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Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Practice Examination Questions With Solutions
Module 4 – Problem 6
Filename: PEQWS_Mod04_Prob06.doc
Note: Units in problem are enclosed in square brackets.
Time Allowed: 20 Minutes
Problem Statement:
Find the Thévenin equivalent of the circuit shown, as seen at terminals A
and B. Draw your equivalent circuit, showing terminals A and B.
iS1=
3iX
iX
A
R1=
20[W]
R2=
15[W]
+
-
vS1=
50[V]
R3=
10[W]
R4=
45[W]
B
Page 4.6.1
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Problem Solution:
Find the Thévenin equivalent of the circuit shown, as seen at terminals A
and B. Draw your equivalent circuit, showing terminals A and B.
iS1=
3iX
iX
A
R1=
20[W]
R2=
15[W]
+
-
vS1=
50[V]
R3=
10[W]
R4=
45[W]
B
The first step in the solution is to find one of the three quantities, opencircuit voltage, short-circuit current, and equivalent resistance. In this case, it
appears that finding the open-circuit voltage would be fairly simple. One reason
for this is that the resistor R2 will have no current through it, and thus will have no
effect, and can be removed. Perhaps more significant is that this will also set the
current iX equal to zero, which will remove the dependent source. Let’s define the
open-circuit voltage, and then solve for it. The voltage is defined in the circuit that
follows.
Page 4.6.2
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
iS1=
3iX
iX
A
R1=
20[W]
R2=
15[W]
+
-
vS1=
50[V]
R3=
10[W]
R4=
45[W]
+
vOC
B
Now, as we noted, this open circuit makes the current iX equal to zero, which
means that the dependent source has a value of zero. In addition, the resistor R2
has no current through it, and so we can remove it as well. With these
simplifications we have the circuit that follows.
A
+
R1=
20[W]
+
-
vS1=
50[V]
R3=
10[W]
R4=
45[W]
vOC
B
Page 4.6.3
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
In this circuit, there is only one connection between the right half of the
circuit and the left half, and there is no source in the right half. Thus, the voltages
and currents in the right half will be zero. We could go ahead as apply KVL and
KCL, but we will find that
vOC  0.
This gives us the Thevenin voltage, which is zero.
Now, we need to find either the equivalent resistance or the short-circuit
current. The short-circuit current will have to be zero, though, since the opencircuit voltage is zero. We can see this from the relationship
vOC  iSC REQ .
This is actually based on the assumption that the equivalent resistance is not
zero, which we will show is true. Thus, if we try to find the REQ from the ratio of
the open-circuit voltage to the short-circuit current, we will have an undefined
expression.
So we will choose to find the equivalent resistance. We set the independent
sources equal to zero, and apply a test source. We will choose a 1[A]-current
source, since that will make finding iX easy. We have the circuit that follows.
iS1=
3iX
iX
A
+
R1=
20[W]
R2=
15[W]
R3=
10[W]
R4=
45[W]
iT=
1[A]
vT
B
Let’s find iX first. We can write
Page 4.6.4
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
iX  1[A]. Thus,
iS1  3iX  3[A].
To make the solution a little easier, we will define a new variable, iR4, as shown in
the figure that follows.
iS1=
3iX
iX
A
+
R1=
20[W]
R2=
15[W]
iR4
R3=
10[W]
R4=
45[W]
iT=
1[A]
vT
B
Now, we write KVL around the right-most mesh, and we get
iR 4 R4  iX R2  vT  0. From KCL, we get
iR 4  iS1  iX  0.
From the second equation, and the previous values, we have
iR 4  iS 1  iX  4[A].
Plugging these values into the first equation, we get
vT  iR 4 R4  iX R2  4[A] 45[W]   1[A] 15[W]  
vT  195[V].
Finally, to get the equivalent resistance, we can write
REQ 
vT 195[V]

 195[W].
iT
1[A]
Page 4.6.5
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Thus, the Thevenin equivalent is just a resistance. We can redraw it as
shown in the figure that follows.
A
RTH=
195[W]
B
Note: In this solution, as in most of the solutions in this project, we have
taken a very slow and careful approach, redrawing the circuit several times. While
it is a good idea to redraw whenever it helps simplify the circuit, it is generally not
necessary to redraw the circuit just to define a new variable. We do it here to be
very clear about the order of steps. When you work a problem, just mark new
variables on your existing diagrams. Only redraw the circuit when you can
simplify it to where it makes it easier for you to look at it. Be careful, but don’t
waste time.
Problem adapted from ECE 2300, Quiz 3, Spring 1992, Department of Electrical and Computer Engineering, Cullen College of Engineering,
University of Houston.
Page 4.6.6