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Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Problems With Assistance
Module 2 – Problem 4
Filename: PWA_Mod02_Prob04.ppt
Go
straight to
the First
Step
Go
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
• Series and Parallel Combinations of
Resistors
• Delta-to-Wye 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 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
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 presentations:
• DPKC_Mod02_Part01.
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Find the resistance
seen at terminals A
and B of the circuit
shown.
Problem Statement
R2=
15[W]
C
R1=
22[W]
R3=
49[W]
R4=
10[W]
R5=
25[W]
A
R6=
39[W]
R7=
56[W]
R9=
27[W]
R8=
33[W]
R10=
82[W]
B
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Solution – First Step – Where to Start?
Find the resistance
seen at terminals A
and B of the circuit
shown.
R2=
15[W]
C
R1=
22[W]
R3=
49[W]
How should we start
this problem? What is
the first step?
R4=
10[W]
R5=
25[W]
A
R6=
39[W]
R7=
56[W]
R9=
27[W]
R8=
33[W]
R10=
82[W]
B
Next slide
Dave Shattuck
University of Houston
Problem Solution – First Step
© Brooks/Cole Publishing Co.
Find the resistance
seen at terminals A
and B of the circuit
shown.
We are asked to find the ratio of voltage to
current at the terminals; this is the
same thing as the resistance. What is
the best first step?
1. Attach a source to terminals A and B.
2. Define currents and voltages for each
of the elements in the circuit.
3. Write a series of KVL and KCL
equations.
4. Simplify the circuit by removing
resistors that do not contribute to the
solution.
5. Combine resistors in series and in
parallel to simplify the circuit.
6. Perform a delta-to-wye
transformation.
R2=
15[W]
C
R1=
22[W]
R3=
49[W]
R4=
10[W]
R5=
25[W]
A
R6=
39[W]
R7=
56[W]
R9=
27[W]
B
R8=
33[W]
R10=
82[W]
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Your Choice for First Step –
Attach a Source to the Terminals
Find the resistance
seen at terminals A
and B of the circuit
shown.
R2=
15[W]
C
R1=
22[W]
R3=
49[W]
R4=
10[W]
R5=
25[W]
This is not the best choice for the first
step.
We could indeed apply a source, with the
intention of then finding the ratio of the
voltage at the terminals to the current
through the terminals. This ratio would be
the resistance. However, this would
require solving the circuit, and we can do
the problem more easily.
A
R6=
39[W]
R7=
56[W]
R9=
27[W]
B
Go back and try again.
R8=
33[W]
R10=
82[W]
Dave Shattuck
University of Houston
Your choice for First Step was –
© Brooks/Cole Publishing Co.
Define Currents and Voltages for each of the Elements in the Circuit
Find the resistance
seen at terminals A
and B of the circuit
shown.
R2=
15[W]
C
R1=
22[W]
R3=
49[W]
R4=
10[W]
R5=
25[W]
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 will 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.
A
R6=
39[W]
R7=
56[W]
R9=
27[W]
B
R8=
33[W]
R10=
82[W]
Dave Shattuck
University of Houston
Your Choice for First Step –
Write a Series of KVL and KCL Equations
© Brooks/Cole Publishing Co.
Find the resistance
seen at terminals A
and B of the circuit
shown.
R2=
15[W]
C
R1=
22[W]
R3=
49[W]
R4=
10[W]
R5=
25[W]
This is not the best choice for the first
step.
We could indeed write a set of KVL and
KCL equations, once we had defined
voltages and currents. However, without
sources all voltages and currents would be
zero. The only way to meaningfully do
this would be to apply a source, and then
solve for the ratio of voltage to current at
the terminals. There is a better approach.
Go back and try again.
A
R6=
39[W]
R7=
56[W]
R9=
27[W]
B
R8=
33[W]
R10=
82[W]
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Your Choice for First Step Was –
Simplify the circuit by removing resistors that do not contribute to the solution.
Find the resistance
seen at terminals A
and B of the circuit
shown.
R2=
15[W]
C
R1=
22[W]
R3=
49[W]
R4=
10[W]
R5=
25[W]
This one way to describe the best choice
for the first step.
The goal is to simplify the circuit, to make
the solution easier and faster. It turns out
that there are resistors in this circuit that
do not contribute to the solution. In other
words, we can take these resistors out, and
the answer will be the same. Certainly,
this is a good thing to do when we can,
since it will simplify the circuit. The key
is to do it correctly.
Let’s begin that process.
A
R6=
39[W]
R7=
56[W]
R9=
27[W]
B
R8=
33[W]
R10=
82[W]
Dave Shattuck
University of Houston
Your Choice for First Step Was –
Combine Resistors in Series and in Parallel to Simplify the Circuit
© Brooks/Cole Publishing Co.
Find the resistance
seen at terminals A
and B of the circuit
shown.
R2=
15[W]
C
R1=
22[W]
R3=
49[W]
R4=
10[W]
R5=
25[W]
This one way to describe the best 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. We can do with
using equivalent circuits, specifically by
repeatedly combining resistors in series
and parallel.
Let’s begin that process.
A
R6=
39[W]
R7=
56[W]
R9=
27[W]
B
R8=
33[W]
R10=
82[W]
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Your Choice for First Step –
Perform a Delta-to-Wye Transformation
Find the resistance
seen at terminals A
and B of the circuit
shown.
R2=
15[W]
C
R1=
22[W]
R3=
49[W]
R4=
10[W]
R5=
25[W]
This is not the best choice for the first
step.
We could indeed perform a delta-to-wye
transformation here. One possible delta
configuration is marked in red in the
circuit here. However, this transformation
is complicated; generally we are best
served by avoiding this unless there is no
other choice. There are several other
choices here. We recommend that you go
back and try again.
A
R6=
39[W]
R7=
56[W]
R9=
27[W]
B
R8=
33[W]
R10=
82[W]
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
How to Begin Simplifying This Circuit
Find the resistance
seen at terminals A
and B of the circuit
shown.
R2=
15[W]
C
R1=
22[W]
R3=
49[W]
We have reached this point, by either of
two different approaches. Each
approach involves simplifying the
circuit, and in particular, the circuit
“seen” by terminals A and B. Which
of the following statements are true?
a) Resistors R2 and R5 are in series.
b) Resistors R8 and R10 are in series.
c) Resistor R1 is in series with an open
circuit.
d) Resistors R8 and R10 are in parallel
with a short circuit.
R4=
10[W]
R5=
25[W]
A
R6=
39[W]
R7=
56[W]
R9=
27[W]
B
R8=
33[W]
R10=
82[W]
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Which Statements Were True?
Find the resistance
seen at terminals A
and B of the circuit
shown.
R2=
15[W]
C
R1=
22[W]
R3=
49[W]
It turns out that all of the following
statements are true. We are going to
start by pointing out what statements
c) and d) mean. Pick one to start
with.
a) Resistors R2 and R5 are in series.
b) Resistors R8 and R10 are in series.
c) Resistor R1 is in series with an open
circuit.
d) Resistors R8 and R10 are in parallel
with a short circuit.
R4=
10[W]
R5=
25[W]
A
R6=
39[W]
R7=
56[W]
R9=
27[W]
B
R8=
33[W]
R10=
82[W]
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
In Series with an Open Circuit
Find the resistance seen at terminals
A and B of the circuit shown.
We said that resistor R1 was in series
with an open circuit. To understand this,
we need to remember that we are finding
the resistance seen at terminals A and B.
In other words, we are thinking about
what would happen if we connected a
source to those two terminals. However,
we are not going to connect anything to
terminal C. This is implied by the
wording of the problem, and is very
important. Thus, no current will flow
through R1, even when a source is
connected.
R2=
15[W]
C
R1=
22[W]
R3=
49[W]
R4=
10[W]
R5=
25[W]
A
R6=
39[W]
R7=
56[W]
R9=
27[W]
R8=
33[W]
R10=
82[W]
B
We could say that resistor R1 can be replaced with the series equivalent resistance of
infinite value. We could also say that the resistor has no current through it, and no voltage
across it, so it will have no effect. Either way, the resistor R1 can be removed.
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
In Parallel with a Short Circuit
Find the resistance seen at terminals
A and B of the circuit shown.
We said that resistors R8 and R10 were
in parallel with a short circuit. Here, there
is a wire across one or more resistors. If
we are looking from outside this
combination, which we are here, then we
can apply the parallel equivalent resistor
formula. A zero resistance in parallel with
any resistance will give a zero resistance.
Stated another way, no voltage will be
across R8 and R10, even when a source is
connected.
R2=
15[W]
C
R1=
22[W]
R3=
49[W]
R4=
10[W]
R5=
25[W]
A
R6=
39[W]
R7=
56[W]
R9=
27[W]
R8=
33[W]
R10=
82[W]
B
We could say that this combination can be replaced with the parallel equivalent
resistance of zero value. We could also say that the resistors have no voltage across them,
and therefore no current through them. Either way, the resistors R8 and R10 can be removed.
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Simplifying the Circuit
R2=
15[W]
Find the resistance
seen at terminals A
and B of the circuit
shown.
R3=
49[W]
Next, we are going to simplify the circuit.
This means that we can remove R1,
R8 and R10. Our simplified circuit is
shown here. Having made this
simplification, we can next work with
statement a).
a) Resistors R2 and R5 are in series. We
can replace them with an equivalent
resistor.
R4=
10[W]
R5=
25[W]
A
R6=
39[W]
R7=
56[W]
R9=
27[W]
B
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Series Combination Replacement
Find the resistance
seen at terminals A
and B of the circuit
shown.
Resistors R2 and R5 were in series. We
have replaced them with an
equivalent resistor, which we call R11.
Next, we recognize that R4 and R11
are in parallel, and that that parallel
combination is in series with R3.
Performing both of these two steps at
once, we can move to the next
equivalent, on the next slide, where
we include the equivalent resistor
R12.
R3=
49[W]
R4=
10[W]
R11=
40[W]
A
R6=
39[W]
R7=
56[W]
R9=
27[W]
B
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Find the resistance
seen at terminals A
and B of the circuit
shown.
We have shown the equivalent, with
resistor R12 in place. Then, the next
step is to find replacements for this
circuit. Which of the following
statements about this circuit are true?
a) Resistor R7 is in series with the
parallel combination of resistors R6
and R12.
b) Resistor R12 is in parallel with
resistor R6.
c) Resistor R6 is in parallel with resistor
R7.
d) The three resistors R6, R7, and R12 are
all in parallel.
Two More Steps Taken
R12=
57[W]
A
R6=
39[W]
R7=
56[W]
R9=
27[W]
B
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
You Said That Resistor R7 Is in Series With the Parallel
Combination of Resistors R6 and R12
Find the resistance
seen at terminals A
and B of the circuit
shown.
R12=
57[W]
A
iA
This is not correct. While resistors
R6 and R12 are indeed in parallel, that
parallel combination is not in series with
R7. This is a fairly subtle point made
more difficult for some students by the
diagonal wire. Still, we need to get this
correct. Here is the point: To be in
series, the same current must flow
through the series combination. That is
not the case here, because the current iA is
not zero. Remember, that a source can be
connected between terminals A and B.
This is not a correct step. Go back and
try again.
R6=
39[W]
R7=
56[W]
R9=
27[W]
B
As you look back, you may notice that we said
that the current through terminal A was not zero, but the
current through terminal C (now removed) was zero.
What’s the difference? The difference is that we are
looking for the resistance between terminals A and B,
which means that we are conceptually placing a source
across those two terminals.
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
You Said That
Resistor R12 Is in Parallel With Resistor R6
Find the resistance
seen at terminals A
and B of the circuit
shown.
This is correct as far as it goes. The
resistor R12 is indeed in parallel with R6.
However, we will save some time if we
recognize that R7 is also in parallel with
each of these two other resistors. We can
combine all three of them in the same
step, and save some time. Remember,
however, that the product-over-sum rule
does not work for more than two
resistors; rather we must use the inverse
of the sum of the inverses rule, at right.
Let’s use this rule to replace these three
resistors by a single resistor, which we
will call R13.
R12=
57[W]
A
R6=
39[W]
R7=
56[W]
R9=
27[W]
B
1
1
1
1



R13 R12 R6 R7
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
You Said That
Resistor R6 Is in Parallel With Resistor R7
Find the resistance
seen at terminals A
and B of the circuit
shown.
This is correct as far as it goes. The
resistor R7 is indeed in parallel with R6.
However, we will save some time if we
recognize that R12 is also in parallel with
each of these two other resistors. We can
combine all three of them in the same
step, and save some time. Remember,
however, that the product-over-sum rule
does not work for more than two
resistors; rather we must use the inverse
of the sum of the inverses rule, at right.
Let’s use this rule to replace these three
resistors by a single resistor, which we
will call R13.
R12=
57[W]
A
R6=
39[W]
R7=
56[W]
R9=
27[W]
B
1
1
1
1



R13 R12 R6 R7
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
You Said That
The Three Resistors R6, R7, and R12 Are All in Parallel
Find the resistance
seen at terminals A
and B of the circuit
shown.
This is correct. We can combine all
three of them in the same step, and save
some time. Remember, however, that the
product-over-sum rule does not work for
more than one resistor; rather we must
use the inverse of the sum of the inverses
rule, at right. Let’s use this rule to
replace these three resistors by a single
resistor, which we will call R13.
R12=
57[W]
A
R6=
39[W]
R7=
56[W]
R9=
27[W]
B
1
1
1
1



R13 R12 R6 R7
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Combining the Three Resistors
R6, R7, and R12 in Parallel
Find the resistance
seen at terminals A
and B of the circuit
shown.
We have combined all three of them
in the same step, replacing these three
resistors by a single resistor, which we
have called R13.
A
R13=
16[W]
R9=
27[W]
B
Now, it is probably clear the R9 and
R13 are in series, and our final answer is
given the next slide.
1
1
1
1



R13 R12 R6 R7
Dave Shattuck
University of Houston
Combining Resistors Yet Again
© Brooks/Cole Publishing Co.
Find the resistance
seen at terminals A
and B of the circuit
shown.
A
R14=
43[W]
B
We have combined the series resistors, and replaced them with an equivalent resistor,
which we called R14. At this point it is clear that the resistance between the terminals
A and B, which we will call REQ, is R14, or
REQ  R14  R9  R13  27[W]  16[W]  43[W].
Go to
Comments
Slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
What if I don’t see that resistors
are in series with open circuits, or
in parallel with short circuits?
• This is a difficult question, because this situation is so
common for beginning students. However, any other way
of combining these resistors will yield the wrong answer.
So, the reply, unfortunately, has to be that you need to
reconsider the rules of series and parallel until you do see
these things. There is only one right answer, and that is the
only answer we want.
• Much of this insight comes with practice.
Therefore, it is important to practice on
simple problems first, and then on more
difficult problems, until these issues
become clear.
Go back to
Overview
slide.