• Study Resource
• Explore

Survey
Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

System of polynomial equations wikipedia, lookup

Signal-flow graph wikipedia, lookup

Elementary algebra wikipedia, lookup

System of linear equations wikipedia, lookup

History of algebra wikipedia, lookup

Equation wikipedia, lookup

Corecursion wikipedia, lookup

Centrality wikipedia, lookup

Transcript
```Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Practice Examination Questions With Solutions
Module 3 – Problem 7
Filename: PEQWS_Mod03_Prob07.doc
Note: Units in problem are enclosed in square brackets.
Time Allowed: 20 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.
vS2=
5[W] iX
vS1=
3[V]
+
-
+
-
-
R1=
1[W]
R2 =
2[W]
vS3=
10 vX
R3 =
3[W]
vS4=
3[V]
vS5=
15[W] iX
vX
+
+
-
+
-
+
-
R6 =
6[W]
R8=
8[W]
iX
R5 =
5[W]
iS1=
7[A]
R7 =
7[W]
Page 3.7.1
R4 =
4[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.
vS2=
5[W] iX
vS1=
3[V]
+
-
+
-
-
R1=
1[W]
R2 =
2[W]
vS3=
10 vX
R3 =
3[W]
vS4=
3[V]
vS5=
15[W] iX
vX
R4 =
4[W]
+
+
-
+
-
+
-
R6 =
6[W]
R8=
8[W]
iX
R5 =
5[W]
iS1=
7[A]
R7 =
7[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 bottom-right node 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 3.7.2
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
vS2=
5[W] iX
vS1=
3[V]
B
A
+
-
+
-
-
+
+
vA
R1=
1[W]
vB
R2 =
2[W]
vS3=
10 vX
R3 =
3[W]
vS4=
3[V]
D
+ vC
-
vX
R4 =
4[W]
vE + +
+
+ vD
-
-
+
-
+
-
-
vS5=
15[W] iX
E
-
C
R6 =
6[W]
R8=
8[W]
iX
R5 =
5[W]
+ vF
R7 =
7[W]
-
iS1=
7[A]
F
Now we need to write the Node-Voltage Method Equations. There will be
six equations plus two more for the dependent source variables iX and vX. We will
take them alphabetically.
For node A, we have two voltage sources connected. The vS1 source has a
series element that we can use to express a current in that branch. However, the vS2
source is between two non-reference essential nodes. We will need a supernode
equation and a constraint equation. The supernode is drawn with a dashed red line
in the circuit diagram that follows.
Page 3.7.3
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
vS2=
5[W] iX
vS1=
3[V]
B
A
+
-
+
-
-
+
+
vA
R1=
1[W]
vB
R2 =
2[W]
vS3=
10 vX
R3 =
3[W]
vS4=
3[V]
D
+ vC
-
vX
R4 =
4[W]
vE + +
+
+ vD
-
-
+
-
+
-
-
vS5=
15[W] iX
E
-
C
R6 =
6[W]
R8=
8[W]
iX
R5 =
5[W]
+ vF
R7 =
7[W]
-
iS1=
7[A]
F
We can write
v v
v v v
v v
v
A+B: A D  A S 1 C  B E  B  0.
R2
R1
R4
R3
The constraint equation is
A+B: vB  v A  vS 2 .
Similarly, for node C we have two voltage sources connected. Again, the vS1
source has the R1 resistor in series, but the vS3 is a voltage source between two nonreference essential nodes. We might think that we would have a supernode here,
but if we look closely, we see that we have yet another voltage source connected to
node D. Since this one is connected to the reference node, this vS4 voltage source
determines the voltage at node D, and therefore all we will need is a constraint
equationf or node C. Let’s write the equations, starting with node C where we
have a constraint equation,
C+D: vD - vC  vS 3.
The voltage at node D is determined by the vS4 voltage source between it and the
reference node. We write
Page 3.7.4
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
D: vD  vS 4 .
Node E is a similar case, with the vS5 voltage source between node E and the
reference node. We can write
E: vE  vS 5 .
Node F is a straightforward application of KCL. We can write
v v
v  vD vF
F: F C  F

 0.
R5
R6
R7
Now, we have to write equations for the dependent sources. The current iX is the
current through R6. We can write
v  vD
iX : iX  F
.
R6
Finally, we need an equation for vX. This comes from the voltage between two
essential nodes, which we can express in terms of the node voltages. We can write
the equation
v X : v X  vE  vB .
This gives us 8 equations in 8 unknowns, and completes the solution that was
requested. Note that although we had three dependent sources, we only needed
two dependent source variable equations, since two dependent sources depended
on the same variable.
Note 1: For clarity in showing this solution, and how it unfolds, we have redrawn
the circuit. 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.
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 and vX, 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:
Page 3.7.5
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
A+B:
v A  vD v A  vS 1  vC vB  vE vB



0
R2
R1
R4
R3
A+B: vB  v A  vS 2
C+D: vD - vC  vS 3
D: vD  vS 4
E: vE  vS 5
F:
vF  vC vF  vD vF


0
R5
R6
R7
iX : iX 
vF  vD
R6
v X : v X  vE  vB
Now, we are going to substitute in the values that were given in the circuit. We get
the following system of equations:
v v
v  3[V]  vC vB  vE
v
A+B: A D  A

 B 0
2[W]
1[W]
4[W] 3[W]
A+B: vB  v A  5[W]iX
C+D: vD - vC  10v X
D: vD  3[V]
E: vE  15[W]iX
vF  vC vF  vD
v

 F 0
5[W]
6[W]
7[W]
v  vD
iX : iX  F
6[W]
v X : v X  vE  vB
F:
Now, we will substitute these equations into MathCAD. The solution is given in a
MathCAD file called PEQWS_Mod03_Prob07_Soln.mcd. The results are given
here:
vA = -4.87958[V]
Page 3.7.6
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
vB = -6.90294[V]
vC = -11.32856[V]
vD = -3[V]
vE = -6.07008[V]
vF = -5.42803[V]
iX = -0.40467[A]
vX = 0.83286[V]
While the node voltage values depend on how you define these variables, the
dependent source variables, iX and vX, should be the same with any approach. You
can use these answers to check your work.
Problem adapted from ECE 2300, Quiz #3, Spring 1998, Department of Electrical and Computer Engineering, Cullen College of Engineering,
University of Houston.
Page 3.7.7
```
Related documents