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Transcript
```Filename: PWA_Mod04_Prob05.ppt
Dave Shattuck
University of Houston
A device can be modeled with sources and resistors,
and has two terminals. When this device is
connected to an ideal voltmeter, as shown in Figure
1, the voltmeter reads vX = 11.4[V]. When this
same device is disconnected from the voltmeter,
and connected to an ideal ammeter, as shown in
iX = 120[mA]. Find the power absorbed by a 50[W]
resistor that is connected to this device.
Problems With Assistance
Module 4 – Problem 5
A
Device
+
vX
A
Voltmeter
Device
iX
Figure 1
B
Figure 2
Go
straight to
the First
Step
Go
Ammeter
straight to
the
Problem
Statement
B
Next slide
Dave Shattuck
University of Houston
Overview of this Problem
In this problem, we will use the following
concepts:
• Equivalent Circuits
• Thévenin’s Theorem
Go
straight to
the First
Step
Go
straight to
the
Problem
Statement
Next slide
Dave Shattuck
University of Houston
Textbook Coverage
The material for this problem is covered in your textbook in
the following sections:
• Circuits by Carlson: Sections #.#
• Electric Circuits 6th Ed. by Nilsson and Riedel: Sections
#.#
• Basic Engineering Circuit Analysis 6th Ed. by Irwin and
Wu: Section #.#
• Fundamentals of Electric Circuits by Alexander and
• Introduction to Electric Circuits 2nd Ed. by Dorf: Sections
#-#
Next slide
Dave Shattuck
University of Houston
Coverage in this Module
The material for this problem is covered in
this module in the following presentation:
• DPKC_Mod04_Part02
Next slide
Next slide
Dave Shattuck
University of Houston
Problem Statement
A device can be modeled with sources and resistors,
and has two terminals. When this device is
connected to an ideal voltmeter, as shown in Figure
1, the voltmeter reads vX = 11.4[V]. When this
same device is disconnected from the voltmeter,
and connected to an ideal ammeter, as shown in
iX = 120[mA]. Find the power absorbed by a 50[W]
resistor that is connected to this device.
A
Device
+
vX
A
Voltmeter
Device
iX
Figure 1
B
Figure 2
B
Ammeter
Dave Shattuck
University of Houston
Solution – First Step – Where to Start?
How should
we start this
problem?
What is the
first step?
A device can be modeled with sources and resistors,
and has two terminals. When this device is
connected to an ideal voltmeter, as shown in Figure
1, the voltmeter reads vX = 11.4[V]. When this
same device is disconnected from the voltmeter,
and connected to an ideal ammeter, as shown in
iX = 120[mA]. Find the power absorbed by a 50[W]
resistor that is connected to this device.
Next slide
A
Device
+
vX
A
Voltmeter
Device
iX
Figure 1
B
Figure 2
B
Ammeter
Dave Shattuck
University of Houston
How should we start this
problem? What is the first
step?
Problem Solution – First Step
A device can be modeled with sources and resistors,
and has two terminals. When this device is
connected to an ideal voltmeter, as shown in Figure
1, the voltmeter reads vX = 11.4[V]. When this
same device is disconnected from the voltmeter,
and connected to an ideal ammeter, as shown in
iX = 120[mA]. Find the power absorbed by a 50[W]
resistor that is connected to this device.
a)
Define the open-circuit
voltage.
b)
Attach a 50[W] resistor to
the device.
c)
Define the short-circuit
current.
d)
Model the device using
Thévenin’s Theorem.
e)
Model the device using
Norton’s Theorem.
A
Device
+
vX
A
Voltmeter
Device
iX
Figure 1
B
Figure 2
B
Ammeter
Dave Shattuck
University of Houston
Your choice for First Step –
Define the open-circuit voltage
A device can be modeled with sources and resistors,
and has two terminals. When this device is
connected to an ideal voltmeter, as shown in Figure
1, the voltmeter reads vX = 11.4[V]. When this
same device is disconnected from the voltmeter,
and connected to an ideal ammeter, as shown in
iX = 120[mA]. Find the power absorbed by a 50[W]
resistor that is connected to this device.
This is not a good choice for the
first step.
Actually, the open-circuit
defined. The ideal voltmeter is
effectively an open circuit.
Thus, vX is the open-circuit
voltage. It does not need to be
and try again.
A
Device
+
vX
A
Voltmeter
Device
iX
Figure 1
B
Figure 2
B
Ammeter
Dave Shattuck
University of Houston
Your choice for First Step –
Attach a 50[W] resistor to the device.
This is not a good choice for the
first step.
A device can be modeled with sources and resistors,
and has two terminals. When this device is
connected to an ideal voltmeter, as shown in Figure
1, the voltmeter reads vX = 11.4[V]. When this
same device is disconnected from the voltmeter,
and connected to an ideal ammeter, as shown in
iX = 120[mA]. Find the power absorbed by a 50[W]
resistor that is connected to this device.
We can attach the 50[W] resistor
to the device, and in fact will
need to do this later. However, if
we do it now, we will not be able
to solve the circuit that results.
We will not know the voltage
across it, as it will not be
11.4[V], and the current through
it will not be 120[mA]. Please
go back and try again.
A
Device
+
vX
A
Voltmeter
Device
iX
Figure 1
B
Figure 2
B
Ammeter
Dave Shattuck
University of Houston
Your choice for First Step –
Define the short-circuit current
A device can be modeled with sources and resistors,
and has two terminals. When this device is
connected to an ideal voltmeter, as shown in Figure
1, the voltmeter reads vX = 11.4[V]. When this
same device is disconnected from the voltmeter,
and connected to an ideal ammeter, as shown in
iX = 120[mA]. Find the power absorbed by a 50[W]
resistor that is connected to this device.
This is not a good choice for the
first step.
Actually, the short-circuit
defined. The ideal ammeter is
effectively an short circuit.
Thus, iX is the short-circuit
current. It does not need to be
and try again.
A
Device
+
vX
A
Voltmeter
Device
iX
Figure 1
B
Figure 2
B
Ammeter
Dave Shattuck
University of Houston
Your choice for First Step –
Model the device using Thévenin’s Theorem
A device can be modeled with sources and resistors,
and has two terminals. When this device is
connected to an ideal voltmeter, as shown in Figure
1, the voltmeter reads vX = 11.4[V]. When this
same device is disconnected from the voltmeter,
and connected to an ideal ammeter, as shown in
iX = 120[mA]. Find the power absorbed by a 50[W]
resistor that is connected to this device.
A
Device
+
vX
Voltmeter
This is a good choice for the first
step. Notice that we are told that
the device has two terminals, and
it is made up of sources and
resistors. Thus, we can use
Thévenin's theorem to model the
device. Once we have the
Thévenin equivalent, it will be
easy to find the voltage across a
50[W] resistor attached to the
device. Let’s find the equivalent.
A
Device
iX
Figure 1
B
Figure 2
B
Ammeter
Your choice for First Step –
Model the device using Norton’s Theorem
Dave Shattuck
University of Houston
This is a good choice for the first
step. Notice that we are told that
the device has two terminals, and it
A device can be modeled with sources and resistors, is made up of sources and resistors.
and has two terminals. When this device is
Thus, we can use Norton’s theorem
connected to an ideal voltmeter, as shown in Figure to model the device. Once we have
the Norton equivalent, it will be
1, the voltmeter reads vX = 11.4[V]. When this
easy to find the voltage across a
same device is disconnected from the voltmeter,
50[W] resistor attached to the
and connected to an ideal ammeter, as shown in
device. For this particular problem,
we have chosen to use Thévenin's
iX = 120[mA]. Find the power absorbed by a 50[W] Theorem. However, your choice
resistor that is connected to this device.
was just as good.
A
A
Device
+
vX
Voltmeter
Device
iX
Figure 1
B
Figure 2
B
Ammeter
Dave Shattuck
University of Houston
Modeling the Device and
Attaching the Voltmeter
A device can be modeled with sources and resistors,
and has two terminals. When this device is
connected to an ideal voltmeter, as shown in Figure
1, the voltmeter reads vX = 11.4[V]. When this
same device is disconnected from the voltmeter,
and connected to an ideal ammeter, as shown in
iX = 120[mA]. Find the power absorbed by a 50[W]
resistor that is connected to this device.
Device
RTH
Now, will this voltage, vTH, be
the voltage across the 50[W]
resistor? Click on yes or no.
A
+
+
-
vTH
vX
-
Figure 3
B
Let’s model the device using a
Thévenin equivalent. We have
done this in the circuit in Figure
3. Note that when we attach the
voltmeter to the device, as in
Figure 3, there is no current
through RTH, in this case. Thus,
vTH = vX = 11.4[V]. In other
the open-circuit voltage.
Voltmeter
vX  vTH  11.4[V].
Dave Shattuck
University of Houston
You Said That: Yes, vTH is the
voltage across the 50[W] resistor
A device can be modeled with sources and resistors,
and has two terminals. When this device is
connected to an ideal voltmeter, as shown in Figure
1, the voltmeter reads vX = 11.4[V]. When this
same device is disconnected from the voltmeter,
and connected to an ideal ammeter, as shown in
iX = 120[mA]. Find the power absorbed by a 50[W]
resistor that is connected to this device.
Device
RTH
When we remove the voltmeter
and connect a 50[W] resistor to
the device, the voltage drop
across RTH will mean that the
voltage across the resistor will
not be vTH. The correct answer is
no, the voltage across the resistor
will depend on RTH.
A
+
+
-
vTH
vX
-
Figure 3
B
Voltmeter
vX  vTH  11.4[V].
Dave Shattuck
University of Houston
You Said That: No, vTH is not the
voltage across the 50[W] resistor
A device can be modeled with sources and resistors,
and has two terminals. When this device is
connected to an ideal voltmeter, as shown in Figure
1, the voltmeter reads vX = 11.4[V]. When this
same device is disconnected from the voltmeter,
and connected to an ideal ammeter, as shown in
iX = 120[mA]. Find the power absorbed by a 50[W]
resistor that is connected to this device.
Device
RTH
vTH
A
vX
-
Figure 3
B
When we remove the voltmeter
and connect a 50[W] resistor to
the device, the voltage drop
across RTH will mean that the
voltage across the resistor will
not be vTH. The correct answer is
no, the voltage across the resistor
will depend on RTH.
Let’s find RTH. To do this, we
will apply the ammeter to the
device. Let’s apply the ammeter.
+
+
-
Voltmeter
vX  vTH  11.4[V].
Dave Shattuck
University of Houston
Modeling the Device and
Attaching the Ammeter
A device can be modeled with sources and resistors,
and has two terminals. When this device is
connected to an ideal voltmeter, as shown in Figure
1, the voltmeter reads vX = 11.4[V]. When this
same device is disconnected from the voltmeter,
and connected to an ideal ammeter, as shown in
iX = 120[mA]. Find the power absorbed by a 50[W]
resistor that is connected to this device.
Device
+
-
RTH
A
iX
vTH=
11.4[V]
Figure 4
B
We have modeled the device
using a Thévenin equivalent.
Then, we attach the ammeter to
the device, as in Figure 4. Our
goal here is to solve for RTH.
Clearly, the current through RTH
in this circuit is iX. The key then
is to find the voltage across RTH.
Remember that an ideal ammeter
has no voltage across it. Thus,
by KVL, the voltage across the
resistor must be vTH. See the next
slide.
Ammeter
vTH  11.4[V].
Dave Shattuck
University of Houston
Voltage Across RTH
A device can be modeled with sources and resistors,
and has two terminals. When this device is
connected to an ideal voltmeter, as shown in Figure
1, the voltmeter reads vX = 11.4[V]. When this
same device is disconnected from the voltmeter,
and connected to an ideal ammeter, as shown in
iX = 120[mA]. Find the power absorbed by a 50[W]
resistor that is connected to this device.
Device
+
+
-
RTH
vRTH
A
iX
+
vAM
iX
vTH=
11.4[V]
Ammeter
Figure 4
B
To find the voltage across RTH,
we write KVL around the loop,
and we get
vTH  vRTH  v AM  0, and
vAM  0, and so
vRTH  vTH  11.4[V].
Now, we have the voltage across
the resistor, and the current
through the resistor, and we have
them in the active sign
convention. Thus, we can write
RTH  
vTH
11.4[V]


iX
120[mA]
RTH  95[W].
Next slide
Dave Shattuck
University of Houston
Attaching 50[W] Resistor
A device can be modeled with sources and resistors,
and has two terminals. When this device is
connected to an ideal voltmeter, as shown in Figure
1, the voltmeter reads vX = 11.4[V]. When this
same device is disconnected from the voltmeter,
and connected to an ideal ammeter, as shown in
iX = 120[mA]. Find the power absorbed by a 50[W]
resistor that is connected to this device.
Device
A
+
v50
RTH= -95[W]
+
-
50[W]
vTH=
11.4[V]
Since we now have the
equivalent circuit for the device,
we can connect the 50[W] resistor
to the equivalent circuit. We get
the circuit shown in Figure 5.
Now, we can use the VDR to find
the voltage across the 50[W]
resistor, v50, as
50[W]
v50  11.4[V]
, or
95[W]  50[W]
v50  12.7[V].
Figure 5
B
Next slide
Dave Shattuck
University of Houston
Power Absorbed by the
50[W] Resistor
A device can be modeled with sources and resistors,
and has two terminals. When this device is
connected to an ideal voltmeter, as shown in Figure
1, the voltmeter reads vX = 11.4[V]. When this
same device is disconnected from the voltmeter,
and connected to an ideal ammeter, as shown in
iX = 120[mA]. Find the power absorbed by a 50[W]
resistor that is connected to this device.
Device
+
v50
50[W]
vTH=
11.4[V]
v50 
12.7[V]




, or
50[W]
50[W]
 3.21[W].
2
A
RTH= -95[W]
+
-
Finally, with the voltage across
the 50[W] resistor, v50, we can
find the power absorbed by the
resistor. We have
pabs ,50[ W ]
pabs ,50[ W ]
2
Figure 5
B
Next slide
Dave Shattuck
University of Houston
The Solution
A device can be modeled with sources and resistors,
and has two terminals. When this device is
connected to an ideal voltmeter, as shown in Figure
1, the voltmeter reads vX = 11.4[V]. When this
same device is disconnected from the voltmeter,
and connected to an ideal ammeter, as shown in
iX = 120[mA]. Find the power absorbed by a 50[W]
resistor that is connected to this device.
Go back to
Problem
Statement
pabs ,50[ W ]  3.21[W].
A
Device
+
vX
A
Voltmeter
Device
iX
Ammeter
Figure 1
B
Figure 2
B
Next slide
Dave Shattuck
University of Houston
What Happened Here?
• This seemed like another strange problem, with the Thévenin
resistance ending up negative. Not only that, but the voltage
from the voltage divider ended up being larger in magnitude
than the source, and with an opposite sign. Is this really
possible?
• The answer is yes, but we should remember that to do this we
generally need to have a dependent source. Remember that we
know almost nothing about the device in this problem. It could
easily have a dependent source (an amplifier) inside of it. In
some cases, this means a negative resistance can occur.
• It is due to the dependent source that we can
have a larger magnitude voltage at the terminals
than we had at the source of the equivalent.
The dependent source can provide positive
power. That’s what the negative resistance
means.
Go back to
Overview
slide.
```