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Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Practice Examination Questions With Solutions
Module 4 – Problem 4
Filename: PEQWS_Mod04_Prob04.doc
Note: Units in problem are enclosed in square brackets.
Time Allowed: 35 Minutes
Problem Statement:
The device in Figure 1 can be modeled with a voltage source in series with a
resistance. The current and voltage for the device are related as shown in the plot
in Figure 2. The device has been connected in a circuit shown in Figure 3. Find iX.
A
iT in [mA]
+
5
vT
Device
-2
2
0
vT in [V]
iT
-
-5
B
Figure 1
Figure 2
iX
A
R1=
1[kW]
Device
+
-
B
vS1=
7[V]
R2=
1.8[kW]
Figure 3
Page 4.4.1
iS1=
8[mA]
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Problem Solution:
The device in Figure 1 can be modeled with a voltage source in series with a
resistance. The current and voltage for the device are related as shown in the plot
in Figure 2. The device has been connected in a circuit shown in Figure 3. Find iX.
A
iT in [mA]
+
5
vT
Device
-2
2
0
vT in [V]
iT
-
-5
B
Figure 1
Figure 2
iX
A
R1=
1[kW]
Device
+
-
B
vS1=
7[V]
R2=
1.8[kW]
iS1=
8[mA]
Figure 3
The first step in the solution is to find the model for the device. As noted in
the problem, it should be clear that the device can be modeled with a Thevenin
equivalent, that is, a voltage source in series with a resistor. So, we need to start by
finding this Thevenin equivalent.
Page 4.4.2
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
To get this we can find the open-circuit voltage and the short-circuit current.
In this case, that is a fairly straightforward process, using the plot in Figure 2.
However, it is very important that we take care about the polarities of the voltage
and current. Let’s take our time and do this carefully. Let’s begin by defining an
open-circuit voltage. We have the circuit that follows.
A
+
vT
+
vOC
Device
iT
-
-
B
We have chosen to define the open-circuit voltage, vOC, so that it will be in
the same polarity as vT. Thus we can look at the plot in Figure 2, and look at the
point where the current iT is zero. This will be the open-circuit voltage value,
which is also the Thevenin voltage. Thus, we have
vOC  2[V].
Next, we define the short-circuit current, iSC, so that it will be in the same
polarity as iT.
Page 4.4.3
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
A
+
vT
Device
iSC
iT
-
-
B
Again, we can get the value from the plot in Figure 2, looking at the point
where the voltage vT is zero. This will be the short-circuit current value. We have
iSC  5[mA].
The next step is very important. The Thevenin resistance will be the ratio of
the open-circuit voltage to the short-circuit current. However, to get the sign
correct, we need the proper relationship between the polarities of these two
quantities. We showed in the DPKC files that if we define the open-circuit voltage
at A with respect to B, and then define the short-circuit current from A through the
short to B, then the equivalent resistance will be
REQ  
vOC
2[V]

 400[W].
iSC 5[mA]
If we don’t do this carefully, we will obtain the wrong sign for the
equivalent resistance, and get the wrong answer.
An alternative approach is sometimes followed to get the
equivalent resistance. This equivalent resistance will be the inverse of
the slope of the line in Figure 2. However, we will again need to be
careful to be able to determine the proper sign for this resistance. The
sign that you obtain is a function of the sign convention used for the
device in Figure 1. Note that we have used the active sign convention
for the relationship between vT and iT in Figure 1. That is why the
equivalent resistance is the negative of the value of the inverse of the
slope. To show why, we will take a brief side trip.
Page 4.4.4
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
The Thevenin equivalent is shown in the figure that follows.
Notice that we have inserted the open-circuit voltage with the proper
polarity and value, but have not yet found the equivalent resistance,
REQ.
REQ
A
+
vT
+
vOC =
2[V]
iT
-
-
B
Now, let’s write KVL around the loop in this circuit. We have
vT  vOC  iT REQ  0. If we solve this for iT , we get
iT 
vT  vOC vT vOC


.
REQ
REQ REQ
This is an equation for a straight line on the iT vs vT axes.
Notice that the slope of this line is the coefficient of vT. Thus, we
have that
m
1
,
REQ
where m is the slope. Since the slope is
Page 4.4.5
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
m
5[mA] 1

, then it is clear that
2[V] REQ
REQ 
2[V]
 400[W].
5[mA]
This is exactly what we said earlier. When we use the active sign
convention for vT and iT, we have equivalent resistance equal to the
negative of the inverse of slope of the iT vs vT plot. You can follow a
similar process to show that if you use the passive sign convention for
vT and iT, the equivalent resistance will be equal to the inverse of slope
of the iT vs vT plot.
Now we are ready to solve for iX. We can now insert our model, the
Thevenin equivalent, for the device in Figure 3. We do this in the circuit that
follows.
iX
REQ = -400[W]
A
R1=
1[kW]
+
+
vOC =
2[V]
-
-
vS1=
7[V]
R2=
1.8[kW]
iS1=
8[mA]
B
We want to find iX, and this is pretty straightforward, now. We can write
that
iX 
vOC  vS1
2[V]  7[V]

, or
REQ  R1 400[W]  1[kW]
iX  8.33[mA].
Page 4.4.6
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
It might be noted that the values of the resistor R2 and the current source iS1
have no effect on the solution to this problem. This is because these two
components are in parallel with the voltage source, and thus have no effect on the
rest of the circuit.
Yet another way to find the value of the equivalent resistance is to draw the
equivalent circuit, as shown in the figure that follows.
REQ
A
+
vT
+
vOC =
2[V]
iT
-
-
B
Now, we know from the plot in Figure 2 that when
vT  0, then iT  5[mA].
This occurs when the two terminals, A and B, are shorted. When this is true, then
we know from KVL around that loop that
vOC  iT REQ , or
2[V]  5[mA]REQ , which can be solved to give
REQ  400[W].
This is yet a third way that gives us the proper sign for the equivalent resistance.
Problem adapted from ECE 2300, Exam 2, Problem 1, Summer 2000, Department of Electrical and Computer Engineering, Cullen College of
Engineering, University of Houston.
Page 4.4.7