
vS1=
5[V]
R3=
39[W]
iX
R 5=
22[W]
Next slide
Dave Shattuck
University of Houston
What Polarity for the Current Source?
© Brooks/Cole Publishing Co.
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.
Use source transformations
to solve for the current iX.
Polarity #1
R 6=
9.2[W]
+
-
Polarity #2
iS4=5[V]/27[W]=
0.185[A]
iS4=5[V]/27[W]=
0.185[A]
R1=
27[W]
R1=
27[W]
R 6=
9.2[W]
R3=
39[W]
vS3=
2.7[V]
+
iX
R 5=
22[W]
-
R3=
39[W]
vS3=
2.7[V]
iX
R 5=
22[W]
Dave Shattuck
University of Houston
You Chose Polarity #1
© Brooks/Cole Publishing Co.
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.
Use source transformations
to solve for the current iX.
Polarity #1
iS4=5[V]/27[W]=
0.185[A]
-
R6=
9.2[W]
R3=
39[W]
vS3=
2.7[V]
iX
R 5=
22[W]
+
-
vS3=
2.7[V]
R 1=
27[W]
-
+
vS1=
5[V]
R1=
27[W]
+
R 6=
9.2[W]
Now combine the parallel resistors.
R3=
39[W]
iX
R 5=
22[W]
Dave Shattuck
University of Houston
You Chose Polarity #2
© Brooks/Cole Publishing Co.
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.
Use source transformations
to solve for the current iX.
Polarity #2
iS4=5[V]/27[W]=
0.185[A]
-
R6=
9.2[W]
R 3=
39[W]
vS3=
2.7[V]
iX
R5=
22[W]
+
-
vS3=
2.7[V]
R 1=
27[W]
-
+
+
R 6=
9.2[W]
vS1=
5[V]
R1=
27[W]
R3=
39[W]
iX
R 5=
22[W]
Dave Shattuck
University of Houston
Combine Parallel Resistors
© Brooks/Cole Publishing Co.
Use source transformations
to solve for the current iX.
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.
R 6=
9.2[W]
iS4= 0.185[A]
Yes, we can replace them.
No, we cannot replace them.
+
-
R7=
16[W]
vS3=
2.7[V]
iX
R 5=
22[W]
Dave Shattuck
University of Houston
You Chose: Yes, We Can Replace Them
© Brooks/Cole Publishing Co.
Use source transformations
to solve for the current iX.
R 6=
9.2[W]
+
-
This choice was not correct.
No, we cannot replace R6 and
R7 with their series
equivalent, because these
two resistors are not in
series.
iS4= 0.185[A]
R7=
16[W]
vS3=
2.7[V]
iX
R 5=
22[W]
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.
Dave Shattuck
University of Houston
You Chose: No, We Cannot Replace Them
© Brooks/Cole Publishing Co.
This choice was correct.
Use source transformations
to solve for the current iX.
R 6=
9.2[W]
+
-
No, we cannot replace R6 and
R7 with their series
equivalent, because these
two resistors are not in
series.
iS4= 0.185[A]
R7=
16[W]
vS3=
2.7[V]
iX
R 5=
22[W]
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.
Dave Shattuck
University of Houston
Replacing Current Source and Resistor
© Brooks/Cole Publishing Co.
Use source transformations
to solve for the current iX.
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]
+
R6=
9.2[W]
vS4=
(0.185[A])(16[W])=
3.0[V]
+
-
vS3=
2.7[V]
-
iX
R5=
22[W]
Dave Shattuck
University of Houston
Series Resistors and Voltage Sources
© Brooks/Cole Publishing Co.
Use source transformations
to solve for the current iX.
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
iX 
R 8=
25.2[W]
vS4=
-0.3[V]
iX  6.4[mA].
+
-
iX
vS 4
0.3[V]


R8  R5 25.2[W]  22[W]
R5=
22[W]
Go to
Comments
slide.
Go back to
Overview
slide.
Dave Shattuck
University of Houston
Was This the Easiest Way to Solve?
© Brooks/Cole Publishing Co.
• 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.
vS1=
5[V]
R 1=
27[W]
-
+
R 2=
56[W]
R 3=
39[W]
vS2=
12[V]
+
iS1=
0.5[A]
R 4=
11[W]
iX
R 5=
22[W]
Go back to
Overview
slide.