Download Problems With Assistance, Module 4, Problem 3

Filename: PWA_Mod04_Prob03.ppt
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Find the Norton Equivalent as seen by resistor R7.
R 2=
30[W]
R1 =
50[W]
Problems With Assistance
i =
R=
Module 45[A]– Problem
3
27[W]
S1
3
vS1=
6[V]
+
-
R4 =
60[W]
R5=
22[W]
-
vS2=
12[V]
+
R7=
10[W]
Go
straight to
the First
Step
iS2=
2[A]
Go
Rstraight
to
6=
the
40[W]
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
• Norton’s Theorem
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_Part03
Next slide
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Problem Statement
Find the Norton Equivalent as seen by resistor R7.
R 2=
30[W]
R1 =
50[W]
iS1=
5[A]
vS1=
6[V]
+
R4 =
60[W]
R5=
22[W]
-
vS2=
12[V]
R 3=
27[W]
+
R7=
10[W]
iS2=
2[A]
R6=
40[W]
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Solution – First Step – Where to Start?
Find the Norton Equivalent as seen by resistor R7.
iS1=
5[A]
vS1=
6[V]
+
R 3=
27[W]
R4 =
60[W]
R5=
22[W]
-
vS2=
12[V]
How should
we start this
problem?
What is the
first step?
R 2=
30[W]
R1 =
50[W]
+
R7=
10[W]
iS2=
2[A]
R6=
40[W]
Next slide
Dave Shattuck
University of Houston
How should we start
this problem?
What is the first
step?
© Brooks/Cole Publishing Co.
Problem Solution – First Step
a)
Define the opencircuit voltage.
b)
Label the
terminals of
resistor R7 and
remove it.
c)
Define the shortcircuit current.
d)
Combine resistors
R2 and R3 in
parallel.
e)
Combine resistors
R6 and R7 in
parallel.
Find the Norton Equivalent as seen by resistor R7.
R 2=
30[W]
R1 =
50[W]
iS1=
5[A]
vS1=
6[V]
+
R4 =
60[W]
R5=
22[W]
-
vS2=
12[V]
R 3=
27[W]
+
R7=
10[W]
iS2=
2[A]
R6=
40[W]
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Your choice for First Step –
Define the open-circuit voltage
The key here is that
we are finding an
equivalent seen by
something; when we
do this, the first step
should always be to
get rid of that
something. Here, we
need to get rid of R7
before doing anything
else.
Find the Norton Equivalent as seen by resistor R7.
R 2=
30[W]
R1 =
50[W]
iS1=
5[A]
vS1=
6[V]
+
R5=
22[W]
-
vS2=
12[V]
R 3=
27[W]
R4 =
60[W]
+
R7=
10[W]
This is not a good
choice for the first
step.
iS2=
2[A]
R6=
40[W]
Remember that in
circuits, a component
or device does not
“see” itself. We
suggest that you go
back and try again.
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Your choice for First Step –
Label the terminals of resistor R7 and remove it
This is the best choice
for this problem.
The key in these kinds
of problems where we
have been asked for an
equivalent “as seen by”
something, is to
remove the something
as the first step.
Find the Norton Equivalent as seen by resistor R7.
R 2=
30[W]
R1 =
50[W]
iS1=
5[A]
vS1=
6[V]
+
R4 =
60[W]
R5=
22[W]
-
vS2=
12[V]
R 3=
27[W]
+
R7=
10[W]
iS2=
2[A]
R6=
40[W]
The something is
assumed to not “see
itself”, and thus it
needs to be removed.
Its terminals become
the place where the
equivalent is found.
Failure to do this can
lead to errors. Let’s go
ahead and remove R7.
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Your choice for First Step –
Define the short-circuit current
This is could be the
first step, since R7 will
not affect the short
circuit current.
However, it is a
dangerous choice. The
key is get rid of R7
right away, and then
we don’t have to worry
about whether it has an
effect or not.
Find the Norton Equivalent as seen by resistor R7.
R 2=
30[W]
R1 =
50[W]
iS1=
5[A]
vS1=
6[V]
+
R4 =
60[W]
R5=
22[W]
-
vS2=
12[V]
R 3=
27[W]
+
R7=
10[W]
So, let’s go back and
try again.
iS2=
2[A]
R6=
40[W]
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Your choice for First Step was –
Combine resistors R2 and R3 in parallel
This is a valid step,
but is not a good
choice for the first
step.
Find the Norton Equivalent as seen by resistor R7.
R 2=
30[W]
R1 =
50[W]
iS1=
5[A]
vS1=
6[V]
+
R 3=
27[W]
R4 =
60[W]
R5=
22[W]
-
vS2=
12[V]
The key here is that
we are finding the
equivalent seen by R7.
Thus, we need to
handle R7 right at the
beginning. Therefore,
although R2 and R3 are
in parallel, we
recommend that you
go back and try again.
+
R7=
10[W]
iS2=
2[A]
R6=
40[W]
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Your choice for First Step was –
Combine resistors R6 and R7 in parallel
This is not a good
choice.
The problem here is
that we are finding the
equivalent as seen by
R7. If we combine R6
and R7 in parallel, then
R7 will be gone, and
we will not be able to
complete the problem
correctly.
Find the Norton Equivalent as seen by resistor R7.
R 2=
30[W]
R1 =
50[W]
iS1=
5[A]
vS1=
6[V]
+
R4 =
60[W]
R5=
22[W]
-
vS2=
12[V]
R 3=
27[W]
+
R7=
10[W]
Please go back and try
again.
iS2=
2[A]
R6=
40[W]
Dave Shattuck
University of Houston
We have named the terminals of R7, and then
removed resistor R7 from the circuit. We
named the terminals A and B.
© Brooks/Cole Publishing Co.
Label the terminals of resistor
R7 and remove it
Let’s go to the next slide and consider what to
solve for first.
Find the Norton Equivalent as seen by resistor R7.
R 2=
30[W]
R1 =
50[W]
iS1=
5[A]
vS1=
6[V]
R 3=
27[W]
+
R4 =
60[W]
R5=
22[W]
A
-
vS2=
12[V]
iS2=
2[A]
+
B
R6=
40[W]
Next slide
Dave Shattuck
University of Houston
We need to consider
what to solve for
first. Click on your
choice from the
choices below.
© Brooks/Cole Publishing Co.
What Should We Solve for
First?
The open-circuit
voltage.
Find the Norton Equivalent as seen by resistor R7.
R 2=
30[W]
R1 =
50[W]
iS1=
5[A]
vS1=
6[V]
The short-circuit
current.
The equivalent
resistance.
R 3=
27[W]
+
The total power
dissipated.
R4 =
60[W]
R5=
22[W]
A
-
vS2=
12[V]
iS2=
2[A]
+
B
R6=
40[W]
You Chose: The open-circuit voltage
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
You said that the first thing to solve for would be the open-circuit voltage. This solution is
not the easiest available to us. For example, this would have five essential nodes, and
would require four simultaneous equations. There are better choices. Please go back
and try again.
Find the Norton Equivalent as seen by resistor R7.
R 2=
30[W]
R1 =
50[W]
iS1=
5[A]
vS1=
6[V]
R 3=
27[W]
+
R4 =
60[W]
R5=
22[W]
A
-
vS2=
12[V]
iS2=
2[A]
+
B
R6=
40[W]
You Chose: The short-circuit current
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
You said that the first thing to solve for would be the short-circuit current. This is a good
choice. Once we apply the short circuit, the R6 resistor could be neglected, we would
have four essential nodes, and three equations, one of which would be very simple.
Let’s find the short-circuit current.
Find the Norton Equivalent as seen by resistor R7.
R 2=
30[W]
R1 =
50[W]
iS1=
5[A]
vS1=
6[V]
R 3=
27[W]
+
R4 =
60[W]
R5=
22[W]
A
-
vS2=
12[V]
iS2=
2[A]
+
B
R6=
40[W]
You Chose: The equivalent resistance
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
You said that the first thing to solve for would be the equivalent resistance. This is a good
choice. However, while this will be the next thing we will choose, there is another
thing that must be found as well; it will be either the open-circuit voltage, or the shortcircuit current. Go back, and determine which you think would be better.
Find the Norton Equivalent as seen by resistor R7.
R 2=
30[W]
R1 =
50[W]
iS1=
5[A]
vS1=
6[V]
R 3=
27[W]
+
R4 =
60[W]
R5=
22[W]
A
-
vS2=
12[V]
iS2=
2[A]
+
B
R6=
40[W]
You Chose: The total power dissipated
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
You said that the first thing to solve for would be the total power dissipated. This is not a
good choice. The total power dissipated is not useful to us at this point. Go back and
try again.
Find the Norton Equivalent as seen by resistor R7.
R 2=
30[W]
R1 =
50[W]
iS1=
5[A]
vS1=
6[V]
R 3=
27[W]
+
R4 =
60[W]
R5=
22[W]
A
-
vS2=
12[V]
iS2=
2[A]
+
B
R6=
40[W]
Finding the Short-Circuit Current
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Let’s find the short-circuit current. The first step is to define the polarity. That is done in
the circuit shown here. We have also defined some voltages to help us get this value.
Find the Norton Equivalent as seen by resistor R7.
R 2=
30[W]
R1 =
50[W]
iS1=
5[A]
vS1=
6[V]
R 3=
27[W]
+
+
vD
R4 =
60[W]
R5=
22[W]
-
+
A
-
vS2=
12[V]
vC
+
-
iS2=
2[A]
iSC
R 6=
40[W]
B
Next slide
Writing the KCL’s (The Node-Voltage Method)
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
We wish to find vC and vD first, and then use those values to get iSC. We write the equations
here:
vC
vC  12[V]

 5[A]  0, and
22[W]
60[W]
vD
v
v  6[V]  12[V]
 D  5[A]  D
 0.
30[W] 27[W]
50[W]
Find the Norton Equivalent as seen
by resistor R7.
R 2=
30[W]
R1 =
50[W]
iS1=
5[A]
vS1=
6[V]
R 3=
27[W]
+
R5=
22[W]
+
-
This simplifies to yield
A
vC
vD  90.4[mS]  5[A]  0.12[A].
-
+
vS2=
12[V]
vC  62.1[mS]  5[A]  0.2[A], and
+
vD
R4 =
60[W]
Solving for vC and vD we get
iS2=
2[A]
iSC
B
R 6=
40[W]
vC  77.3[V], and
vD  56.6[V].
Next slide
Finding iSC
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Using the values for vC and vD that are shown here, we can find iSC. We write a KCL for the
red dashed closed surface:
vC  77.3[V], and

vD  56.6[V].
 77.3[V ]   56.6[V]   56.6[V] 



  2[A]  iSC .
 22[W]   27[W]   30[W] 
Find the Norton Equivalent as
seen by resistor R7.
R 2=
30[W]
R1 =
50[W]
Solving for iSC we get
iS1=
5[A]
vS1=
6[V]
R 3=
27[W]
+
R5=
22[W]
-
+
A
vC
+
-
iSC  1.53[A].
+
vD
R4 =
60[W]
vS2=
12[V]
vC
v
v
 D  D  2[A]  iSC  0, or
22[W] 27[W] 30[W]
iS2=
2[A]
iSC
R 6=
40[W]
B
Next slide
Next Step
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
We have found one of the three possible quantities. The best choice for the next quantity to
find is probably the equivalent resistance. To find this, we set the independent sources
equal to zero, which will simplify the circuit a great deal. Let’s do this in the next
slide.
Find the Norton Equivalent as seen by resistor R7.
iS1=
5[A]
vS1=
6[V]
iSC  1.53[A].
R 2=
30[W]
R1 =
50[W]
R 3=
27[W]
+
R4 =
60[W]
R5=
22[W]
A
-
vS2=
12[V]
iS2=
2[A]
+
B
R6=
40[W]
Next slide
Dave Shattuck
University of Houston
Finding the Equivalent Resistance, Step 1
© Brooks/Cole Publishing Co.
To find the equivalent resistance, we set the independent sources equal to zero. With this we
have the circuit below. We note that R4 and R5 are in series, and R2 and R3 are in
parallel. Let’s combine these and redraw.
Find the Norton Equivalent as seen by resistor R7.
R 3=
27[W]
R4=
60[W]
iSC  1.53[A].
R2 =
30[W]
R1=
50[W]
next slide
R 5=
22[W]
A
R6=
40[W]
B
Next slide
Dave Shattuck
University of Houston
Finding the Equivalent Resistance, Step 2
© Brooks/Cole Publishing Co.
Having combined these resistors, it should be clear that the series combination of R1 and R9
is in parallel with R8, which is in parallel with R6. Note that each is connected between
terminals A and B. Thus we have the equivalent resistance, as seen by A and B, as
Find the Norton Equivalent as seen by resistor R7.
R9 =
14.2[W]
R1=
50[W]
REQ  ( R1  R9 ) || R8 || R6 
REQ  (50[W]  14.2[W]) || 82[W] || 40[W] 
REQ  19.0[W].
iSC  1.53[A].
R 8=
82[W]
A
R6=
40[W]
B
Next slide
The Norton Equivalent
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
The Norton equivalent is a current source equal to the short-circuit
current, in parallel with the equivalent resistance. Thus, we
have as the answer the circuit drawn below.
Find the Norton Equivalent as seen by resistor R7.
iS1=
5[A]
vS1=
6[V]
REQ  19.0[W].
R 2=
30[W]
R1 =
50[W]
iSC  1.53[A].
R 3=
27[W]
+
R4 =
60[W]
R5=
22[W]
Check this
solution
A
-
vS2=
12[V]
+
iS2=
2[A]
R7=
10[W]
Go back to
Problem
Statement
R6=
40[W]
B
A
RN=
19[W]
iN=
1.53[A]
R7=
10[W]
B
Checking the Solution
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Let’s check our answer to see if it works. Let’s find the current
through the R7 resistor for the Norton equivalent. We will call
this current iX. We have
A
RN=
19[W]
iN=
1.53[A]
iX
R7=
10[W]
B
19[W]
iX  1.53[A]

19[W]  10[W]
iX  1.00[A].
Next slide
Checking the Solution, Original Circuit
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Let’s check our answer to see if it works. Let’s find the current
through the R7 resistor for the original circuit. Again, we will
call this current iX. We will also define node voltages, to allow
us to solve. We have the circuit below.
R 2=
30[W]
R1 =
50[W]
iS1=
5[A]
vS1=
6[V]
+
+
R4 =
60[W]
R5=
22[W]
A
+
+
vG
-
vS2=
12[V]
R 3=
27[W]
vE
R7=
10[W]
+
-
-
iX
B
vF
iS2=
2[A]
R 6=
40[W]
Next slide
Checking the Solution, Original Circuit Equations
Dave Shattuck
University of Houston
vE  12[V]
v v
 5[A]  E F  0;
60[W]
22[W]
v v
v v
vF
v
v v
 2[A]  F  F E  F G  F G  0; and
40[W]
10[W] 22[W] 27[W] 30[W]
vG  6[V]  12[V] vG  vF vG  vF


 5[A]  0.
50[W]
30[W] 27[W]
© Brooks/Cole Publishing Co.
The node-voltage
equations are:
iS1=
5[A]
vS1=
6[V]
+
R5=
22[W]
A
+
vE
R7=
10[W]
+
-
-
iX
B
vF
vF  10.0[V]; and
vG  48.8[V].
Using this, we can
find iX, which is
vF
10.0[V]


10[W] 10[W]
iX  1.00[A].
iX 
+
vG
-
vE  84.6[V];
This is
the same
answer
as
before.
R 3=
27[W]
+
R4 =
60[W]
vS2=
12[V]
The solutions are:
R 2=
30[W]
R1 =
50[W]
iS2=
2[A]
R 6=
40[W]
-
Go to Comments Slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Was This Worth It?
• This is a good question. Again, the best answer is, “It depends.”
• We have gone through a fair amount of work, but by doing so
we have a simpler circuit. Whether it was worth the work
depends on what we were going to use the circuit for.
• For example, if we were to connect the circuit to 12 different
resistors, or to 12 different current sources, it would be much
easier to solve the simpler circuit each time, and in the end it
would be worth it. For one resistor, it was probably not a good
use of our time.
• Note, though, that Norton’s Theorem also has benefits as a way
of thinking about a circuit. This will pay off in many areas,
among them when we are designing circuits.
Go back to
Overview
slide.