Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Problems With Assistance
Module 3 – Problem 4
Filename: PWA_Mod03_Prob04.ppt
Go
straight to
the First
Step
Go
straight to
the
Problem
Statement
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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 Mesh-Current 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
This material is covered in your textbook in the following
sections:
• Circuits by Carlson: Sections 4.2 & 4.3
• Electric Circuits 6th Ed. by Nilsson and Riedel: Sections
4.1, & 4.5 through 4.7
• Basic Engineering Circuit Analysis 6th Ed. by Irwin and
Wu: Section 3.2
• Fundamentals of Electric Circuits by Alexander and
Sadiku: Sections 3.4 & 3.5
• Introduction to Electric Circuits 2nd Ed. by Dorf: Sections
4-5 & 4-6
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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_Part03 and
DPKC_Mod03_Part04
This same problem is solved with the NodeVoltage Method in
• PWA_Mod03_Prob01
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Problem Statement
Use the meshcurrent method to
solve for the voltage
vX.
R
33 4 =
[W
]
R2=
27[W]
vS=
5[V]
+
R1=
22[W]
R3=
39[W]
+ vX
iS1=
0.5[A]
R5=
10[W]
-
iS2=
2[S] vX
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Solution – First Step – Where to Start?
Use the meshcurrent method to
solve for the voltage
vX.
R
33 4 =
[W
]
R2=
27[W]
vS=
5[V]
+
R1=
22[W]
R3=
39[W]
+ vX
iS1=
0.5[A]
R5=
10[W]
-
How should
we start this
problem?
What is the
first step?
iS2=
2[S] vX
Next slide
Dave Shattuck
University of Houston
Problem Solution – First Step
© Brooks/Cole Publishing Co.
Use the meshcurrent method to
solve for the voltage
vX.
How should we start this
problem? What is the first
step?
R
33 4 =
[W
]
R2=
27[W]
vS=
5[V]
+
R1=
22[W]
R3=
39[W]
+ vX
iS1=
0.5[A]
R5=
10[W]
-
iS2=
2[S] vX
a)
Write KVL for each mesh
b)
Identify the meshes and
define the mesh currents
c)
Write KCL for each node
d)
Combine resistors in
parallel or series
Dave Shattuck
University of Houston
Your choice for First Step –
Write KVL for each mesh
© Brooks/Cole Publishing Co.
Use the meshcurrent method to
solve for the voltage
vX.
This is not a good choice for
the first step, although we will
write KVL equations for most
meshes soon.
R2=
27[W]
R
33 4 =
[W
]
vS=
5[V]
+
R1=
22[W]
R3=
39[W]
+ vX
iS1=
0.5[A]
R5=
10[W]
One purpose of the meshcurrent method is to find a
systematic way of writing the
correct number of equations.
It is important, then, to know
how many equations we are
going to write.
-
iS2=
2[S] vX
Go back and try again.
Dave Shattuck
University of Houston
Your choice for First Step –
Write KCL for each node
© Brooks/Cole Publishing Co.
Use the meshcurrent method to
solve for the voltage
vX.
This is not a good choice.
The mesh-current method
involves writing KVL
equations, not KCL equations.
While we may write KCL
equations as needed for
constraint equations, it is not
the systematic step that we
take in using the mesh-current
method. This is not the way to
start this method.
R
33 4 =
[W
]
R2=
27[W]
vS=
5[V]
+
R1=
22[W]
R3=
39[W]
+ vX
iS1=
0.5[A]
R5=
10[W]
-
iS2=
2[S] vX
Go back and try again.
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Your choice for First Step was –
Combine resistors in parallel or series
Use the meshcurrent method to
solve for the voltage
vX.
This might be helpful, but is
not the best choice for the
first step.
R
33 4 =
[W
]
R2=
27[W]
vS=
5[V]
+
R1=
22[W]
R3=
39[W]
+ vX
iS1=
0.5[A]
R5=
10[W]
-
iS2=
2[S] vX
Generally, it is a good thing to
simplify a circuit, where
we can do so. Here, you
may have noted that R4
and R3 are in parallel, and
can be combined into a
single resistor. We will
not even lose any
dependent source
variables. However, the
mesh-current method
does not require that we
simplify the circuits, and
sometimes we cannot do
so. Therefore, we
recommend that you go
back and try again.
Your choice for First Step was –
© Brooks/Cole Publishing Co.
Identify the meshes and define the mesh currents
Dave Shattuck
University of Houston
Use the meshcurrent method to
solve for the voltage
vX.
This is the best choice.
The first step is to make sure
that we have identified all
the meshes and defined
the mesh currents.
R2=
27[W]
R
33 4 =
[W
]
vS=
5[V]
How many meshes are there in
this circuit? Your answer
is:
+
R1=
22[W]
R3=
39[W]
+ vX
iS1=
0.5[A]
R5=
10[W]
-
iS2=
2[S] vX
a)
3 meshes
b)
4 meshes
c)
5 meshes
d)
6 meshes
e)
7 meshes
Dave Shattuck
University of Houston
Your choice for the number of essential nodes – 3
© Brooks/Cole Publishing Co.
Use the meshcurrent method to
solve for the voltage
vX.
This is not correct. Try again.
R
33 4 =
[W
]
R2=
27[W]
vS=
5[V]
+
R1=
22[W]
R3=
39[W]
+ vX
iS1=
0.5[A]
R5=
10[W]
-
iS2=
2[S] vX
Dave Shattuck
University of Houston
Your choice for the number of essential nodes – 4
© Brooks/Cole Publishing Co.
Use the meshcurrent method to
solve for the voltage
vX.
This is not correct. Try again.
R
33 4 =
[W
]
R2=
27[W]
vS=
5[V]
+
R1=
22[W]
R3=
39[W]
+ vX
iS1=
0.5[A]
R5=
10[W]
-
iS2=
2[S] vX
Dave Shattuck
University of Houston
Your choice for the number of essential nodes – 5
© Brooks/Cole Publishing Co.
Use the meshcurrent method to
solve for the voltage
vX.
This is correct. Let’s define
the mesh currents.
R
33 4 =
[W
]
R2=
27[W]
vS=
5[V]
+
R1=
22[W]
R3=
39[W]
+ vX
iS1=
0.5[A]
R5=
10[W]
-
iS2=
2[S] vX
Dave Shattuck
University of Houston
Your choice for the number of essential nodes – 6
© Brooks/Cole Publishing Co.
This is not correct. Try again.
Use the meshcurrent method to
solve for the voltage
vX.
R
33 4 =
[W
]
R2=
27[W]
vS=
5[V]
+
R1=
22[W]
R3=
39[W]
+ vX
iS1=
0.5[A]
R5=
10[W]
-
iS2=
2[S] vX
Dave Shattuck
University of Houston
Your choice for the number of essential nodes – 7
© Brooks/Cole Publishing Co.
This is not correct. Try again.
Use the meshcurrent method to
solve for the voltage
vX.
R
33 4 =
[W
]
R2=
27[W]
vS=
5[V]
+
R1=
22[W]
R3=
39[W]
+ vX
iS1=
0.5[A]
R5=
10[W]
-
iS2=
2[S] vX
Dave Shattuck
University of Houston
Defining the Mesh Currents
© Brooks/Cole Publishing Co.
Use the meshcurrent method to
solve for the voltage
vX.
R2=
27[W]
iB
iC
R4=
33[W]
vS=
5[V]
The next step is to
define the mesh currents.
iD
+
R1=
22[W]
R3=
39[W]
+ vX
iS1=
0.5[A]
iA
R5=
10[W]
iE
-
iS2=
2[S] vX
We have done so here.
Now, we are ready to write
the Mesh-Current Method
Equations. Even before we
do, we can predict that we
will need to write six
equations, one for each
mesh (5) and one for the
dependent source variable
vX.
Dave Shattuck
University of Houston
Writing the Mesh-Current Equations – 1
© Brooks/Cole Publishing Co.
Use the meshcurrent method to
solve for the voltage
vX.
The equation for Mesh A is obtained from the current source:
A: iA  0.5[A]
R2=
27[W]
iB
iC
R4=
33[W]
vS=
5[V]
iD
+
R1=
22[W]
R3=
39[W]
+ vX
iS1=
0.5[A]
iA
R5=
10[W]
iE
-
iS2=
2[S] vX
Next equation
Dave Shattuck
University of Houston
Writing the Mesh-Current Equations – 2
© Brooks/Cole Publishing Co.
Use the meshcurrent method to
solve for the voltage
vX.
The equation for Mesh B is obtained by writing KVL around mesh
B.
B:  5[V]  (iB  iE )22[W]  0
R2=
27[W]
iB
iC
R4=
33[W]
vS=
5[V]
iD
+
R1=
22[W]
R3=
39[W]
+ vX
iS1=
0.5[A]
iA
R5=
10[W]
iE
-
iS2=
2[S] vX
Next equation
Dave Shattuck
University of Houston
Writing the Mesh-Current Equations – 3
© Brooks/Cole Publishing Co.
Use the meshcurrent method to
solve for the voltage
vX.
The equation for Mesh C is:
C: 5[V]  iC 27[W]  (iC  iD )33[W]  0
R2=
27[W]
iB
iC
R4=
33[W]
vS=
5[V]
iD
+
R1=
22[W]
R3=
39[W]
+ vX
iS1=
0.5[A]
iA
R5=
10[W]
iE
-
iS2=
2[S] vX
Next equation
Dave Shattuck
University of Houston
Writing the Mesh-Current Equations – 4
© Brooks/Cole Publishing Co.
Use the meshcurrent method to
solve for the voltage
vX.
The equation for Mesh D is:
D: (iD  iC )33[W]  (iD  iE )39[W]  0
R2=
27[W]
iB
iC
R4=
33[W]
vS=
5[V]
iD
+
R1=
22[W]
R3=
39[W]
+ vX
iS1=
0.5[A]
iA
R5=
10[W]
iE
-
iS2=
2[S] vX
Next equation
Dave Shattuck
University of Houston
Writing the Mesh-Current Equations – 5
© Brooks/Cole Publishing Co.
Use the meshcurrent method to
solve for the voltage
vX.
The equation for Mesh E is obtained from the dependent current
source, which determines the mesh current.
E: iE  2[S]vX
R2=
27[W]
iB
iC
R4=
33[W]
vS=
5[V]
iD
+
R1=
22[W]
R3=
39[W]
+ vX
iS1=
0.5[A]
iA
R5=
10[W]
iE
-
iS2=
2[S] vX
Next equation
Dave Shattuck
University of Houston
Writing the Mesh-Current Equations – 6
© Brooks/Cole Publishing Co.
Use the meshcurrent method to
solve for the voltage
vX.
The equation for the dependent source variable vX is:
vX : vX  (iE  iD )39[W]
R2=
27[W]
iB
iC
R4=
33[W]
vS=
5[V]
iD
+
R1=
22[W]
R3=
39[W]
+ vX
iS1=
0.5[A]
iA
R5=
10[W]
iE
-
Make sure that you agree
with this equation for the
dependent source variable
vX . It can be obtained by
writing Ohm’s Law for the
39[W] resistor. The branch
current through this
resistor, going from left to
right, is iE-iD.
iS2=
2[S] vX
Next step
A: iA  0.5[A]
Dave Shattuck
University of Houston
B:  5[V]  (iB  iE )22[W]  0
© Brooks/Cole Publishing Co.
Writing the Node-Voltage
Equations – All
Use the mesh-
C: 5[V]  iC 27[W]  (iC  iD )33[W]  0
current method to
solve for the voltage
vX.
D: (iD  iC )33[W]  (iD  iE )39[W]  0
E: iE  2[S]v X
v X : v X  (iE  iD )39[W]
R2=
27[W]
iB
iC
R4=
33[W]
vS=
5[V]
iD
+
R1=
22[W]
R3=
39[W]
+ vX
iS1=
0.5[A]
iA
R5=
10[W]
iE
The next step is to solve the
equations. We can do this by
various approaches. We will
choose to use MathCAD for
this module.
-
iS2=
2[S] vX
Next step
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Solving the Mesh- Current
Equations
A: iA  0.5[A]
iA  0.5[A]
B:  5[V]  (iB  iE )22[W]  0
iB  0.033[A]
C: 5[V]  iC 27[W]  (iC  iD )33[W]  0
iC  0.189[A]
D: (iD  iC )33[W]  (iD  iE )39[W]  0
iD  0.192[A]
E: iE  2[S]v X
iE  0.194[A]
v X : v X  (iE  iD )39[W]
v X  0.097[V]
The next step is to solve the
equations. We can do this by
various approaches. We will
choose to use MathCAD for
this module.
See Note
Go back to
Overview
slide.
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
What if I like the Mesh-Current Method
much more than the Node-Voltage Method?
• If you like the Mesh-Current Method more than the NodeVoltage Method, you are in the majority of beginning
circuit-analysis students.
• However, if you note that this problem was the same one as
was solved in PWA_Prob01 in this module, you will find
that here the solution required 6 equations (really only 5,
when we ignore the A mesh equation that was not used).
When it was done with the Node-Voltage Method, there
were only 4 equations, and one of those was not used, and
we could solve it easily by hand.
• Even if you prefer one method, learn them both!
Go back to
Overview
slide.