PWA_Mod01_Prob03_v05
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Transcript PWA_Mod01_Prob03_v05
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Problems With Assistance
Module 1 – Problem 3
Filename: PWA_Mod01_Prob03.ppt
This problem is adapted from:
Problem 1.34, page 38 in Circuits by A. Bruce Carlson
Brooks/Cole Thomson Learning
2000
ISBN: 0-534-37097-7
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:
• Kirchhoff’s Voltage Law
• Kirchhoff’s Current Law
• Ohm’s Law
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 1.3 & 1.4
• Electric Circuits 6th Ed. by Nilsson and Riedel: Sections
2.2 & 2.4
• Basic Engineering Circuit Analysis 6th Ed. by Irwin and
Wu: Section 2.1 & 2.2
• Fundamentals of Electric Circuits by Alexander and
Sadiku: Sections 2.2 & 2.4
• Introduction to Electric Circuits 2nd Ed. by Dorf: Sections
3-2 & 3-3
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 presentation:
• DPKC_Mod01_Part04
Next slide
Dave Shattuck
University of Houston
Problem Statement
© Brooks/Cole Publishing Co.
Find vb and R3 in Fig. P1.34, given that vs=12[V] and va = 6[V].
Figure P1.34
12[W]
R3
4[W]
+
vS
+
va
1[A]
+
vb
3[A]
-
-
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Solution – First Step – Where to Start?
Find vb and R3 in Fig. P1.34, given that vs=12[V] and va = 6[V].
Figure P1.34
12[W]
R3
How should we start
this problem? What
is the first step?
4[W]
+
vS
+
va
1[A]
+
vb
3[A]
-
-
Next slide
Dave Shattuck
University of Houston
Problem Solution – First Step
© Brooks/Cole Publishing Co.
Find vb and R3 in Fig. P1.34, given that vs=12[V] and va = 6[V].
Figure P1.34
How should we start this
problem? What is the
first step?
12[W]
Write a series of KVL,
KCL and Ohm’s Law
Equations
b)
Fill in the given
information in the
circuit schematic
c)
Define variables for
voltages and currents
needed to solve the
problem
d)
Convert the sources to
their equivalent
resistances
R3
4[W]
+
a)
vS
+
va
1[A]
+
vb
3[A]
-
-
Dave Shattuck
University of Houston
Your choice for First Step was –
Write a series of KVL, KCL and Ohm’s Law Equations
© Brooks/Cole Publishing Co.
Find vb and R3 in Fig. P1.34, given that vs=12[V] and va = 6[V].
Figure P1.34
This is a reasonable thing to do, but
it is too soon to do it.
12[W]
R3
4[W]
+
vS
+
va
1[A]
+
vb
3[A]
It is important to define variables
before writing equations using
them.
It is also useful to include the
information given in this
problem directly into the
circuit schematic. It is a good
idea to get used to thinking in
terms of these diagrams.
Go back and try again.
-
-
Dave Shattuck
University of Houston
Your choice for First Step was –
Fill in the given information in the circuit schematic
© Brooks/Cole Publishing Co.
Find vb and R3 in Fig. P1.34, given that vs=12[V] and va = 6[V].
Figure P1.34
12[W]
4[W]
+
vS=
12[V]
R3
+
va=6[V]
1[A]
+
vb
3[A]
-
This is a good first step. Generally,
it is a reasonable first step to
include the information that is
given in the problem in a way
that helps us move forward.
Here, this means putting the
information about vs and va in
the schematic. This has been
done here.
Now what is the second step?
a)
Write a series of KVL, KCL and
Ohm’s Law Equations
c)
Define variables for voltages and
currents needed to solve the
problem
d)
Convert the sources to their
equivalent resistances
-
Dave Shattuck
University of Houston
Your choice for Second Step was –
Write a series of KVL, KCL and Ohm’s Law Equations
© Brooks/Cole Publishing Co.
Find vb and R3 in Fig. P1.34, given that vs=12[V] and va = 6[V].
Figure P1.34
This is a reasonable thing to do, but
it is too soon to do it.
12[W]
4[W]
+
vS=
12[V]
R3
+
va=6[V]
1[A]
Go back and try again.
+
vb
3[A]
-
It is important to define variables
before writing equations using
them.
-
Your choice for Second Step was –
© Brooks/Cole Publishing Co.
Define variables for voltages and currents needed to solve
the problem
Dave Shattuck
University of Houston
Find vb and R3 in Fig. P1.34, given that vs=12[V] and va = 6[V].
Figure P1.34
+
i12
v12
12[W]
R3
i3
4[W]
+
vS=
12[V]
+
v3
-
-
+
va=6[V]
1[A]
+
vb
3[A]
-
-
This is the best choice. It is
important to define variables
before writing equations using
them.
Which ones should be included?
There is no single answer to this
question. The key is to recognize
that we must define any variable
before using it in an equation. We
have included a few that we think
that we will need in the diagram
here. If we missed any, we just
need to add them before writing the
equation that uses it. The next step
is to actually write the equations.
Dave Shattuck
University of Houston
Your choice for Second Step was –
Convert the sources to their equivalent resistances
© Brooks/Cole Publishing Co.
Find vb and R3 in Fig. P1.34, given that vs=12[V] and va = 6[V].
Figure P1.34
12[W]
4[W]
+
vS=
12[V]
R3
+
va=6[V]
1[A]
+
vb
3[A]
-
-
This is a bad choice. There is not
an equivalent resistance for an
ideal source. Sources and
resistances are completely
different things. Try again.
Your choice for First Step was –
© Brooks/Cole Publishing Co.
Define variables for voltages and currents needed to solve
the problem
Dave Shattuck
University of Houston
Find vb and R3 in Fig. P1.34, given that vs=12[V] and va = 6[V].
Figure P1.34
+
i12
+
v3
v12
12[W]
R3
i3
4[W]
+
-
vS
-
+
va
1[A]
+
vb
3[A]
This is a good first step. It is important to
define variables before writing
equations using them.
Which ones should be included?
There is no single answer to this question.
The key is to recognize that we must
define any variable before using it in an
equation. We have included a few that we
think that we will need in the diagram here.
If we missed any, we just need to add them
before writing the equation that uses it.
The next step is to actually write the
equations.
Now what is the second step?
-
a) - Write a series of KVL, KCL and Ohm’s Law
Equations
b)
Fill in the given information in the circuit schematic
d)
Convert the sources to their equivalent resistances
Dave Shattuck
University of Houston
Your choice for Second Step was –
Write a series of KVL, KCL and Ohm’s Law Equations
© Brooks/Cole Publishing Co.
Find vb and R3 in Fig. P1.34, given that vs=12[V] and va = 6[V].
Figure P1.34
+
i12
+
v3
v12
12[W]
R3
i3
4[W]
+
-
vS
-
+
va
1[A]
+
vb
3[A]
-
-
This is a reasonable thing to do, but
it is too soon to do it.
Generally, it is a reasonable first
step to include the information
that is given in the problem in
a way that helps us move
forward. Here, this means
putting the information about
vs and va in the schematic. Go
back and try again.
Dave Shattuck
University of Houston
Your choice for Second Step was –
Fill in the given information in the circuit schematic
© Brooks/Cole Publishing Co.
Find vb and R3 in Fig. P1.34, given that vs=12[V] and va = 6[V].
Figure P1.34
+
i12
v12
12[W]
R3
i3
4[W]
+
vS=
12[V]
+
v3
-
-
+
va=6[V]
1[A]
+
vb
3[A]
-
-
This is a good choice. Generally, it
is reasonable to include the
information that is given in the
problem in a way that helps us
move forward. Here, this
means putting the information
about vs and va in the
schematic. This has been done
here.
Next, we can write the equations.
Dave Shattuck
University of Houston
Your choice for Second Step was –
Convert the sources to their equivalent resistances
© Brooks/Cole Publishing Co.
Find vb and R3 in Fig. P1.34, given that vs=12[V] and va = 6[V].
Figure P1.34
+
i12
+
v3
v12
12[W]
R3
i3
4[W]
+
-
vS
-
+
va
1[A]
+
vb
3[A]
-
-
This is a bad choice. There is not
an equivalent resistance for an
ideal source. Sources and
resistances are completely
different things. Try again.
Dave Shattuck
University of Houston
Your choice for First Step was –
Convert the sources to their equivalent resistances
© Brooks/Cole Publishing Co.
Find vb and R3 in Fig. P1.34, given that vs=12[V] and va = 6[V].
Figure P1.34
12[W]
R3
4[W]
+
vS
+
va
1[A]
+
vb
3[A]
-
-
This is a bad choice. There is not
an equivalent resistance for an
ideal source. Sources and
resistances are completely
different things. Try again.
Dave Shattuck
University of Houston
Writing the Equations
© Brooks/Cole Publishing Co.
Find vb and R3 in Fig. P1.34, given that vs=12[V] and va = 6[V].
Figure P1.34
+
i12
v12
12[W]
R3
i3
4[W]
+
vS=
12[V]
+
v3
-
-
+
va=6[V]
1[A]
+
vb
3[A]
-
-
Now, we have agreed that we are
ready to write equations. We
can use KVL, KCL and Ohm’s
Law. How many do we write?
Which ones first?
There are many answers. While
there is some benefit to simply
writing equations wherever we
can, it is better to try to see
small steps that take us to the
answer. It is easier to write
and solve a series of 6 single
equations in a single unknown,
than to writing all 6 equations
and attempting to solve them
simultaneously. Let’s try this.
Next slide
Dave Shattuck
University of Houston
Writing the Equations – Picking the First One to Write
© Brooks/Cole Publishing Co.
Find vb and R3 in Fig. P1.34, given that vs=12[V] and va = 6[V].
Figure P1.34
+
i12
v12
12[W]
R3
i3
4[W]
+
vS=
12[V]
+
v3
-
-
+
va=6[V]
1[A]
+
vb
3[A]
We would like to write an equation
for a single unknown variable,
that has only that unknown in
it. The variable should also be
useful to me in solving the
problem.
Note that we can write a KVL
around the left hand loop, and
the only unknown will be v12.
The loop is shown in red.
Let’s write the equation:
vs v12 va 0
-
-
12[V] v12 6[V] 0
v12 6[V]
Next slide
Dave Shattuck
University of Houston
Writing the Equations – Picking the Second One to Write
© Brooks/Cole Publishing Co.
Find vb and R3 in Fig. P1.34, given that vs=12[V] and va = 6[V].
Figure P1.34
+
i12
R3
i3
4[W]
+
vS=
12[V]
v12=6[V]
12[W]
+
v3
-
-
+
va=6[V]
1[A]
+
vb
3[A]
-
-
So, now we know that v12 = 6[V].
Let’s look for another useful
equation.
Note that we can write Ohm’s Law
for the 12[W] resistor, and get
i12. Let’s write the equation:
v12
12[W],
i12
or solving, we get
v12
i12
0.5[A]
12[W]
Next slide
Dave Shattuck
University of Houston
Writing the Equations – Picking the Next One to Write
© Brooks/Cole Publishing Co.
Find vb and R3 in Fig. P1.34, given that vs=12[V] and va = 6[V].
Figure P1.34
+
i12
vS=
12[V]
v12=6[V]
12[W]
+
v4 4[W]
+
-
-
+
va=6[V]
1[A]
+
v3
R3
i3
i4
+
vb
3[A]
-
-
Next, knowing i12, we can write
KCL at the middle node to get
the current through the 4[W]
resistor.
However, if we try to write this
KCL, we find that we have not
yet defined the current through
the 4[W] resistor. This needs
to be our next step.
We have done this here. In
anticipation of needing it
shortly, we have also defined
the voltage across the 4[W]
resistor.
Next slide
Dave Shattuck
University of Houston
Writing the Equations – Writing the Next One
© Brooks/Cole Publishing Co.
Find vb and R3 in Fig. P1.34, given that vs=12[V] and va = 6[V].
Figure P1.34
We will write the KCL at the
middle node, in red.
+
i12
i3
v4 4[W]
+
-
+
vS=
12[V]
v12=6[V]
12[W]
+
va=6[V]
1[A]
i4
+
vb
3[A]
-
-
+
v3
i12 i4 1[A] 0
R3
i4 1 0.5 1.5[A]
-
Note that our answer for i4 came out
negative. Some students are
tempted to say that they have
defined i4 incorrectly, and need
to go back and redefine it.
Don’t! You can’t define a
reference polarity incorrectly.
This is a fine answer.
Next slide
Dave Shattuck
University of Houston
Writing the Equations – Getting Ohm’s Law Right
© Brooks/Cole Publishing Co.
Find vb and R3 in Fig. P1.34, given that vs=12[V] and va = 6[V].
Figure P1.34
Next, we write Ohm’s Law, and get
+
i12
vS=
12[V]
v12=6[V]
12[W]
+
v4 4[W]
+
-
-
+
va=6[V]
1[A]
+
v3
R3
i3
i4
+
vb
3[A]
-
-
Next slide
v4
4[W]
i4
v4 i4 4[W]
1.5 4 6[V]
Note that when we defined v4 and
i4, we defined them in the
active convention for the
resistor. Again, we can’t
define reference polarities
incorrectly; we just need to use
the definitions that result
correctly. Here, this means
writing Ohm’s Law with a
minus sign.
Dave Shattuck
University of Houston
Writing the Equations – One Answer
© Brooks/Cole Publishing Co.
Find vb and R3 in Fig. P1.34, given that vs=12[V] and va = 6[V].
Figure P1.34
+
i12
vS=
12[V]
v12=6[V]
12[W]
+
v4 4[W]
+
-
-
+
va=6[V]
1[A]
+
v3
R3
i3
i4
+
vb
3[A]
-
-
Next, we can write KVL for the
bottom right hand loop, and
get one of our desired answers,
va v 4 vb 0
vb va v 4
vb 6 6 0
Interestingly, the voltage vb is zero.
This is the voltage across the
right hand current source.
Remember that the voltage
across a current source is not
always zero; see va for
example.
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Writing the Equations – Another KVL
Find vb and R3 in Fig. P1.34, given that vs=12[V] and va = 6[V].
Figure P1.34
Next, we can write KVL for the
upper right hand loop, and get
+
i12
vS=
12[V]
v12=6[V]
12[W]
+
v4 4[W]
+
-
-
+
va=6[V]
1[A]
+
v3
R3
i3
i4
+
vb
v3 v12 v 4
v3 6 6 12[V].
Note that the positive answer is the
only correct answer.
3[A]
-
v12 v 3 v4 0
-
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Writing the Equations – Another KCL
Find vb and R3 in Fig. P1.34, given that vs=12[V] and va = 6[V].
Figure P1.34
Next, we can write KCL for the
right hand node, and get
+
i12
vS=
12[V]
v12=6[V]
12[W]
+
v4 4[W]
+
-
-
+
va=6[V]
1[A]
+
v3
R3
i3
i4
+
vb
i3 i4 3[A]
i3 1.5 3 1.5[A].
Note that the positive answer is the
only correct answer.
3[A]
-
i3 i4 3[A] 0
-
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Writing the Equations – Second Answer
Find vb and R3 in Fig. P1.34, given that vs=12[V] and va = 6[V].
Figure P1.34
+
i12
vS=
12[V]
v12=6[V]
12[W]
+
v4 4[W]
+
-
-
+
va=6[V]
1[A]
+
v3
R3
i3
i4
+
vb
Finally, we can write Ohm’s Law
for R3, and get
v3
R3
i3
12[V]
R3
8[W].
1.5[A]
3[A]
-
-
Note that here we had defined the
reference polarities for i3 and
v3 in the passive convention,
and the positive sign in Ohm’s
Law resulted from that.
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
How did we know which equations to
write, and in what order?
• This is an excellent question. In this problem, we looked at the
voltages and currents that were given, and saw an equation that we
could write for a single unknown. We just started from there.
• This will not always happen. When problems become complex, we
will want to find a systematic way of knowing what equations we have
to write, and write only that minimum number of equations. We will
call these systematic ways The Node-Voltage Method and The MeshCurrent Method. We will learn these methods later.
• For now, we have no choice but to try writing equations,
and seeing what we need to solve. Try to use some insight.
Go to Next
Note
Go back to
Overview
slide.
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Why do we have to worry about the sign
in Everything?
• This is one of the central themes in circuit analysis. The polarity, and
the sign that goes with that polarity, matters. The key is to find a way to
get the sign correct every time.
• This is why we need to define reference polarities for every voltage and
current.
• This is why we need to take care about what relationship we have used
to assign reference polarities (passive sign convention and active sign
convention).
An analogy: Suppose I was going to give you $10,000. This
would probably be fine with you. However, it will matter a
great deal which direction the money flows. You will care a
great deal about the sign of the $10,000 in this transaction.
If I give you -$10,000, it means that you are giving $10,000 to
me. This would probably not be fine with you!
Go back to
Overview
slide.