Download Practice Examination Module 3 Problem 6 - Courses

Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Practice Examination Questions With Solutions
Module 3 – Problem 6
Filename: PEQWS_Mod03_Prob06.doc
Note: Units in problem are enclosed in square brackets.
Time Allowed: 30 Minutes
Problem Statement:
Use the mesh-current 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.
iX
R1 =
1[W]
+
R3 =
7[W]
vS1=
3[V]
R4 =
10[W]
iS1=
6[S]vX
R2 =
2[W]
R6 =
8[W]
R5 =
4[W]
R7 =
9[W]
-
iS2=
5iX
R8 =
12[W]
iS3=
11[A]
vX
R10=
14[W]
vS2=
15[V]
R11=
16[W]
+
+
Page 3.6.1
R9 =
13[W]
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Problem Solution:
The problem statement was:
Use the mesh-current 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.
iX
R1 =
1[W]
+
R3 =
7[W]
vS1=
3[V]
R4 =
10[W]
iS1=
6[S]vX
R2 =
2[W]
R6 =
8[W]
R5 =
4[W]
R7 =
9[W]
-
iS2=
5iX
R8 =
12[W]
iS3=
11[A]
vX
R10=
14[W]
vS2=
15[V]
R9 =
13[W]
R11=
16[W]
+
+
-
The first step in the solution is to identify the meshes, and define the mesh
currents. This is done in the circuit schematic that follows. The mesh currents are
marked in red.
Page 3.6.2
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
iX
R1 =
1[W]
iA
+
R3 =
7[W]
iB
vS1=
3[V]
iC
iE
iS1=
6[S]vX
R2 =
2[W]
iD
R6 =
8[W]
R5 =
4[W]
R4 =
10[W]
R7 =
9[W]
-
iF
iS2=
5iX
vX
iS3=
11[A]
R10=
14[W]
iG
R8 =
12[W]
vS2=
15[V]
iH
R9 =
13[W]
R11=
16[W]
+
+
-
Now we need to write the Mesh-Current Method equations. There will be
eight equations plus two more for the dependent source variables iX and vX. We will
take them alphabetically.
For mesh A, we have a fairly straightforward application of KVL. We write
A: iA R1  vS1  iA R2  0.
For mesh B, we have a similar case to that we had in mesh A, and we can just write
B:  vS1  (iB  iC ) R3  (iB  iF ) R5  0.
Mesh C is a supermesh case, with the current source iS1 as a part of the C mesh and
the D mesh. The supermesh is drawn using a dashed black line in the circuit
diagram that follows.
Page 3.6.3
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
iX
R1 =
1[W]
iA
+
R3 =
7[W]
iB
vS1=
3[V]
iC
iE
iS1=
6[S]vX
R2 =
2[W]
iD
R6 =
8[W]
R5 =
4[W]
R4 =
10[W]
R7 =
9[W]
-
iF
iS2=
5iX
vX
iS3=
11[A]
R10=
14[W]
iG
R8 =
12[W]
vS2=
15[V]
iH
R9 =
13[W]
R11=
16[W]
+
+
-
We have the supermesh equation,
C+D: (iC  iB ) R3  (iD  iH ) R7  (iC  iG ) R6  0,
and we have the constraint equation,
C+D: iD  iC  iS1.
Now, for mesh E, there is a fairly simple equation,
E:  iE  R4  0.
Next, we have the F mesh. We have two current sources in this mesh; one is a part
of two meshes and one is a part of only this mesh. The current source that is a part
of only this mesh, iS2, determines the mesh current, and we can write
F: iF  iS 2 .
Now, for mesh G, we can write the constraint equation given by the value of the iS3
current source,
F+G: iG  iF  iS 3 .
The H mesh is fairly straightforward. We write
H: (iH  iG ) R8  (iH  iD ) R7  iH R9  iH R11  0.
Now, we have to write equations for the dependent sources. The current iX
is equal to the current in the B mesh, and we can write
iX : iB  iX .
Page 3.6.4
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Finally, we need to write an equation for the voltage, vX. We can pick a closed
loop and write a KVL equation. Generally, we want to avoid closed loops that
include current sources, since we don’t know the voltage across a current source.
We pick a closed loop that is shown in the following circuit with a red dashed line.
iX
R1 =
1[W]
iA
+
R3 =
7[W]
iB
vS1=
3[V]
iC
iE
iS1=
6[S]vX
R2 =
2[W]
iD
R6 =
8[W]
R5 =
4[W]
R4 =
10[W]
R7 =
9[W]
-
iF
iS2=
5iX
vX
iS3=
11[A]
R10=
14[W]
iG
R8 =
12[W]
vS2=
15[V]
iH
R9 =
13[W]
R11=
16[W]
+
+
-
We can write
v X : v X  (iG  iC ) R6  (iG  iH ) R8  vS 2  iF R10  0.
This is 10 equations in 10 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 a couple of 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.
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 different directions for the mesh currents. While it is not requested in this
problem, a numerical solution for iX and vX is given here. If you are in doubt about
Page 3.6.5
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
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:
A: iA R1  vS 1  iA R2  0
B:  vS 1  (iB  iC ) R3  (iB  iF ) R5  0
C+D: (iC  iB ) R3  (iD  iH ) R7  (iC  iG ) R6  0
C+D: iD  iC  iS 1
E:  iE  R4  0
F: iF  iS 2
F+G: iG  iF  iS 3
H: (iH  iG ) R8  (iH  iD ) R7  iH R9  iH R11  0
iX : iB  iX
v X : v X  (iG  iC ) R6  (iG  iH ) R8  vS 2  iF R10  0
Now, we are going to substitute in the values that were given in the circuit. We get
the following system of equations:
A: iA1[W]  3[V]  iA 2[W]  0
B:  3[V]  (iB  iC )7[W]  (iB  iF )4[W]  0
C+D: (iC  iB )7[W]  (iD  iH )9[W]  (iC  iG )8[W]  0
C+D: iD  iC  6[S]v X
E:  iE 10[W]  0
F: iF  5iX
F+G: iG  iF  11[A]
H: (iH  iG )12[W]  (iH  iD )9[W]  iH 13[W]  iH 16[W]  0
iX : iB  iX
v X : v X  (iG  iC )8[W]  (iG  iH )12[W]  15[V]  iF 14[W]  0
Now, we will substitute these equations into MathCAD. The solution is given in a
MathCAD file called PEQWS_Mod03_Prob06_Soln.mcd. The results are given
here:
Page 3.6.6
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
iA = -1[A]
iB = -1.20387[A]
iC = 1.11926[A]
iD = 3.44002[A]
iE = 0[A]
iF = -6.01935[A]
iG = 4.98065[A]
iH = 1.81456[A]
iX = -1.20387[A]
vX = 0.38679[V]
While the mesh current 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, Final Exam, Problem 1, Summer 1998, Department of Electrical and Computer Engineering, Cullen College of
Engineering, University of Houston.
Page 3.6.7