Download PWA_Mod02_Prob05_v04

Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Problems With Assistance
Module 2 – Problem 5
Filename: PWA_Mod02_Prob05.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.
Problem Statement
Find an expression for the
power delivered by the source
in this circuit as a function of
the value of the source, iS.
R1=
50[W]
R2=
33[W]
R3=
47[W]
Rm=
5[W]
iS
R4=
22[W]
RX=
10[W]
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
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, iS.
R1=
50[W]
R2=
33[W]
R3=
47[W]
Rm=
5[W]
iS
R4=
22[W]
RX=
10[W]
How should we start this
problem? What is the first
step?
Next slide
Dave Shattuck
University of Houston
Problem Solution – First Step
© Brooks/Cole Publishing Co.
Find an expression for the
power delivered by the source
in this circuit as a function of
the value of the source, iS.
How should we start this
problem? What is the best
first step?
1. Write a series of KVL or
KCL equations.
2. Combine resistors in series
and in parallel to simplify
the circuit.
3. Use delta-to-wye
transformations to simplify
this circuit.
4. Define currents and voltages
for each of the elements in
the circuit.
R1=
50[W]
R2=
33[W]
R3=
47[W]
Rm=
5[W]
iS
R4=
22[W]
RX=
10[W]
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
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, iS.
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.
R1=
50[W]
R2=
33[W]
R3=
47[W]
Rm=
5[W]
iS
R4=
22[W]
RX=
10[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 an expression for the
power delivered by the source
in this circuit as a function of
the value of the source, iS.
This is not a 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 voltage across
the current source, we can get this
by converting the circuit
connected to the source to a single
resistor. However, there are no
series resistors in this circuit, and
no parallel resistors.
Go back and try again.
R1=
50[W]
R2=
33[W]
R3=
47[W]
Rm=
5[W]
iS
R4=
22[W]
RX=
10[W]
Verify for yourself that there are no series resistors in this
circuit, and that there are no parallel resistors in this circuit.
Notice as well that this is not given to us as being a “balanced
bridge”, as in a Wheatstone Bridge. Thus, there is current
through Rm. The resistors R2 and R4 are not in series. The
resistors R3 and RX are not in series.
Dave Shattuck
University of Houston
Your Choice for First Step Was –
Use Delta-to-wye Transformations to Simplify This Circuit
© Brooks/Cole Publishing Co.
Find an expression for the
power delivered by the source
in this circuit as a function of
the value of the source, iS.
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 voltage across
the current source, we can get this
by converting the circuit
connected to the source to a single
resistor. We can use the delta-towye transformations, or wye-todelta transformations, as the first
step.
Let’s go ahead and do this.
R1=
50[W]
R2=
33[W]
R3=
47[W]
Rm=
5[W]
iS
R4=
22[W]
RX=
10[W]
There are at least two different delta configurations in this
circuit, and at least three different wye configurations. Of
these, any transformation would make the circuit subject to
further simplification by replacing series combinations or
parallel combinations of resistors with their equivalents.
Dave Shattuck
University of Houston
Your Choice for First Step Was –
Define Currents and Voltages for Each of the Elements in the Circuit
© Brooks/Cole Publishing Co.
Find an expression for the
power delivered by the source
in this circuit as a function of
the value of the source, iS.
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.
R1=
50[W]
R2=
33[W]
R3=
47[W]
Rm=
5[W]
iS
R4=
22[W]
RX=
10[W]
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Using Delta-to-Wye Transformations
Find an expression for the
power delivered by the source
in this circuit as a function of
the value of the source, iS.
We have decided to simplify
this circuit by performing deltato-wye transformations. In this
problem, it really does not matter
which one we choose. Therefore,
we will simply pick one, and
make the transformation. We
will transform the delta
configuration that is marked in
red in this figure. Let’s prepare
to make the transformation.
R1=
50[W]
R2=
33[W]
R3=
47[W]
Rm=
5[W]
iS
R4=
22[W]
RX=
10[W]
Generally, we might want to redraw the equivalent after the
transformation, for example with a quick sketch, just to be
sure that this transformation will be of benefit. This could be
done quickly, and help us choose the right approach. You
might even be able to do this in your mind.
Next Slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Preparing to Make a Delta-to-Wye Transformations
A
Find an expression for the
power delivered by the source
in this circuit as a function of
the value of the source, iS.
We have decided to simplify
this circuit by performing deltato-wye transformations. To
prepare to make the
transformation, we want to name
the terminals. The point is this:
we are going to be taking
resistors out and putting others in
their place, and there will be
three different values. We need
to be careful to get them all in the
proper places. Using appropriate
names for the terminals and
resistors will help us with this.
R1=
50[W]
R2=
33[W]
R3=
47[W]
Rm=
5[W]
iS
B
C
R4=
22[W]
RX=
10[W]
In these modules, we have labeled the terminals A, B, and C
when we were defining the equations for the delta-to-wye
transformation. Thus, we follow that rule. If possible, it is a
good idea to have slides 36 and 37 from DPKC Part 1 Module
2 available, or other versions available, when working
problems.
Next Slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Renaming Resistors
A
Find an expression for the
power delivered by the source
in this circuit as a function of
the value of the source, iS.
We also choose to rename
the resistors. We do this so that
we can use the formulas we have
derived, and never substitute the
wrong value in. We leave the
existing names in place as well.
Since there were no resistors
with these same names before,
this works satisfactorily.
R1=
50[W] R2=RC=
33[W]
R3=RB=
47[W]
Rm=RA=
5[W]
iS
B
C
R4=
22[W]
RX=
10[W]
We have used the rule that the resistor opposite to node A is
called resistor RA, the resistor opposite to node B is called
resistor RB, the resistor opposite to node C is called resistor RC.
This is the same rule we used when we derived the equations
for the delta-to-wye transformation.
A
Dave Shattuck
University of Houston
R1=
50[W]
© Brooks/Cole Publishing Co.
R1'
The Formulas
Find an expression for the
power delivered by the source
in this circuit as a function of
the value of the source, iS.
iS
R2'
Now, we use the formulas
below to make the
transformation. The transformed
circuit is shown.
B
C
R4=
22[W]
RB RC
R1 ' 
RA  RB  RC
RA RC
R2 ' 
RA  RB  RC
RA RB
R3 ' 
RA  RB  RC
R3'
Next Slide
RX=
10[W]
Notice that we have used primes (') in the
names, R1', R2', and R3', since there will be
confusion with the names of resistors that
were already in the circuit. Other names
would be possible, of course, but these names
make keeping things straight a little easier,
since they are close to the ones in our
formulas.
A
Dave Shattuck
University of Houston
R1=
50[W]
© Brooks/Cole Publishing Co.
The Values for the
Transformed Circuit
Find an expression for the
power delivered by the source
in this circuit as a function of
the value of the source, iS.
R1'=
18[W]
iS
Substituting values into
these equations, we get
R1 ' 
RB RC
47[W]33[W]

 18[W]
RA  RB  RC 5[W]  47[W]  33[W]
RA RC
5[W]33[W]
R2 ' 

 1.9[W]
RA  RB  RC 5[W]  47[W]  33[W]
R3 ' 
RA RB
5[W]47[W]

 2.8[W]
RA  RB  RC 5[W]  47[W]  33[W]
R2'=
1.9[W]
R3'=
2.8[W]
B
C
R4=
22[W]
RX=
10[W]
Now, we are ready to make
series and parallel combinations. At
this point, it is probably clear that R1
and R1' are in series, R2' and R4 are in
series, and R3' and RX are in series.
Let's combine these next.
Next Slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Combining Series Resistors
Find an expression for the
power delivered by the source
in this circuit as a function of
the value of the source, iS.
We replace the series
combination R1 and R1' with their
equivalent, which we will call R5.
We replace the series
combination R2' and R4 with their
equivalent, which we will call R6.
We replace the series
combination R3' and RX with their
equivalent, which we will call R7.
R5=
68[W]
iS
R6=
24[W]
R7=
13[W]
The next step is to replace the parallel
combination of R6 and R7 with their
equivalent resistor which we will call R8.
Next Slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Combining Parallel Resistors
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 now going to replace
the parallel combination of R6 and
R7 with their equivalent resistor
which we will call R8. It is clear
now that R5 and R8 are in series,
and this combination can be
replaced with an equivalent
resistor, which we will call R9.
R5=
68[W]
iS
R8=
8.4[W]
Next Slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
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.
R9=
76[W]
iS
We are now going to replace
the series combination of R5 and
R8 with an equivalent resistor,
which we will call R9.
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 R9, so we can just find the
power absorbed by R9.
pDel ,iS  pAbs , R9  iS R9  iS 76[W].
2
2
Go to
Comments Slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
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.
• While this is true, we still recommend that you learn this
approach to solving circuits. There are
many ways in which equivalent circuits
can help us, and they are crucial tools.
They are worth the time it take to
understand them.
Go back to
Overview
slide.