Download PEQWS_Mod03_Prob10_v02

Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Practice Examination Questions With Solutions
Module 3 – Problem 10
Filename: PEQWS_Mod03_Prob10.doc
Note: Units in problem are enclosed in square brackets.
Time Allowed: 30 Minutes
Problem Statement:
Find iX and the power delivered by the independent voltage source in the circuit
below.
7 iX
1[kW]
3.3[kW]
4.7[kW]
3[kW]iX
+
+
-
12[mA]
1.9[kW]
10[V]
5[mA]
3.4[kW]
2.2[kW]
iX
Page 3.10.1
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Problem Solution:
The problem statement was:
Find iX and the power delivered by the independent voltage source in the circuit
below.
7 iX
1[kW]
3.3[kW]
4.7[kW]
3[kW]iX
+
+
-
12[mA]
1.9[kW]
10[V]
5[mA]
3.4[kW]
2.2[kW]
iX
The first step in this problem is to pick the best approach to the problem.
We can certainly solve this by simply writing KVL and KCL and Ohm’s Law
expressions until we can solve, but it is better to take a systematic approach. Our
two primary systematic approaches are the Node-Voltage Method, and the MeshCurrent Method.
Let’s consider these two possibilities. If we were to pick the Node-Voltage
Method, we have five essential nodes, and therefore four equations, plus one for
the dependent source variable. If we were to pick the Mesh-Current Method, we
have four meshes, and therefore four equations, plus one for the dependent source
variable. There is no benefit to either method in terms of the number of equations.
Considering that our goal is to get the current iX, and the current through the
Page 3.10.2
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
independent voltage source, let’s use the Mesh-Current Method, which yields
currents. Let’s assign the mesh currents.
7 iX
1[kW]
3.3[kW]
iB
3[kW]iX
12[mA]
+
+
10[V]
4.7[kW]
-
iA
1.9[kW]
iC
iD
5[mA]
3.4[kW]
2.2[kW]
iX
Now we need to write the Mesh-Current Method equations. There will be
four equations plus one more for the dependent source variable iX. We will take
them alphabetically.
For mesh A, we have a current source as a part of two meshes. Therefore,
we will need to write a supermesh equation, and a constraint equation. The path
for the supermesh is shown with a black, dashed line in the schematic given below.
Page 3.10.3
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
7 iX
1[kW]
3.3[kW]
iB
3[kW]iX
12[mA]
+
+
10[V]
4.7[kW]
-
iA
1.9[kW]
iC
iD
5[mA]
3.4[kW]
2.2[kW]
iX
We can write the KVL for this closed path, and get
A+C: iA1[kW]  (iA  iB )3.3[kW]  (iC  iD )1.9[kW]  iC 3.4[kW]  10[V]  0.
For the constraint equation, we have
A+C: iA  iC  12[mA].
For mesh B, we have a current source as a part of only that one mesh. Thus,
we can write
B: iB  7iX .
For mesh D, we again have a current source as a part of only that one
mesh. Thus, we can write
D: iD  5[mA].
Now, we have to write equations for the dependent sources. The current iX
is equal to one of the mesh currents, so we can write
iX : iX  iC .
This gives us five equations in five unknowns,
Page 3.10.4
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
A+C: iA1[kW]  (iA  iB )3.3[kW]  (iC  iD )1.9[kW]  iC 3.4[kW]  10[V]  0
A+C: iA  iC  12[mA]
B: iB  7iX
D: iD  5[mA]
iX : iX  iC
and we can solve.
This problem can be solved fairly easily by substituting the results of the
later equations into the first. However, since we have a computer here, we will
choose to insert the equations into MathCAD, in a file called
PEQWS_Mod03_Prob10_Soln.mcd. The results are given here:
iA = 11[mA]
iB = 6.87[mA]
iC = -982[A]
iD = 5[mA]
iX = -982[A]
The current iX is one answer that was requested.
Next, we want to find the power delivered by the independent voltage
source. Note that iC is the current through the voltage source. Thus, we can write
pdel ,voltage _ source  iC (10[V])  9.82[mW] .
Notice that the sign in this expression is positive, because the voltage and current
polarities are related by the active sign convention in this situation.
Note: 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.
Page 3.10.5