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Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Problems With Assistance
Module 3 – Problem 2
Filename: PWA_Mod03_Prob02.ppt
This problem is adapted from:
Problem 4.6, page 183 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
• The Node-Voltage Method
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 4.1 & 4.3
• Electric Circuits 6th Ed. by Nilsson and Riedel: Sections
4.2 through 4.4
• Basic Engineering Circuit Analysis 6th Ed. by Irwin and
Wu: Section 3.1
• Fundamentals of Electric Circuits by Alexander and
Sadiku: Sections 3.2 & 3.3
• Introduction to Electric Circuits 2nd Ed. by Dorf: Sections
4-2 through 4-4
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_Mod03_Part01 and DPKC_Mod03_Part02
A similar problem is worked in:
• PWA_Mod03_Prob01
Next slide
Dave Shattuck
University of Houston
Problem Statement
© Brooks/Cole Publishing Co.
Find v2, v4, and the power supplied by each source.
R1=
10[kW]
R3=
18[kW]
+
v4
+
i1
i2
+
vS=
60[V]
v2
-
i3
R2=
15[kW]
i4
R4=
12[kW]
iS=
3[mA]
-
-
Next slide
Figure P4.6
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Solution – First Step – Where to Start?
How should
we start this
problem?
What is the
first step?
Find v2, v4, and the power supplied by each source.
R1=
10[kW]
R3=
18[kW]
+
v4
+
i1
i2
+
vS=
60[V]
v2
-
i3
R2=
15[kW]
i4
R4=
12[kW]
iS=
3[mA]
-
Figure P4.6
Next slide
How should we start this
problem? What is the first
step?
Dave Shattuck
University of Houston
Problem Solution – First Step
© Brooks/Cole Publishing Co.
Find v2, v4, and the power supplied by each source.
R1=
10[kW]
Write KCL for each node
b)
Identify the essential
nodes
c)
Write KVL for each loop
d)
Pick the reference node
e)
Combine resistors in
parallel or series
R3=
18[kW]
+
v4
+
i1
i2
+
vS=
60[V]
a)
v2
-
i3
R2=
15[kW]
i4
R4=
12[kW]
-
Figure P4.6
iS=
3[mA]
This is not the best choice for the
first step, although we will write
KCL equations for most nodes
soon.
Dave Shattuck
University of Houston
Your choice for First Step –
Write KCL for each node
© Brooks/Cole Publishing Co.
It is generally worth while to spend
some time looking at the problem
and choosing an approach before
beginning to write equations. Note
that we have six variables defined
already, but will not need that
many.
Find v2, v4, and the power supplied by each source.
R1=
10[kW]
Go back and try again.
R3=
18[kW]
+
v4
+
i1
i2
+
vS=
60[V]
v2
-
i3
R2=
15[kW]
i4
R4=
12[kW]
-
Figure P4.6
iS=
3[mA]
This is not the best choice for
the first step.
Dave Shattuck
University of Houston
Your choice for First Step –
Write KVL for each loop
© Brooks/Cole Publishing Co.
It is generally worth while to
spend some time looking at the
problem and choosing an
approach before beginning to
write equations. Note that we
have six variables defined
already, but will not need that
many.
Find v2, v4, and the power supplied by each source.
R1=
10[kW]
R3=
18[kW]
+
v4
+
i1
i2
+
vS=
60[V]
Go back and try again.
v2
-
i3
R2=
15[kW]
i4
R4=
12[kW]
-
Figure P4.6
iS=
3[mA]
This will be helpful, but is not the
best choice for the first step.
Dave Shattuck
University of Houston
Your choice for First Step was –
Pick the reference node
© Brooks/Cole Publishing Co.
Find v2, v4, and the power supplied by each source.
R1=
10[kW]
R3=
18[kW]
+
v4
+
i1
i2
+
vS=
60[V]
v2
-
i3
R2=
15[kW]
The node-voltage method indeed
requires that we pick, and label,
the reference node. However, it
is usually wise to be sure that
we know where all the essential
nodes are, and how many
connections they have, before
making this choice. Thus,
while it may not be necessary
in simple problems like this, we
recommend that you go back
and try again.
i4
R4=
12[kW]
-
Figure P4.6
iS=
3[mA]
This might be helpful, but is not the
best choice for the first step.
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Your choice for First Step was –
Combine resistors in parallel or
series
Generally, it is a good thing to
simplify a circuit, where we can
do so. Here, we cannot do so
since, there are no resistors in
series or parallel. Therefore,
we recommend that you go
back and try again.
Find v2, v4, and the power supplied by each source.
R1=
10[kW]
R3=
18[kW]
+
v4
+
i1
i2
+
vS=
60[V]
v2
-
i3
R2=
15[kW]
Note to advanced students: We could
use delta-to-wye or wye-to-delta
transformations, but we are going to
take a different approach here.
i4
R4=
12[kW]
-
Figure P4.6
iS=
3[mA]
This is the best choice. By making
sure that we have identified the
essential nodes, we can determine
how many equations will be needed in
the node-voltage method.
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Your choice for First Step was –
Identify the essential nodes
How many essential nodes are there
in this circuit? Your answer is:
a) 3 essential nodes
b) 4 essential nodes
Find v2, v4, and the power supplied by each source. c) 5 essential nodes
R1=
10[kW]
R3=
18[kW]
+
v4
+
i1
i2
+
vS=
60[V]
v2
-
i3
R2=
15[kW]
i4
R4=
12[kW]
-
Figure P4.6
iS=
3[mA]
Dave Shattuck
University of Houston
Your choice for the number of essential nodes – 4
© Brooks/Cole Publishing Co.
This is not correct. Remember
that essential nodes must
have at least 3
connections. In addition,
remember that two nodes
connected by a wire were
really only one node.
Try again.
Find v2, v4, and the power supplied by each source.
R1=
10[kW]
R3=
18[kW]
+
v4
+
i1
i2
+
vS=
60[V]
v2
-
i3
R2=
15[kW]
i4
R4=
12[kW]
-
Figure P4.6
iS=
3[mA]
Dave Shattuck
University of Houston
Your choice for the number of essential nodes – 3
© Brooks/Cole Publishing Co.
This is correct. The
essential nodes are marked
with red in this schematic.
There is a non-essential node,
which is marked with green.
With only 3 essential nodes, the nodevoltage method is a good choice, since we will
have only 2 simultaneous equations. The next
step is to pick one of them as the reference node.
Which one should we pick?
Find v2, v4, and the power supplied by each source.
R1=
10[kW]
R3=
18[kW]
+
v4
+
i1
i2
+
vS=
60[V]
v2
-
i3
R2=
15[kW]
i4
R4=
12[kW]
-
Figure P4.6
iS=
3[mA]
Dave Shattuck
University of Houston
Your choice for the number of essential nodes – 5
© Brooks/Cole Publishing Co.
This is not correct. Try again.
Remember that two nodes
connected by a wire were
really only one node.
Find v2, v4, and the power supplied by each source.
R1=
10[kW]
R3=
18[kW]
+
v4
+
i1
i2
+
vS=
60[V]
v2
-
i3
R2=
15[kW]
i4
R4=
12[kW]
-
Figure P4.6
iS=
3[mA]
Dave Shattuck
University of Houston
Choosing the Reference Node
© Brooks/Cole Publishing Co.
The next step is to pick one of them as the reference node. We
have chosen the node at the bottom as the reference node. This is
considered to be the best choice, since it has 4 connections to it. The
equations will probably be easier to write with this as reference node.
In addition, the two node voltages that result are the voltages we
were asked to find. Next, we define the node-voltages.
Find v2, v4, and the power supplied by each source.
R1=
10[kW]
R3=
18[kW]
+
v4
+
i1
i2
+
vS=
60[V]
v2
-
i3
R2=
15[kW]
i4
R4=
12[kW]
-
Figure P4.6
iS=
3[mA]
The next step is to define the node-voltages.
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Defining the Node-Voltages
We have done so here. Now, we are ready to
write the Node-Voltage Method
Equations. Even before we do, we can
predict that we will need to write two
equations, one for each non-reference
essential node.
Find v2, v4, and the power supplied by each source.
R1=
10[kW]
R3=
18[kW]
2
4
+
v4
+
i1
i2
+
vS=
60[V]
v2
-
i3
R2=
15[kW]
i4
R4=
12[kW]
-
Figure P4.6
iS=
3[mA]
Dave Shattuck
University of Houston
Writing the Node-Voltage Equations – 1
© Brooks/Cole Publishing Co.
The equation for Node 2 is given here.
Note that we used an expression for
the current in R1 to express the current
in the voltage source. The resistor R1
and the voltage source are in series.
v2 vS v2 v4
v2
Node 2:
0, or
10[kW] 18[kW] 15[kW]
v2 60[V] v2 v4
v2
Node 2:
0
10[kW]
18[kW] 15[kW]
Find v2, v4, and the power supplied by each source.
R1=
10[kW]
R3=
18[kW]
2
4
+
v4
+
i1
i2
+
vS=
60[V]
v2
-
i3
R2=
15[kW]
i4
R4=
12[kW]
iS=
3[mA]
-
Figure P4.6
Next equation
Dave Shattuck
University of Houston
Writing the Node-Voltage Equations – 2
© Brooks/Cole Publishing Co.
The equation for Node 4 is given here.
v4
v v
4 2 3[mA] 0
12[kW] 18[kW]
Node 4:
Find v2, v4, and the power supplied by each source.
R1=
10[kW]
R3=
18[kW]
2
4
+
v4
+
i1
i2
+
vS=
60[V]
v2
-
i3
R2=
15[kW]
i4
R4=
12[kW]
iS=
3[mA]
-
Figure P4.6
Next step
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Writing the Node-Voltage
Equations – All
v2 60[V] v2 v4
v2
0
10[kW]
18[kW] 15[kW]
v4
v v
Node 4:
4 2 3[mA] 0
12[kW] 18[kW]
Node 2:
The next step is to solve the
equations. Let’s solve.
Find v2, v4, and the power supplied by each source.
R1=
10[kW]
R3=
18[kW]
2
4
+
v4
+
i1
i2
+
vS=
60[V]
v2
-
i3
R2=
15[kW]
i4
R4=
12[kW]
iS=
3[mA]
-
Figure P4.6
Next step
Dave Shattuck
University of Houston
We have used MathCAD to solve the
two simultaneous equations. This is
shown in a MathCAD file called
© Brooks/Cole Publishing Co.
Solving the Node-Voltage
Equations
PWA_Mod03_Prob02_Soln.mcd
When we solve, we find that
which should be available in this
module.
v2 = 36[V], and
v4 = 36[V].
Find v2, v4, and the power supplied by each source.
R1=
10[kW]
R3=
18[kW]
2
4
+
v4
+
i1
i2
+
vS=
60[V]
v2
-
i3
R2=
15[kW]
i4
R4=
12[kW]
iS=
3[mA]
-
Figure P4.6
Next step
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Using the Node-Voltages to Solve for Desired Quantities – Part 1
We found that
We can use this to find the power supplied by the current
source directly. Note that v4 is the voltage across the
current source. Note also that v4 and iS are in the active
convention for this source. Therefore, we can write:
v2 = 36[V], and
v4 = 36[V].
pdel,iS = v4 iS = 36[V]3[mA] = 108[mW]
Find v2, v4, and the power supplied by each source.
R1=
10[kW]
R3=
18[kW]
2
4
+
v4
+
i1
i2
+
vS=
60[V]
v2
-
i3
R2=
15[kW]
i4
R4=
12[kW]
iS=
3[mA]
-
Figure P4.6
Next step
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Using the Node-Voltages to Solve for Desired Quantities – Part 2
We found that
v2 = 36[V], and
v4 = 36[V].
Next, we want to find the power supplied by the voltage
source. For this, we need to find the current through the
voltage source, which has already been labeled as i1. We
write the expression for this just as we had when writing the
KCL expression for node 2.
60[V] v2
i1
10[kW]
60[V] 36[V]
2.4[mA] i1
10[kW]
Find v2, v4, and the power supplied by each source.
R1=
10[kW]
R3=
18[kW]
2
4
+
v4
+
i1
i2
+
vS=
60[V]
v2
-
i3
R2=
15[kW]
i4
R4=
12[kW]
iS=
3[mA]
-
Figure P4.6
Next step
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Using the Node-Voltages to Solve for Desired Quantities – Part 3
We found that
Now, we can find the power supplied by the voltage source.
Note that vS and i1 are in defined in the active convention
for the voltage source. Thus, we can write:
v2 = 36[V], and
v4 = 36[V].
pdel ,vS vS i1 60[V]2.4[mA] 144[mW]
Find v2, v4, and the power supplied by each source.
R1=
10[kW]
R3=
18[kW]
2
4
+
v4
+
i1
i2
+
vS=
60[V]
v2
-
i3
R2=
15[kW]
i4
R4=
12[kW]
iS=
3[mA]
-
Figure P4.6
See Note
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
What happened? The two node-voltages
were the same!
• It is true that in this problem, the two node-voltages were the same.
This occurred because Carlson chose the values of vS, iS, R1, R2 and R4 to
make this happen. You can prove to yourself that R3 makes no
difference in this case by varying its value, and solving again. For all
nonzero values of R3, the solution will be the same.
• This raises yet another important point. The Node-Voltage Method
gives us general equations which apply for the way the circuit is laid
out, called the topology. Once you have the equations, you could set
v2=v4, and solve for values of vS, iS, R1, R2 and R4 to make this happen.
The node-voltage technique, once in hand, has many uses.
Go back to
Overview
slide.