Download PEQWS_Mod04_Prob01_v03 - Courses

Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Practice Examination Questions With Solutions
Module 4 – Problem 1
Filename: PEQWS_Mod04_Prob01.doc
Note: Units in problem are enclosed in square brackets.
Time Allowed: 30 Minutes
Problem Statement:
For the circuit shown, calculate the Thévenin equivalent resistance with respect to
terminals a and b.
vS2=
iX
R3 =
30[]
a
+
R1 =
10[]
iX
+
-
R2 =
20[]
iS1=
5iX
vS1=
100[V]
-
R4 =
40[]
b
Page 4.1.1
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Problem Solution:
For the circuit shown, calculate the Thévenin equivalent resistance with respect to
terminals a and b.
vS2=
iX
R3 =
30[]
a
+
R1 =
10[]
iX
+
-
R2 =
20[]
iS1=
5iX
vS1=
100[V]
-
R4 =
40[]
b
The first step in the solution is to decide what items should be solved for.
Note that the only thing we have been asked for is the Thévenin resistance.
Specifically, we could solve for the open-circuit voltage and short-circuit current,
and then take the ratio to get the solution. Another possibility is to find the
Thévenin resistance directly.
Here, and generally when we are asked for only the equivalent resistance, it
is fastest to solve for it directly. This is true even when there are dependent
sources present, as here.
Therefore, we will set the independent sources equal to zero, and then apply
a test source. We will choose a 1[V]-voltage source, since it looks as though that
would allow us to find the solutions quickly. We have done so in the circuit that
follows.
Page 4.1.2
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
+
iX
vS2=
iX
R3 =
30[]
R2 =
20[]
a
+
R1 =
10[]
iT
vC
iS1=
5iX
+
R4 =
40[]
-
vT=
1[V]
b
-
Notice that in addition, we have defined a current iT that we will solve, to
find the Thévenin resistance. Also, we have defined a reference node, and a node
voltage vC, that we will use to find the solution. In fact, we will start by writing the
KCL for node C, and the dependent source variable equation for iX. We have
vC vC vC  vS 2  vT
 
 iS 1  0, and
R2 R1
R3
vC
 iX .
R1
Now, we can substitute in values,
vC
v
v  10[]iX  1[V]
 C  C
 5iX  0, and
20[] 10[]
30[]
vC
 iX .
10[]
We plug the second equation into the first, and get
 v 
vC  10[]  C   1[V]
 v 
vC
v
 10[] 
 C 
 5  C   0.
20[] 10[]
30[]
 10[] 
Solving for vC, we get
 v 
vC
v
1[V]
 C 
 5  C   0, or
20[] 10[] 30[]
 10[] 
vC 650[mS]  33.3[mA], which means that
vC  51.3[mV].
From this, it is clear that
Page 4.1.3
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
vC
51.3[mV]

 5.13[mA].
10[]
10[]
Now, we can solve for iT by applying KCL at the a node, and get
vT vT  vS 2  vC

 iT  0, which is
R4
R3
iX 
1[V] 1[V]  10[] 5.13[mA]  51.3[mV]

 iT .
40[]
30[]
Thus we have that
1[V]
1[V]

 iT  58.3[mA].
40[] 30[]
Finally, we note that the equivalent resistance, which is the Thévenin
resistance, is the ratio of vT/iT, and so we have
v
1[V]
REQ  RTH  T 
 17.1[].
iT 58.3[mA]
Thus, our answer is
RTH  17.1[].
We could have found the Thévenin voltage, which is the open-circuit
voltage. Without showing the work here, we note that this would have been
57.1[V]. Then, we could have found the short-circuit current, which would have
been 3.33[A]. We leave the work for this to the interested student. Taking the
ratio of these, we would get the same answer as above. In our opinion, though, the
approach that we took was quicker, and therefore better.
Problem adapted from ECE 2300, Exam 2, Problem 1, Spring 2001, Department of Electrical and Computer Engineering, Cullen College of
Engineering, University of Houston.
Page 4.1.4