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Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Practice Examination Questions With Solutions
Module 8 – Problem 6
Filename: PEQWS_Mod08_Prob06.doc
Note: Units in problem are enclosed in square brackets.
Time Allowed: 20 Minutes
Problem Statement:
It is known that to maximize the average power transferred to a load, the
impedance of that load should be the complex conjugate of the Thevenin
impedance of the circuit connected to the load. Find the load impedance ZL for this
circuit that will maximize the average power transferred to that load.
ZR1=
3[W]
ZL1=
4j[W]
Vs2=
6[W] Ix
Vs1(w)=
100[V]
+
-
+
Ix(w)
-
Page 8.6.1
ZC1=
-8j[W]
Load
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Problem Solution:
The problem statement was:
It is known that to maximize the average power transferred to a load, the
impedance of that load should be the complex conjugate of the Thevenin
impedance of the circuit connected to the load. Find the load impedance ZL for this
circuit that will maximize the average power transferred to that load.
Vs2=
6[W] Ix
ZL1=
4j[W]
ZR1=
3[W]
Vs1(w)=
100[V]
+
-
+
ZC1=
-8j[W]
Ix(w)
Load
-
From the statement in the problem, it is clear that the goal is to find the
Thevenin impedance of the circuit connected to the load. There are different
approaches that can be taken for such a problem. Since there is a dependent source
present, we will use the test-source method.
We are already in the phasor domain, so we do not need to redraw. To use
the test-source method, we need to set the independent sources equal to zero, and
then apply a test source in place of the load.
ZR1=
3[W]
Vs2=
6[W] Ix
ZL1=
4j[W]
ZC1=
-8j[W]
Ix(w)
Page 8.6.2
+
-
Test
Source
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
For this problem, we will choose to apply a current source. A voltage source
would be almost as good, but by examining the circuit, it seems likely that a
current source will make the solution a little bit easier. The reason is that we will
want to find Ix(w) in this problem. By using a current source, we can use this
current to write a Current Divider Rule expression for Ix(w). Remember, either
type of source will work. If you do not understand this reason, or have a reason to
choose differently, you will still be able to solve, and your answer should still be
the same. We will use a value of 1[A], just for simplicity. We can redraw the
circuit as shown in the figure that follows.
ZL1=
4j[W]
ZR1=
3[W]
Vs2=
6[W] Ix
+
-
+
Vt
Ix(w)
I t=
1[A]
ZC1=
-8j[W]
-
The goal is to find the voltage across the test source, Vt. Then the Thevenin
impedance will be the ratio of Vt to It.
The current source enters the parallel combination of ZC1 in parallel with the
series combination of ZR1 and ZL1. We can write
I x (w )  It
Z R1  Z L1
3 4 j
 1[A]
 1106[A].
Z R1  Z L1  ZC1
3  4 j   8 j
Now, we can write KVL around the right-hand loop, and we get an
expression for Vt,
Vt  I x Z C1  Vs 2  1106[A]  8 j[W]  1106[A]  6[W] 
Vt  1106[A]  6  8 j[W]  1053.1[V].
Now, the Thevenin impedance is Vt/It, so
ZThev =
Vt
 1053.1[W]  6  8 j.
It
We want the complex conjugate of this, so the answer is
Z Load =  ZThev  = 10  53.1[W]  6  8 j.
*
Page 8.6.3
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Remember that we get a complex conjugate by taking the opposite sign for the
phase, or the opposite sign for the imaginary part.
Problem adapted from ECE 2300, Final Examination, Problem #6, in the fall of 1994, Department of Electrical and Computer Engineering,
Cullen College of Engineering, University of Houston.
Page 8.6.4