Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

Th´venin’s, Norton’s, and Maximum Power Transfer theorems
e
This worksheet and all related files are licensed under the Creative Commons Attribution License,
version 1
...
To view a copy of this license, visit http://creativecommons
...
0/, or send a
letter to Creative Commons, 559 Nathan Abbott Way, Stanford, California 94305, USA
...

Resources and methods for learning about these subjects (list a few here, in preparation for your
research):

1

Questions
Question 1
Ideal voltage sources and ideal current sources, while both being sources of electrical power, behave very
differently from one another:

Voltage sources

Current sources

V

Explain how each type of electrical source would behave if connected to a variable-resistance load
...
That is, a perfect
voltage source will hold its output voltage constant regardless of the load imposed upon it:

Ideal voltage sources assumed

Esource

10 V

1 kΩ

Rload

Esource

10 V

1 mΩ

E = 10 V

E = 10 V

I = 10 mA

Rload

I = 10 kA

In real life, there is no such thing as a perfect voltage source, but sources having extremely low internal
resistance come close
...
A common symbol for a current source is a circle with an arrow
inside (always pointing in the direction of conventional flow, not electron flow!)
...
Why don’t the calculated figures from the previous paragraph agree with the actual measurement?
file 00106
Question 4
Calculate the voltage dropped across the load resistor, and the current through the load resistor, for
the following load resistance values in this circuit:

1
...

file 00389

4

Question 5
Calculate the voltage dropped across the load resistor, and the current through the load resistor, for
the following load resistance values in this circuit:

1
...

file 00390

5

Question 6
A very common sort of graph used in electronics work is the load line, showing all possibilities of load
voltage and load current that a particular power source is able to supply to a load:

12
11
10

Each point on the load line represents
the output voltage and current for a
unique amount of load resistance
...
At high currents, the output voltage will be very low (upper-left end of load line)
...
If all internal
components of the power source are linear in nature, the load line will always be perfectly straight
...
2 kΩ
...
Hint: it
only takes two points to define a line!
file 03513

6

Question 7
In the following circuit, an adjustable voltage source is connected in series with a resistive load and
another voltage source:

Vconstant

Vadjust

1 kΩ

Rload

Determine what will happen to the current in this circuit if the adjustable voltage source is increased
...

One way to define electrical resistance is by comparing the change in applied voltage (∆V ) to the change
in resultant current (∆I)
...

Determine the amount of voltage output by an open-circuited (ideal) current source
...
Finally, draw an equivalent circuit
showing an ideal current source somehow connected to a resistance in such a way that its open-circuited
output voltage is identical to the practical current source
...
Inside this box, you were told, was a voltage source (an ideal voltage source connected
in series with a resistance):

Box

Terminals

How would you experimentally determine the voltage of the ideal voltage source inside this box, and
how would you experimentally determine the resistance of the series resistor? By ”experimentally,” I mean
determine voltage and resistance using actual test equipment rather than assuming certain component values
(remember, this ”black box” is sealed, so you cannot look inside!)
...
The output voltage of real batteries, though, always ”sags” to
some degree under the influence of a load
...
How do you suggest the internal resistance of a chemical battery be experimentally measured?
file 03223
Question 11
Suppose you were handed a black box with two metal terminals on one side, for attaching electrical
(wire) connections
...

file 01038

9

Question 12

Load lines are special types of graphs used in electronics to characterize the output voltage and current
behavior of different power sources:

12
11
Each point on the load line represents
the output voltage and current for a
unique amount of load resistance
...
And, if we know the plot will be a straight line, all we need in
order to plot a complete load line are two data points
...
In other words,
we see how much voltage the source will output with no load connected (Iload = 0 milliamps) and then we
see how much current the source will output into a direct short (Vload = 0 volts):
10

Open-circuit condition

12
11

Rinternal

10

Vinternal

9

VOC = ???

ISC

8
7
Iload
6
(mA)
5

Short-circuit condition
Rinternal

4

Vinternal

3
2

ISC = ???

VOC

1
0
0 1

2

3

4

5

6

7

8

9 10 11 12

Vload (volts)
Suppose we have two differently-constructed power sources, yet both of these sources share the same
open-circuit voltage (VOC ) and the same short-circuit current (ISC )
...
In other words, explain how we would know just from
the limited data of VOC and ISC that these two power sources will behave exactly the same when connected
to the same load resistance, whatever that load resistance may be
...
5 volts

VOC = 7
...
75 mA

ISC = 8
...
Inside this box, you were told, was a voltage source connected in series with a resistance
...
From your experimental data you then sketched a circuit with the following
component values:

5 kΩ
6V

However, you later discovered that you had been tricked
...

file 02020

12

Question 14
Suppose you were handed a black box with two metal terminals on one side, for attaching electrical
(wire) connections
...


Box

???

Terminals

Your task was to experimentally determine the values of the voltage source and the resistor inside the
box, and you did just that
...
33 kΩ
5V

However, you later discovered that you had been tricked
...

file 03227

13

Question 15
Examine this circuit, consisting of an ideal voltage source and several resistors:

1 kΩ

24 V

1 kΩ

Load terminals

1 kΩ

1 kΩ

First, calculate the voltage seen at the load terminals with a voltmeter directly connected across them
(an open-circuit condition):

V

A

V

A

1 kΩ

1 kΩ

OFF

A

24 V

1 kΩ

COM

1 kΩ

Next, calculate the current seen at the load terminals with an ammeter directly connected across them
(a short-circuit condition):
14

V

A

V

A

1 kΩ

1 kΩ

OFF

A

24 V

1 kΩ

COM

1 kΩ

Based on these open- and short-circuit calculations, draw a new circuit consisting of a single voltage
source and a single (series) resistor that will respond in the exact same manner
...

file 03228

15

Question 16
Suppose you were handed a black box with two metal terminals on one side, for attaching electrical (wire)
connections
...


Box

???

Terminals

Your task was to experimentally determine the values of the current source and the resistor inside the
box, and you did just that
...
Instead of containing a current source and a
resistor, the circuit inside the box was actually a voltage source connected in series with a resistor:

470 Ω
5
...

file 02024

16

Question 17
Suppose you had an AC/DC power supply, which performed as follows (open-circuit and loaded test
conditions):

Off

Switch off:
Vout = 14
...
8 volts DC
Iout = 845 mA DC

Lamp

Switch

Draw a Th´venin equivalent circuit to model the behavior of this power supply
...
Take for example these measurements, under loaded and no-load conditions:

Clamp-on
ammeter

arc

Metal plates being
welded together

V

A

V

A
OFF

A

18

COM

Clamp-on
ammeter

(no arc)

V

A

V

A
OFF

A

COM

Based on these measurements, draw a Norton equivalent circuit for the arc welder
...
Note the voltage and current measurements taken for this particular welder:

Welding

Not welding

Clamp-on
ammeter

Clamp-on
ammeter

(arc)

V

V

A

V

A

V

A

A
OFF

OFF

A
A

COM

COM

Determine two Th´venin equivalent circuits for the arc welder
...
The second circuit will be a voltage source and internal impedance
connected through an ideal transformer with a step-down ratio of 8 to 1:

Simple equivalent circuit
ZTh

Equivalent circuit with transformer
ZTh
8:1

VTh

VTh

VTh =

VTh =

ZTh =

ZTh =

file 03665

20

Question 20
An electric arc welder is a low-voltage, high-current power source used to generate hot arcs capable of
melting metal
...
The first circuit will simply be an AC
e
voltage source and an internal impedance
...
2 kΩ
file 02021
Question 23
Give a step-by-step procedure for reducing this circuit to a Norton equivalent circuit (one current source
in parallel with one resistor):

1 kΩ

30 V

1 kΩ

5 kΩ

2
...

e
• Step #1:
• Step #2:
file 02456
Question 25
Give a step-by-step procedure for ”Nortonizing” any circuit: finding the Norton equivalent current
(VN orton ) and Norton equivalent resistance (RN orton )
...
2 kΩ resistor; then calculate the voltage across the 1 kΩ load:

10 V

5 kΩ

2
...
However, there is definite value in determining a Th´venin equivalent circuit for
e
the voltage source, 5 kΩ resistor, and 2
...
Explain why Th´venin’s Theorem becomes the more efficient way to predict
e
load voltage if multiple load resistor values are considered
...
3 kΩ resistor; then calculate the voltage across the 1 kΩ load:

3
...

file 03243

25

Question 30
Convert the following Th´venin equivalent circuit into a Norton equivalent circuit:
e

30 Ω
15 V

file 03240
Question 31
Convert the following Norton equivalent circuit into a Th´venin equivalent circuit:
e

2
...
However, we may
approximate the behavior of an ideal current source with a high-voltage source and large series resistance:

Thevenin approximation
Ideal current source
high R

(approximately
equivalent to)

high V

Such a Thevenin equivalent circuit, however imperfect, will maintain a fairly constant current through
a wide range of load resistance values
...

Thankfully, though, it is not difficult to build voltage sources that are relatively close to perfect: circuits with
very low internal resistance such that the output voltage sags only a little under high-current conditions
...
Draw a Norton equivalent circuit showing
how to approximate an ideal voltage source using an ideal (perfect) current source and a shunt resistance
...
However, it should be realized that their
output voltages ”sag” under load:

Unloaded

Loaded

1 kΩ

12 V

1 kΩ

1 kΩ

1 kΩ

12 V

Less than

8V

8V
1 kΩ

1 kΩ

4V

Less than

4V

Load

Just how much a voltage divider’s output will sag under a given load may be a very important question
in some applications
...
However, if we were asked to predict the voltage across the sensor supply terminals for a
variety of different sensor current conditions, we would be faced with a much more complex problem:
Sensor current (Isensor )
0 mA
1 mA
2 mA
3 mA
4 mA
5 mA

Sensor supply voltage
8 volts

28

One technique we could use to simplify this problem is to reduce the voltage divider resistor network
into a Th´venin equivalent circuit
...

Show how this could be done, then complete the table of sensor supply voltages shown above
...

Later that day, her instructor assigns a quick lab exercise: measure the current through a parallel resistor
circuit with an applied voltage of 3 volts, as shown in the following schematic diagram
...

This will be a great opportunity to use the new power source circuit, as 3 volts is well within the voltage
adjustment range!
She first sets up her circuit to output 3 volts precisely (turning the 10 kΩ potentiometer to the 50%
position), measuring with her voltmeter as she did when initially testing the circuit
...
This is a very large discrepancy between her prediction and the measured value for current!
Use Th´venin’s Theorem to explain what went wrong in this experiment
...
What is the equivalent source voltage and resistance, as seen from
the load terminals?

AC voltage source
50 Ω

10:1

24 VAC

Rload

file 01034

31

Question 36
Observe the following circuit:

9V

1

kΩ

kΩ

2
2
...
7 kΩ

1
...


Note that it is not reducible to a single resistance and power source
...
And, while it is a bridge circuit, you are not able to simply analyze the resistor
ratios because it is obviously not in a state of balance!
If you were asked to calculate voltage or current for any component in this circuit, it would be a difficult
task
...
2 kΩ resistor (the resistor
in the upper-right corner of the bridge)
...
2 kΩ resistor as the load in a Th´venin or
e
Norton equivalent circuit
...

file 01881
Question 39
At what load resistance value is the power dissipation maximized for the source’s internal resistance?

Rsource

Vsource

+
−

Rload

file 02028
Question 40
Explain why it is important for optimum performance to connect speakers of the proper impedance to
an audio power amplifier
...
Explain how we could experimentally determine the optimum internal resistance of
the electrolysis cell, prior to actually building it, using nothing but the solar panel, a rheostat, and a DMM
(digital multimeter)
...


+V
RC
Zout = ???

R1

Rsource

R2
RE

When in its active mode, a transistor operates like a current regulator
...
Draw an equivalent
circuit for the amplifier during this Th´venizing/Nortonizing process to show how the output impedance is
e
determined
...
Without considering the presence of the transistor or the
emitter resistance, calculate the impedance as ”seen” from the input terminal resulting from the two resistors
R1 and R2 in the following common-collector amplifier circuit:

-V
39 kΩ

R1

10 kΩ

R2

Zin
Vout
RE

Remember, what you are doing here is actually determining the Th´venin/Norton equivalent resistance
e
as seen from the input terminal by an AC signal
...

Next, calculate the input impedance of the same circuit, this time considering the presence of the
transistor and emitter resistor, assuming a current gain (β or hf e ) of 60, and the following formula for
impedance at the base resulting from β and RE :
ZB ≈ (β + 1)RE

-V
39 kΩ

R1
β = 60

Zin
10 kΩ

R2

Vout

2
...

file 03127

36

Question 45
Sometimes you will see amplifier circuits expressed as collections of impedances and dependent sources:

Amplifier
Zout
Dependent
voltage source

Zin

Vin

AVVin

Vout

With this model, the amplifier appears as a load (Zin ) to whatever signal source its input is connected
to, boosts that input voltage by the gain factor (AV ), then outputs the boosted signal through a series
output impedance (Zout ) to whatever load is connected to the output terminals:

Amplifier

Signal source

Vsource

Load

Zout

Zsource
Vin

Dependent
voltage source

Zin

AVVin

Vout

Zload

Explain why all these impedances (shown as resistors) are significant to us as we seek to apply amplifier
circuits to practical applications
...
Typically, students
practice by working through lots of sample problems and checking their answers against those provided by
the textbook or the instructor
...

You will learn much more by actually building and analyzing real circuits, letting your test equipment
provide the ”answers” instead of a book or another person
...

2
...

4
...

Draw the schematic diagram for the circuit to be analyzed
...

Check the accuracy of the circuit’s construction, following each wire to each connection point, and
verifying these elements one-by-one on the diagram
...
Mathematically analyze the circuit, solving for all values of voltage, current, etc
...
Carefully measure those quantities, to verify the accuracy of your analysis
...
If there are any substantial errors (greater than a few percent), carefully check your circuit’s construction
against the diagram, then carefully re-calculate the values and re-measure
...

I recommend resistors between 1 kΩ and 100 kΩ, unless, of course, the purpose of the circuit is to illustrate
the effects of meter loading!
One way you can save time and reduce the possibility of error is to begin with a very simple circuit and
incrementally add components to increase its complexity after each analysis, rather than building a whole
new circuit for each practice problem
...
This way, you won’t have to measure any component’s value more
than once
...
An ideal current source will output as
much or as little voltage as necessary to maintain a constant current through it, for any given load resistance
...

Answer 3
Ohm’s Law would suggest an infinite current (current = voltage divided by zero resistance)
...

If you think that the wire used in the experiment is not resistance-less (i
...
it does have resistance),
and that this accounts for the disparity between the predicted and measured amounts of current, you are
partially correct
...
However, if you re-calculate current with a wire resistance of 0
...

Follow-up question #1: explain why wire resistance alone does not explain the modest short-circuit
current
...

Answer 4
I’ll let you do the calculations on your own! Hint: there is a way to figure out the answer without having
to calculate all five load resistance scenarios
...


39

Answer 5
I’ll let you do the calculations on your own! Hint: there is a way to figure out the answer without having
to calculate all five load resistance scenarios
...

Follow-up question #2: although it is difficult to find real devices that approximate ideal current sources,
there are a few that do
...
Describe which scenario
would be the safest from a perspective of shock hazard: an open-circuited current transformer, or a shirtcircuited current transformer
...
2 kΩ

7
Iload
6
(mA)
5

11 V

Rload

4
3
2
1
0
0 1

2

3

4

5

6

7

8

9 10 11 12

Vload (volts)

Hint: the easiest points on find on this load line are the points representing open-circuit and short-circuit
conditions (i
...
Rload = ∞ Ω and Rload = 0 Ω, respectively)
...
In the
second circuit, current will remain constant as Vadjust is increased, yielding an infinite total resistance
...
The practical current source shown
in the diagram outputs 24 volts
...
Compare this with the internal resistance of an ideal voltage source
...

The voltage source’s value is measured, while the resistor’s value is calculated using Ohm’s Law
...

The current source’s value is measured, while the resistor’s value is calculated using Ohm’s Law
...
This
means that any load resistance, when connected to each of the power sources, will experience the exact same
voltage and current
...
The equivalence you see here is an application of Th´venin’s Theorem
...
The equivalence you see here is an application of Th´venin’s Theorem
...
667 kΩ

8V

Follow-up question: is this circuit truly equivalent to the original shown in the question? Sure, it
responds the same under extreme conditions (open-circuit and short-circuit), but will it respond the same
as the original circuit under modest load conditions (say, with a 5 kΩ resistor connected across the load
terminals)?

42

Answer 16
A good way to demonstrate the electrical equivalence of these circuits is to calculate their responses
to identical load resistor values
...

Follow-up question: give a step-by-step procedure for converting a Th´venin equivalent circuit into a
e
Norton equivalent circuit, and visa-versa
...
775 Ω

14
...
102 Ω

Answer 19
Simple equivalent circuit
VT h = 45
...
2478 Ω
Equivalent circuit with transformer
VT h = 366
...
86 Ω
Answer 20
Simple equivalent circuit
VT h = 48
...
230 Ω
Equivalent circuit with transformer
VT h = 388 volts
ZT h = 14
...
89 A

Answer 21
Yes, because all the constituent components are linear and bilateral
...
Here is the equivalent circuit for the circuit given in the question:

2
...
293 V

Load terminals

Answer 23
I will let you research the procedure for determining Norton equivalent circuits, and explain it in your
own words
...
198 mA

2
...
I’ll leave you to it!
Follow-up question: describe the difference in how one must consider voltage sources versus current
sources when calculating the equivalent circuit’s resistance (RT hevenin ) of a complex circuit containing both
types of sources?
Answer 25
This is easy enough for you to look up in any electronics textbook
...
209 volts

44

Answer 27
Vload = 2
...
Explain why this is
...
53

8
...
What
e
safety issues might be raised by the parallel connection of large batteries such as these?
Answer 30

0
...
7 Ω
RTh
8V

VTh

Sensor current (Isensor )
0 mA
1 mA
2 mA
3 mA
4 mA
5 mA

Sensor supply voltage
8 volts
7
...
667 volts
6 volts
5
...
667 volts

Follow-up question: if we cannot allow the sensor supply voltage to fall below 6
...

e

46

Answer 34
With the 10 kΩ potentiometer set in the 50% position, this student’s power source circuit resembles a
3 volt source in series with a 5 kΩ resistance (the Th´venin equivalent circuit) rather than an ideal 3 volt
e
source as assumed when she made her prediction for circuit current
...

Follow-up question #2: identify at least one circuit failure which would result in zero measured
(ammeter) current
...
4 VAC; equivalent source resistance = 0
...

Answer 36
V2
...
6624 volts
Hint: the Th´venin equivalent circuit looks like this (with the 2
...
64 Ω

7
...
2 kΩ

47

Answer 37

5 kΩ
Rinternal
24 V

Rload

30

Pload
(mW)

20

10

0
1k

3k

5k

7k

9k

11k 13k 15k 17k 19k

Rload (Ω)
Answer 38
This I leave to you to research!
Answer 39
Source dissipation will be maximized at Rload = 0 Ω
...

Answer 40
Improper speaker impedance may result in low power output (excessive Z) or overheating of the amplifier
(insufficient Z)
...
Surprised at this answer? Expecting power to
be maximized at Rsource = Rload , perhaps? If so, you have misunderstood the Maximum Power Transfer
Theorem
...

Follow-up question: assuming that the open-circuit voltage of this solar panel were high enough to pose
a shock hazard, describe a procedure you might use to safely connect a ”test load” to the panel
...
959 kΩ
Zin (complete circuit) ≈ 7
...

Typically, the values of Zsource and Zload are fixed by the nature of the source and load devices, respectively,
and the only impedances we have the freedom to alter are those within the amplifier
...
Have them invent voltage and current values for these voltage and current sources,
respectively, then calculate all other circuit parameters given several different values of load resistance
...
However, there are devices the closely approximate ideal current sources
(current transformers in AC circuits and ”current mirror” DC transistor circuits, for example)
...
A student of
mine once stuffed a 6-volt ”lantern” battery in his tool pouch, only to have it discharge smoke an hour later,
after the battery terminals had been shorted together by a wrench handle!
No, Ohm’s Law is not being cheated here: shorting a voltage source with a 0 Ω conductor will not
result in infinite current, because there are other sources of resistance in such a circuit
...

Notes 4
Ask your students to make a general prediction about the internal resistance of voltage sources
...
Is
this a relative determination, or an absolute determination?
Notes 5
Ask your students to make a general prediction about the internal resistance of current sources
...
Is
this a relative determination, or an absolute determination?
Notes 6
The purpose of this question is to lend an analytical geometric perspective to the subject of power
source behavior, by showing how the output voltage and current may be plotted on a graph
...

Notes 7
If students are unable to analyze the two circuits qualitatively as suggested in the question, the follow-up
question should clear things up
...

In case anyone should ask, the proper definition for resistance is expressed as a derivative
...

∆I
dI
Notes 8
A point of difficulty with some students is the word infinite
...
If any of your students are
confused in the same manner, it will become evident when they try to explain the open-circuit output voltage
of an ideal current source
...

Notes 10
Although real chemical batteries do not respond as simply as this equivalent circuit would suggest, the
model is accurate enough for many purposes
...

Notes 12
This is a ”poor man’s proof” of Th´venin’s and Norton’s theorems: that we may completely characterize
e
a power source in a simple, equivalent circuit by finding the original circuit’s open-circuit voltage and shortcircuit current
...

Notes 13
Ask your students to clearly state Th´venin’s Theorem, and explain how it may be applied to the
e
two-resistor circuit to obtain the one-resistor circuit
...

Notes 15
The purpose of this question is to get students thinking about Th´venin equivalent circuits from the
e
perspective of how the original circuit responds to extreme variations in load resistance
...

Notes 16
Ask your students to clearly state both Th´venin’s and Norton’s Theorems, and also discuss why both
e
these theorems are important electrical analysis tools
...
Discuss also the limitations
of this modeling, especially in light of the condition of linearity for the proper application of Th´venin’s
e
theorem
...
Of course, we must make certain assumptions when
modeling in this fashion: we assume, for instance, that the arc welder is a linear device, which may or may
not be true
...
AC clamp-on meters are simpler, cheaper, and thus more popularly
known, but devices using the Hall effect are capable of inferring DC current by the strength of an unchanging
magnetic field, and these Hall-effect devices are available at modest expense
...
Of course, we must make certain assumptions when
modeling in this fashion: we assume, for instance, that the arc welder is a linear device, which may or may
not be true
...
Of course, we must make certain assumptions when
modeling in this fashion: we assume, for instance, that the arc welder is a linear device, which may or may
not be true
...
” Can they give examples of components that are nonlinear,
and/or unilateral?
Notes 22
It should be easy for your students to research an algorithm (step-by-step procedure) for determining a
Th´venin equivalent circuit
...
Let them do the work, and explain it to you and their classmates!
Notes 24
I really mean what I say here about looking this up in a textbook
...

The follow-up question is very important, because some circuits (especially transistor amplifier circuits)
contain both types of sources
...
When performing this analysis on transistor amplifiers,
the circuit often becomes much simpler than its original form with all the voltage sources shorted and current
sources opened!

52

Notes 25
I really mean what I say here about looking this up in a textbook
...

The follow-up question is very important, because some circuits (especially transistor amplifier circuits)
contain both types of sources
...
When performing this analysis on transistor amplifiers,
the circuit often becomes much simpler than its original form with all the voltage sources shorted and current
sources opened!
Notes 26
Ask your students to show how (step-by-step) they arrived at the equivalent circuit, prior to calculating
load voltage
...

Notes 27
Ask your students to show how (step-by-step) they arrived at the equivalent circuit, prior to calculating
load voltage
...

Notes 28
Nothing but practice here
...

Notes 29
Ask your students whether they used Th´venin’s Theorem or Norton’s theorem to solve for the fault
e
current
...

Notes 30
Nothing special here, just practice converting between Thevenin and Norton sources
...

Notes 31
Nothing special here, just practice converting between Thevenin and Norton sources
...

Notes 32
This question is not so much a practical one as it is designed to get students to think a little deeper
about the differences between ideal voltage and current sources
...


53

Notes 33
Students are known to ask, ”When are we ever going to use Th´venin’s Theorem?” as this concept is
e
introduced in their electronics coursework
...
This question does exactly that: demonstrate how to predict voltage ”sag” for a loaded
voltage divider in such a way that is much easier than using Ohm’s Law and Kirchhoff’s Laws directly
...
Nothing significant about this choice –
just an opportunity for students to see other ways of drawing schematics
...

Notes 34
This challenges students to apply Th´venin’s Theorem to a practical scenario: a loaded voltage divider
...

Notes 35
Ask your students to explain how they obtained the equivalent voltage and current figures for this
transformer-coupled source
...
Be prepared to help them through this step during
discussion time
...
While it may not be intuitive, at least it is useful and easy to remember!
Notes 39
The wrong answer anticipated in the ”Answer” section reflects a common student misconception, and
a tendency to memorize simple rules rather than think and analyze circuit behavior
...

Notes 40
I find it typical that students of electronics new to basic electrical theory are fascinated by discussions
related to audio technology, as it is difficult to find someone who does not in some way appreciate music
...


54

Notes 41
The wrong answer anticipated in the ”Answer” section reflects a common student misconception, and
a tendency to memorize simple rules rather than think and analyze circuit behavior
...

Notes 42
Students should at this point understand the maximum power transfer theorem, and also the concept
of a voltage source having a certain amount of internal resistance
...
Do not be surprised if a student suggests using the meter
to measure the panel’s resistance directly (though this will not work with a real photovoltaic panel)
...

Notes 43
The main problem students usually have when Th´venizing or Nortonizing this circuit is what to do
e
with the current source
...

Remind your students if necessary that each source is to be replaced by its respective internal impedance
...
For
current sources (with infinite internal impedance, ideally) it means replacing them with open circuits
...
The most
e
confusing point of this for most students seems to be how to regard the DC power supply
...

e
′
To be proper, the transistor’s dynamic emitter resistance (re ) could also be included in this calculation,
but this just makes things more complex
...
With an emitter resistor value of 1500 ohms, the dynamic emitter resistance is
negligibly small anyway
...
Many
interesting things to discuss here!

55

Notes 46
It has been my experience that students require much practice with circuit analysis to become proficient
...
While this approach makes students proficient in
circuit theory, it fails to fully educate them
...
They also need real, hands-on practice building circuits
and using test equipment
...
This way, the mathematical theory ”comes alive,” and students gain practical proficiency
they wouldn’t gain merely by solving equations
...
Students
will also develop real troubleshooting skills as they occasionally make circuit construction errors
...
Discuss these issues with your students in the same Socratic manner you would normally discuss
the worksheet questions, rather than simply telling them what they should and should not do
...
If your goal is to educate theoretical physicists, then stick with abstract analysis, by all means!
But most of us plan for our students to do something in the real world with the education we give them
...

Furthermore, having students build their own practice problems teaches them how to perform primary
research, thus empowering them to continue their electrical/electronics education autonomously
...
Nuclear physics, biology, geology, and chemistry professors would just love to be able to have their
students apply advanced mathematics to real experiments posing no safety hazard and costing less than a
textbook
...
Exploit the convenience inherent to your science, and get those students
of yours practicing their math on lots of real circuits!

56



---