Download PEQWS_Mod08_Prob09_v06

Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Practice Examination Questions With Solutions
Module 8 – Problem 9
Filename: PEQWS_Mod08_Prob09.doc
Note: Units in problem are enclosed in square brackets.
Time Allowed: 35 Minutes
Problem Statement:
In the circuit below, find the nonzero frequency at which the impedance between
terminals a and b is purely resistive. Find the impedance at this frequency.
a
R=
180[W]
C=
1[mF]
L=
50[mH]
b
Page 8.9.1
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Problem Solution:
The problem statement was:
In the circuit below, find the nonzero frequency at which the impedance between
terminals a and b is purely resistive. Find the impedance at this frequency.
a
R=
180[W]
C=
1[mF]
L=
50[mH]
b
The first step here is to convert to the phasor domain, so we redraw the
circuit as shown in the figure that follows. This figure is given assuming that  is
given with units of [rad/s].
a
ZR=R=
180[W]
ZC=1/jC=
(1/j10-6)[W]
ZL=jL=
j50x10-3[W]
b
Page 8.9.2
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
We want the equivalent impedance between terminals a and b. The
inductive and resistive impedances are in series, and that combination is in parallel
with the capacitive impedance. We get the impedance Zab,

Z ab ( )   Z R  Z L  || Z C =  R  j L  || 1
1

j C 
( ) 
.
1
 R  j L    j C 
j C

 R  j L 
Z ab
In many problems we would go ahead and substitute in the values for R, L,
and C. However, in this kind of problem, where we may have to perform some
complex algebra, we prefer to wait and substitute in near the end, to keep the
notation simple and clear.
Now, the goal is to find a frequency, , where the impedance is purely
resistive. One key to this problem is understanding what is meant by purely
resistive. When an impedance is purely resistive, that means that the impedance is
real, or has no imaginary part. Another way of saying this is that the phase of the
impedance is zero.
Let’s apply this to the impedance, Zab. First, let’s multiply the top and the
bottom of the expression by jC, to get the fractions out of the numerator and
denominator,


R  j L  1
j C
 R  j L 
j C 
Z ab ( ) 



j C  R  j L   1
 R  j L  jC   1
j C
Z ab ( ) 


 R  j L 
 R  j L 
 R   j  L  .


2 2
2
 jCR   j  LC   1 1  j  LC  jCR 1   2 LC   j CR 
Notice that we have isolated the real and imaginary parts of the numerator and
denominator in parentheses in this last expression.
If this expression for Zab is going to be real, that means that the imaginary
part must be zero. However, getting an expression for the Im{Zab} is not easy; we
would need to multiply the top and bottom of the expression by the complex
conjugate of the denominator, which is fairly complicated. Once we do that, we
need to set that complicated expression for Im{Zab} equal to zero. It can be done,
but it will take some time. For this problem, we can solve more easily by taking
the other approach, that is, recognizing that for Zab to be real, the phase must be
zero.
Page 8.9.3
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
The phase of Zab, or Zab, is the phase of the numerator minus the phase of
the denominator, or


Z ab ( )   R   j  L    1   2 LC   j CR  .
Now, the phase of a complex number is the arctangent of the imaginary part
divided by the real part, so we can write
Z ab ( )  tan 1
 L   tan 1 CR  .
 R
1   2 LC 
This phase must be equal to zero, so we can write that
 L   tan 1  CR   0, or
 R
1   2 LC 
 L   tan 1  CR  .
tan 1
 R
1   2 LC 
tan 1
Now, if the arctangents are equal, the simplest solution is where the quantities are
equal. Thus, we can say that
 L   CR  ,
 R  1   2 LC 
and solve for . We note, without showing it here, that this is the same equation
we would have gotten by finding an expression for the imaginary part of Zab and
setting it equal to zero.
Let’s solve for . Cross multiplying, we get
 L  1   2 LC    R  CR  , or
  L   2 L2C    CR 2 .
One solution of this equation is  = 0, but we have been asked for the nonzero
solution. Therefore we can divide both sides by , and solve for . We have
L 
L C   CR 2 , and solving we have
2 2
 2 L2C  CR 2  L, or
CR 2  L
 
.
L2C
2
Finally, we take the square root of both sides, rejecting the negative frequency
value as physically meaningless. We can now substitute in the values to obtain
 106  180    50 103 
CR 2  L


.
2
L2C
50 103  106 
2
Page 8.9.4
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
This can be solved to get
  2, 650[rad/s].
The impedance at this frequency can be found by plugging back in, and we
get
Z ab (2, 650[rad/s]) 
 R   j  2, 650[rad/s]L 
.
2
1

2,
650[rad/s]
LC

j
2,
650[rad/s]
CR






Note, that while the imaginary part is zero, this does not mean that the imaginary
part of the numerator is zero, or that the imaginary part of the denominator is zero.
They are not zero. However, their phases will be zero. Let’s substitute in and
show this,
Z ab (2, 650[rad/s]) 
Z ab (2, 650[rad/s]) 

180  j 2, 650  50 103 

 

1   2, 650   50 103 106   j 2, 650 106  180 
2
180  j 132.5 
 649 10   j  477 10 
3
3



223.536.36
[W].
805 103 36.32
The small difference between the phase of the numerator and the phase of the
denominator is due to round off errors. The answer, then, is that
Zab (2,650[rad/s])  278[W].
Note 1: While in some cases the impedance, when purely resistive, is equal to the
resistance of the resistor, it is not always true. Clearly, it is not true here, where the
resistor was 180[W], but the impedance was 278[W]. Do not assume this without
being sure it is true.
Note 2: If you look back, you will find the expression for the frequency was
CR 2  L

.
L2C
Note that we needed to have L > CR2, or else the quantity under the radical would
have been negative or zero. If L = CR2, then we would have had a zero solution,
which was not what was requested. If L < CR2, then we would have had an
imaginary solution. In this case, there would be no nonzero frequency where the
impedance was purely resistive.
Page 8.9.5