Download PEQWS_Mod01_Prob08_v02 - Courses

Dave Shattuck
University of Houston
Practice Examination Questions With Solutions
Module 1 – Problem 8
Filename: PEQWS_Mod01_Prob08.doc
Note: Units in problem are enclosed in square brackets.
Time Allowed: 30 minutes*.
Problem Statement:
In the circuit shown below, find the voltage vX.
R1 =
1.0[kW]
R3 =
3.3[kW]
R4 =
2.7[kW]
R5 =
2.2[kW]
iS=
5[mA]
+
R7 =
4.7[kW]
vX
R6 =
5.9[kW]
-
R8 =
8.2[kW]
R2 =
1.2[kW]
* This “time allowed” figure is based on the student’s circuit analysis skills
at this point. Once the student has learned the material in the second module, this
problem should be completed in significantly less time than what is shown here.
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Dave Shattuck
University of Houston
Problem Statement:
In the circuit shown below, find the voltage vX.
R1 =
1.0[kW]
R3 =
3.3[kW]
R4 =
2.7[kW]
R5 =
2.2[kW]
iS=
5[mA]
+
R7 =
4.7[kW]
vX
R6 =
5.9[kW]
-
R8 =
8.2[kW]
R2 =
1.2[kW]
Solution:
The first step in the solution of such problems is to define the variables that
we are going to need. However, beginning students often have the problem of
trying to decide what to name. It is not always clear which voltages and which
currents will be of interest in a problem. In addition, we can waste time if we
always label everything, everywhere, in a problem.
One approach is to just start somewhere, and name (define) variables as
needed. This is the approach that we will use here. I will begin by defining the
voltages across resistors R5 and R6, and the currents through these same two
resistors. I have chosen a particular polarity for each of these voltages, but this
choice is arbitrary. How did I know that I needed to define these two voltages?
The key is that I looked at this circuit, and decided that I was going to need to write
Page 1.8.2
Dave Shattuck
University of Houston
KVL around the loop that is marked with a red dashed line in the figure that
follows.
R1 =
1.0[kW]
R3 =
3.3[kW]
R4 =
2.7[kW]
R5 =
2.2[kW]
iS=
5[mA]
+
v5
+
-
i5
R7 =
4.7[kW]
vX
+
R6 =
5.9[kW]
v6
i6
R8 =
8.2[kW]
R2 =
1.2[kW]
When we look at this circuit, we may notice that these two resistors, R5 and
R6, are connected to the rest of the circuit at only one end. Stated another way, the
right hand side of R5 is not connected to anything. If we were to write the KCL for
the right hand “node” of R5, we would get
i5  0,
which is equivalent to
i5  0.
We have put “node” in quotation marks because it is not really a connection point
between two parts of the circuit. However, it does not matter whether this is a
“node” or not; we can write KCL for it anyway. We can write KCL for any closed
surface.
Using Ohm’s law, then, we can show that v5 is also zero. A similar set of
arguments holds for R6, and we can show that v6 is zero as well.
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Dave Shattuck
University of Houston
For our next step, we want to work towards finding v7 and v8, which are
defined in the circuit diagram that follows. If we can find v7 and v8, we will be
able to find vX. We will find them by first finding i7 and i8.
Note that i7 is also the current through the R3 resistor, because i5 is zero.
Similarly, we recognize that i8 is the current through the R4 resistor because i6 is
zero. Note also that the current through the R1 resistor is iS, since this is the only
path available for the current. Thus, we can write KCL for the closed surface
shown as a red dashed line below, and write
iS  i7  i8  0, or
i7  i8  5[mA].
(Eq. 1)
R1 =
1.0[kW]
R3 =
3.3[kW]
R4 =
2.7[kW]
R5 =
2.2[kW]
iS=
5[mA]
+
v5
+
-
vX
+
v6
i5
+
v7
+
i6
i8
i7
R7 =
4.7[kW]
R2 =
1.2[kW]
R6 =
5.9[kW]
v8
R8 =
8.2[kW]
-
-
This is one equation with two unknowns. If we can write another equation
using just these two unknowns, we could solve. There is a tendency to figure that
since KCL worked so well, we should just do it again. However, when we try this,
by writing KCL for the bottom node, a closed surface shown as a red dashed line in
the circuit below, we get the following equation,
Page 1.8.4
Dave Shattuck
University of Houston
iS  i7  i8  0, or
i7  i8  5[mA].
This equation is simply the same equation we had before, after multiplying both
sides by minus one. This is not an independent equation, and it will not help us.
Later in this course we will show ways to systematically write independent
equations. Until then, we simply continue to write equations until we have the
same number of independent equations and unknowns.
R1 =
1.0[kW]
R3 =
3.3[kW]
R4 =
2.7[kW]
R5 =
2.2[kW]
iS=
5[mA]
+
v5
+
-
vX
+
v6
i5
+
v7
+
i6
i8
i7
R7 =
4.7[kW]
R2 =
1.2[kW]
R6 =
5.9[kW]
v8
R8 =
8.2[kW]
-
-
It makes sense that if we use KVL, we might get an independent equation.
Therefore, we write KVL for the closed loop drawn in the circuit diagram that
follows. The closed loop is shown as a red dashed line.
Page 1.8.5
Dave Shattuck
University of Houston
R1 =
1.0[kW]
+
+
i7
i8
v4
R3 =
3.3[kW]
R4 =
2.7[kW]
v3
iS=
5[mA]
R5 =
2.2[kW]
-
+
v5
+
-
vX
R6 =
5.9[kW]
+
-
v6
i5
i6
+
+
v7
i8
i7
v8
R7 =
4.7[kW]
R2 =
1.2[kW]
R8 =
8.2[kW]
-
-
Writing the equation, we have
v7  v3  v4  v8  0.
This doesn’t seem to help, since we have just introduced four more unknowns.
However, we can use Ohm’s law to express each of these voltages in terms of a
current, and get
i7 R7  i7 R3  i8 R4  i8 R8  0, or with values,
i7 4.7[kW]  i7 3.3[kW]  i8 2.7[kW]  i8 8.2[kW]  0.
(Eq. 2)
Taken together, the two equations Eq. 1 and Eq. 2 are two equations in two
unknowns. I can solve these, and get
i7  2.88[mA], and
i8  2.12[mA].
Using these two values, I can find the voltages v7 and v8, as
Page 1.8.6
Dave Shattuck
University of Houston
v7  2.88[mA]R7  13.5[V], and
v8  2.12[mA]R8  17.4[V].
With this information, I am ready to solve for vX. Using KVL around the loop
shown with a red dashed line in the figure that follows, I can write
v7  v5  vX  v6  v8  0, which upon substituting gives
13.5[V]  0  vX  0  17.4[V ]  0.
R1 =
1.0[kW]
+
+
i7
i8
v4
R3 =
3.3[kW]
R4 =
2.7[kW]
v3
R5 =
2.2[kW]
-
iS=
5[mA]
+
v5
+
-
vX
R6 =
5.9[kW]
+
-
v6
i5
i6
+
+
v7
i8
i7
v8
R7 =
4.7[kW]
R2 =
1.2[kW]
R8 =
8.2[kW]
-
-
We can solve for vX and get
vX  3.9[V].
Some notes are worth making about this solution. You may have noticed
that several of the resistor values were not used in the solution. Specifically,
resistors R1, R2, R5 and R6 do not affect the solution. If you put in different values
for these resistors, the solution does not change. This is common. It might be
asked why these resistors are included if they do not affect the solution. There are
Page 1.8.7
Dave Shattuck
University of Houston
many possible answers to this question. One answer is that they are used for other
applications of the circuit. The resistors R5 and R6 in this circuit may be used to
limit the current when something is placed across the terminals of the voltage vX.
Another answer is that they do affect other values that were not solved for in this
problem. The resistors R1 and R2 do affect the amount of power delivered by the
current source. In any case, our major concern is to be able to solve the circuits as
they are given, which generally includes the ability to recognize what parameters
we will use, and what parameters we will not use. You will become better at this
as your skill in solving circuits improves.
Problem adapted from Final Exam, Problem 1, Summer 2000, University of Houston, Department of Electrical and Computer
Engineering, Cullen College of Engineering.
Page 1.8.8