Download PEQWS_Mod03_Prob03_v04 - Courses

Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Practice Examination Questions With Solutions
Module 3 – Problem 3
Filename: PEQWS_Mod03_Prob03.doc
Note: Units in problem are enclosed in square brackets.
Time Allowed: 25 Minutes
Problem Statement:
Use the node-voltage method to write a complete set of independent equations that
could be used to solve this circuit. Do not attempt to solve the equations. Do not
attempt to simplify the circuit.
R1 =
47[W]
R2 =
7[W]
R3 =
10[W]
+
iS1=
2[A]
iS3=
9[A]
iS2=
8[A]
vS4=
11[V]
R4 =
12[W]
-
vS1=
3[W] iX
vX
iX
+
iS5=
5iQ
R6 =
13[W]
-
R5 =
18[W]
vS3=
4 vQ
+
+
R7 =
20[W]
vS2=
2 vX
R9 =
6[W]
-
-
+
+
vQ
iQ
R10=
14[W]
iS4=
16[A]
R11=
15[W]
R12=
17[W]
-
Page 3.3.1
R8 =
19[W]
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Problem Solution:
The problem statement was:
Use the node-voltage method to write a complete set of independent equations that
could be used to solve this circuit. Do not attempt to solve the equations. Do not
attempt to simplify the circuit.
R1 =
47[W]
R2 =
7[W]
R3 =
10[W]
+
iS1=
2[A]
iS3=
9[A]
iS2=
8[A]
vS4=
11[V]
R4 =
12[W]
-
vS1=
3[W] iX
vX
iX
+
iS5=
5iQ
R6 =
13[W]
-
R5 =
18[W]
vS3=
4 vQ
+
R8 =
19[W]
+
R7 =
20[W]
vS2=
2 vX
R9 =
6[W]
-
-
+
+
vQ
iQ
R10=
14[W]
iS4=
16[A]
R11=
15[W]
R12=
17[W]
-
The first step in the solution is to identify the essential nodes, and pick one
of them as the reference node. This is done in the circuit schematic that follows.
The essential nodes are marked in red.
We pick the node with the diagonal wires as the reference node, since that
node has the most connections. The other essential nodes are named with letters,
and these letters are used to name the node voltages.
Page 3.3.2
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
R1 =
47[W]
R2 =
7[W]
A
R3 =
10[W]
+
vA
+
iS1=
2[A]
iS3=
9[A]
iS2=
8[A]
vS4=
11[V]
R4 =
12[W]
-
vS1=
3[W] iX
vX
+
iX
iS5=
5iQ
R6 =
13[W]
-
- -
-
vS3=
4 vQ
+
vB
R9 =
6[W]
R5 =
18[W]
B
R8 =
19[W]
+
R7 =
v
20[W] C
vS2=
2 vX
+
C
+
-
+
+
vQ
iQ
R10=
14[W]
iS4=
16[A]
R11=
15[W]
vD
D -
R12=
17[W]
+
Now we need to write the Node-Voltage Method Equations. There will be
four equations plus four more for the dependent source variables iX, vX, iQ, and vQ.
We will take them alphabetically.
For node A, we have a voltage source in one branch, but it has a current
source in series with it, so we can use the current of the current source. We can
write
v
A: A  iS 2  iS1  0.
R2
For node B, there is a voltage source between two non-reference essential nodes.
This fits the situation that we described as a supernode. However, there is a special
case here. We will find it difficult to write the supernode equation because the
current in the vS3 voltage source is not known. However, it turns out that we don’t
need to write the supernode equation, since vS3 voltage source is a voltage source
Page 3.3.3
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
between an essential node and the reference node, and we can just use it to write
the needed equation. This is a special case, but if we understand the way we
handled the supernode case, and why, it should be clear that we can just write the
two equations that follow,
B: vB - vC  vS 2 , and
C: vC  vS 3 .
Node D is a relatively straightforward application of KCL. We get
v v
v v
D: D B  iS 4  D C  0.
R10  R11
R12
Now, we have to write equations for the dependent sources. The current iX
is the current in the wire in the middle of the reference node. We need to draw a
closed surface, and write a KCL equation for that closed surface. Let’s draw this
closed surface here, as a dashed line.
R1 =
47[W]
R2 =
7[W]
A
R3 =
10[W]
+
vA
+
iS1=
2[A]
iS3=
9[A]
iS2=
8[A]
vS4=
11[V]
R4 =
12[W]
-
vS1=
3[W] iX
vX
+
iX
iS5=
5iQ
R6 =
13[W]
-
- -
-
vS3=
4 vQ
+
vB
R9 =
6[W]
R5 =
18[W]
B
+
R7 =
v
20[W] C
vS2=
2 vX
+
C
+
-
+
+
vQ
iQ
R10=
14[W]
iS4=
16[A]
R11=
15[W]
vD
D -
Page 3.3.4
+
R12=
17[W]
R8 =
19[W]
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
We have enclosed the iS3 current source and the R4 resistor, since the node shorts
them. However, this is not necessary. What is necessary is to include every
current that crosses the closed surface. Using this closed surface, we can write
0  vB
0  vA 0  vS 4
iX : iX  iS 5 
 iS 2 

 0.
R6
R2
R3
In this equation, we have written terms like (0 – vB) to make it clear that this is the
voltage across the R6 resistor. Usually we write these terms as the difference
between node voltages; the node voltage at the reference node is zero. Obviously,
the zero is not needed in these expressions, and is only shown for clarity.
Next, to get an equation for iQ, we will write a KCL equation for node B. Note that
we did not do this yet, because of the vS2 voltage source. We can write
v v
v
iQ : -iQ  B D  iS1  B  0.
R10  R11
R6
Next, we write an equation for vQ. This is the difference between node voltages, so
we can write
vQ : vC  vD  vQ .
Finally, we need an equation for vX. We will write KVL around the loop marked
with the dashed red line to get this equation.
Page 3.3.5
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
R1 =
47[W]
R2 =
7[W]
A
R3 =
10[W]
+
vA
+
iS1=
2[A]
iS3=
9[A]
iS2=
8[A]
vS4=
11[V]
R4 =
12[W]
-
vS1=
3[W] iX
vX
+
iX
iS5=
5iQ
R6 =
13[W]
-
- -
-
vS3=
4 vQ
+
B
R8 =
19[W]
+
R7 =
v
20[W] C
vS2=
2 vX
vB
R9 =
6[W]
R5 =
18[W]
+
C
+
-
+
iQ
+
vQ
R10=
14[W]
iS4=
16[A]
R11=
15[W]
vD
D -
R12=
17[W]
+
We can write the equation
v X :  v X  vB  iS 1R9  vS 1  0.
This gives us 8 equations in 8 unknowns, and completes the solution that was
requested.
Note 1: For clarity in showing this solution, and how it unfolds, we have redrawn
the circuit several times. In solving this problem on an examination, we would not
redraw each time, but rather make marks on the original circuit. In addition, we
would not include all of the text that is present here. With this, it should be
possible to complete the problem in the allotted time.
Page 3.3.6
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Note 2: Some students have difficulty trying to determine whether their solution
was a valid one, particularly if they have taken a slightly different approach such as
picking a different reference node. While it is not requested in this problem, a
numerical solution for the dependent source variables iX, vX, iQ, and vQ is given
here. If you are in doubt about the validity of your solution, solve for these
quantities, and compare with this solution. If your solution is significantly
different, then something must be wrong.
Our equations were:
v
A: A  iS 2  iS 1  0
R2
B: vB - vC  vS 2
C: vC  vS 3
D:
vD  vB
v v
 iS 4  D C  0
R10  R11
R12
iX : iX  iS 5 
iQ : -iQ 
0  vB
0  v A 0  vS 4
 iS 2 

0
R6
R2
R3
vB  vD
v
 iS 1  B  0
R10  R11
R6
vQ : vC  vD  vQ
v X :  v X  vB  iS 1R9  vS 1  0
Now, we are going to substitute in the values that were given in the circuit. We get
the following system of equations:
Page 3.3.7
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
vA
 8[A]  2[A]  0
7[W]
B: vB - vC  2v X
A:
C: vC  4vQ
D:
vD  vB
v v
 16[A]  D C  0
17[W]
14  15 [W]
 vB
v A 11[V]
 8[A] 

0
13[W]
7[W] 10[W]
v v
v
iQ : -iQ  B D  2[A]  B  0
14  15[W]
13[W]
vQ : vC  vD  vQ
iX : iX  5iQ 
v X :  v X  vB  2[A]6[W]  3[W]iX  0
Now, we will substitute these equations into MathCAD. The solution is given in a
MathCAD file called PEQWS_Mod03_Prob03_Soln.mcd. The results are given
here:
vA = -70[V]
vB = 274.54299[V]
vC = -585.59439[V]
vD = -439.19579[V]
iX = 238.87056[A]
iQ = 43.73037[A]
vQ = -146.3986[V]
vX = 430.06869[V]
While the node voltage values depend on how you define these variables, the
dependent source variables iX, vX, iQ, and vQ should be the same with any approach.
You can use these answers to check your work.
Problem adapted from ECE 2300, Quiz 4, Summer 1998, Department of Electrical and Computer Engineering, Cullen College of Engineering,
University of Houston.
Page 3.3.8