Download PEQWS_Mod08_Prob05_v03 - Courses

Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Practice Examination Questions With Solutions
Module 8 – Problem 5
Filename: PEQWS_Mod08_Prob05.doc
Note: Units in problem are enclosed in square brackets.
Time Allowed: 30 Minutes
Problem Statement:
Find the steady-state value of i(t).
C1=
10[mF]
L1=
2.2[mH]
R1=
20[W]
iX(t)
i(t)
+
vS1(t)
iS=
4 iX
R2=
30[W]
-
+
vS2(t)
-
vS1(t) = 50[V]
vS2(t) = 100 cos (10,000[rad/s] t - 30°)[V]
Page 8.5.1
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Problem Solution:
The problem statement was:
Find the steady-state value of i(t).
C1=
10[mF]
L1=
2.2[mH]
R1=
20[W]
iX(t)
i(t)
+
vS1(t)
iS=
4 iX
R2=
30[W]
-
+
vS2(t)
-
vS1(t) = 50[V]
vS2(t) = 100 cos (10,000[rad/s] t - 30°)[V]
Since only the steady-state solution is desired, and there are only sinusoidal
sources, this problem can be solved using the phasor transform approach. There is
one important variation here, in that vS1(t) is a dc source. This can be thought of as
a sinusoid with frequency equal to zero, which is how we can make the first
statement. If we choose to think of it in this way, then we have two frequencies,
and we need to apply superposition to use the phasor transform method. Even if
we do not choose to think of dc sources as sinusoids with zero frequency, we will
still need to apply superposition to be able to use the phasor technique for the vS2(t)
source.
Remember that the iS current source is a dependent source, and therefore it is
not set equal to zero when using superposition; instead, it is left in place when
each of the independent sources is taken one at a time.
Thus, the first step in this problem is to apply superposition. Let’s take the
vS1(t) source first, and set the vS2(t) source equal to zero. We can redraw the circuit
as shown in the figure that follows.
Page 8.5.2
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
C1=
10[mF]
L1=
2.2[mH]
R1=
20[W]
iX1(t)
i1(t)
+
vS1(t)
iS=
4 iX1
R2=
30[W]
-
vS1(t) = 50[V]
Now, at dc, an inductor is a short circuit, and a capacitor is an open circuit.
We can simplify this circuit using this, and draw the following circuit.
R1=
20[W]
iX1(t)
i1(t)
+
vS1(t)
iS=
4 iX1
R2=
30[W]
-
vS1(t) = 50[V]
It is pretty clear that iX1 = 0. Thus, we can set the dependent source equal to zero,
and the circuit reduces to a voltage source connected to two series resistors. We
can write
i1 (t )  
vS1
50[V]

 1[A].
R1  R2  20  30  [W]
Don’t be deceived by the apparent lack of contribution from the dependent
source. This is a special case, since it turned out that iX1 was zero. In general, the
dependent source can have a contribution, and should not be neglected.
Page 8.5.3
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Next, we take the vS2(t) source, and set the vS1(t) source equal to zero.
Again, we can redraw, and we get the following figure.
C1=
10[mF]
L1=
2.2[mH]
R1=
20[W]
iX2(t)
i2(t)
+
iS=
4 iX2
R2=
30[W]
vS2(t)
-
vS2(t) = 100 cos (10,000[rad/s] t - 30°)[V]
Now, to solve for i2(t) we want to convert the circuit to the phasor domain.
This is done in the circuit schematic that follows.
ZC1=
-10j[W]
ZL1=
22j[W]
ZR1=
20[W]
A
Ix2(w)
+
Va
I2(w)
Is=
4 Ix2
ZR2=
30[W]
Vs2(w)
+
-
Vs2(w) = 100Ð-30°)[V]
Now, the goal is to solve for I2(w), using circuit analysis techniques and complex
arithmetic. We will choose to solve using the node-voltage method, applied in the
Page 8.5.4
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
phasor domain. We have already defined the node voltage Va, which we will solve
for first. Let’s write the node voltage equations,
Va
V
V  Vs 2
 a  a
 I s  0, which upon substituting in values, gives
Z R2 Z R1 Z L1  ZC1
Va
V
V  100Ð  30[V]
 a  a
 4 I x 2  0.
30[W] 20[W]
22 j[W]  10 j[W]
We also need an equation for I x2 ,
100Ð  30[V]  Va
I x2 
.
22 j[W]  10 j[W]
We can substitute this value for Ix2 back into the first equation, which leaves us
with one equation in one unknown. We have
Va
V
V  100Ð  30[V]
100Ð  30[V]  Va
 a  a
4
 0, which simplifies to
30[W] 20[W]
22 j[W]  10 j[W]
22 j[W]  10 j[W]
Va
V
V  100Ð  30[V]
 a 5 a
 0.
30[W] 20[W]
22 j[W]  10 j[W]
Collecting terms on each side of the equation, we get
 1
1
5  500Ð  30[V]
Va 


.

12 j[W]
 30[W] 20[W] 12 j[W] 
This can be solved for Va,
500Ð  30[V]
12 j[W]
Va 
 98.1Ð  41.3[V].
 1
1
5 
 30[W]  20[W]  12 j[W] 


From this, we can solve for the desired quantity, I2,
Va
98.1Ð  41.3[V]
I2 =

 4.90Ð  41.3[A].
20[W]
20[W]
Remember that this is not the answer. To get this part of the solution, we need to
perform the inverse transform, and get
i2 (t )  4.90cos(10,000[rad/s]t  41.3)[A].
The answer then, is the sum of these two responses, or
i(t )  i1 (t )  i2 (t )  1  4.90cos(10,000[rad/s]t  41.3)[A].
Page 8.5.5
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Note 1: In this problem, as a part of the solution we have redrawn the circuit in the
phasor domain. We recommend that you do this, even if you are solving this
problem on an examination. This is part of what is expected for you to complete
the problem, and the allotted time has been adjusted to account for this. It is a
good habit to avoid mixing the time domain and the phasor domain, and this
should be avoided in diagrams as well as in equations. On a related note, it is also
important to complete the problem, which means performing the inverse transform.
Many students are so elated at completing the solution for I2(w), that they fail to
transform it back to i2(t). Don’t neglect this important, while simple, step.
On the other hand, in this solution we have redrawn the circuits several more
times after the transformation to the phasor domain. If you can solve these circuits
without redrawing, this may save you some time. Just be careful not to do so many
things at once that you become confused. Use techniques such as redrawing the
circuit to help you keep what you are doing clear in your mind.
Note 2: Notice that the dependent source was not set equal to zero when using
superposition. The superposition approach only applies to independent sources.
The dependent source remains in place for each independent source that is used.
Problem adapted from ECE 2300, Final Examination, Problem #3, in the summer of 1999, Department of Electrical and Computer Engineering,
Cullen College of Engineering, University of Houston.
Page 8.5.6