Upcoming Schedule
Nov. 17
review
Nov. 19
Exam 3
Chap. 20-21
Nov. 21
21.6-21.7
18.8
Nov. 24
Chapter 22
Quiz 9
Nov. 26
vacation
Nov. 28
vacation
Dec. 3
boardwork
Dec. 5
23.6-23.9
Quiz 10a
Dec. 1
23.1, 23.4,
23.5
True Fact you didn’t know: all electrical motors operate on smoke. Every
motor has the correct amount of smoke sealed inside it at the factory. If
this smoke ever gets out, the motor is no longer functional.
http://www.xs4all.nl/~jcdverha/scijokes/2_16.html#subindex
Exam 3
Average score = 184.8 (very high!).
High score = 200.
Nice job.
End-of-semester quiz schedule is here. Let me explain it!
Grades spreadsheet is here. Let’s look at it for a minute.
Announcements page and Powerpoint lectures contain
“official” announcements. Pdf versions of lectures are for your
printing convenience, and may not contain the latest updates.
A note on attendance policy. The syllabus says:
“The lowest score of the four exams (3 semester, 1 final) will be dropped.
However, you will not be allowed to drop the final exam unless you
attend at least 2/3 of the scheduled class meetings after exam 3. If
you attend less than 2/3 of the scheduled class meetings during that time, I
will drop the lowest of the three in-semester exam scores.”
This means exactly what it says. If you attend at least 2/3 of
the final classes and skip the final exam, a “0” goes in your final
exam score, and will be the dropped score.
If you attend less than 2/3 of the final classes and skip the final
exam, a “0” goes in your final exam score. The dropped score
will be the lowest of your 3 exams. The “0” for the final will kill
your grade!
There are 8 class meetings after exam 3. You need to attend
5.4 of them to drop the final.
Because 5.4 classes doesn’t make sense, I’ll count exam 3. You
need to attend 6 of the 9 class meetings after (and including)
exam 3.
If you plan to take the final, you need to attend all classes!
You “should” attend all classes because this is such wonderful
stuff. (Actually, optics really is interesting and potentially
useful.)
Now for something completely off-topic…
Ozark Christmas
Castleman Hall, Saturday, December 13, 7:30 pm.
Christmas bluegrass
story dance
Tickets: $10 adult, $5 student
(including UMR), children free.
Sponsored by Rolla Kiwanis Club; every
penny goes to service projects in Rolla.
“The Night Before Christmas”
21-6 Counter emf
A changing magnetic field in wire produces a current. A
constant magnetic field does not.
In section 21.5 we saw how changing the magnetic field
experienced by a coil of wire produces ac current.
But the electrical current produces a magnetic field, which by
Lenz’s law, opposes the change in flux which produced the
current in the first place.
http://campus.murraystate.edu/tsm/tsm118/Ch7/Ch7_4/Ch7_4.htm
The effect is “like” that of friction.
The counter emf is “like” friction that opposes the original
change of current.
Motors have many coils of wire, and thus generate a large
counter emf when they are running.
Good—keeps the motor from “running away.” Bad—”robs”
you of energy.
If your house lights dim when an appliance starts up, that’s
because the appliance is drawing lots of current and not
producing a counter emf.
When the appliance reaches operating speed, the counter emf
reduces the current flow and the lights “undim.”
Motors have design speeds their engineers expect them to run
at. If the motor runs at a lower speed, there is less-thanexpected counter emf, and the motor can draw more-thanexpected current.
If a motor is jammed or overloaded and slows or stops, it can
draw enough current to melt the windings and burn out. Or
even burn up.
21-7 Transformers
No, no, no…
A transformer is a device for increasing or
decreasing an ac voltage.
Pole-mounted
transformer
Power Substation
ac-dc
converter
A transformer is basically two coils of wire wrapped around
each other, or wrapped around an iron core.
When an ac voltage is applied to the primary coil, it induces an
ac voltage in the secondary coil.
A “step up” transformer increases the output voltage in the
secondary coil; a “step down” transformer reduces it.
The ac voltage in the primary coil causes a magnetic flux
change given by
ΔφB
VP = NP
.
Δt
The changing flux (which is efficiently “carried” in the
transformer core) induces an ac voltage in the secondary coil
given by
ΔφB
VS = NS
.
Δt
Dividing the two equations gives the transformer equation
OSE:
VS
NS
=
.
VP
NP
For a step-up transformer, NS > NP and VS > VP (the voltage is
stepped up).
For a step-down transformer, NS < NP and VS < VP (the voltage
is stepped down).
Transformers only work with ac voltages; a dc voltage does not
produce the necessary changing flux.
A step-up transformer increases the voltage. Is this an
example of “getting something for nothing?”
No, because even though transformers are extremely efficient,
some power (and therefore energy) is lost.
If no power is lost, we can use P = IV to get
OSE:
IS
NP
=
.
IP
NS
flipped!
If transformers only work on ac, how come you showed a
picture of an ac-dc converter a few slides back?
Ever wanted to cut open one of those ac-dc converters and see
what they look like inside?
An ac-dc converter first steps down the 120 volt line voltage,
and then converts the voltage to dc:
fewer turns in the
secondary coil
a diode is a device that
lets current flow one
way only (dc)
Pictures of ac-dc converter came from
http://www.howstuffworks.com/inside-transformer.htm
Example 21-9 A transformer for home use of a portable radio
reduces 120 V ac to 9 V dc. The secondary contains 30 turns
and the radio draws 400 mA. Calculate (a) the number of
turns in the primary; (b) the current in the primary; and (c) the
power transformed.
VS
NS
=
VP
NP
VP
NP
=
VS
NS
VP
NP = NS
VS
120 V 

NP =  30 turns 
9 V 
NP = 400 turns
IS
NP
=
IP
NS
NS
IP
=
IS
NP
NS
IP = IS
NP
30 turns 

IP =  0.400 A 
 400 turns 
IP = 0.030 A
The power output to the secondary coil is
PS = IS VS
PS = 0.400 A   9 V 
PS = 3.6 W .
This is the same as the power input to the primary coil because
our transformer equation derivation assumed 100% efficient
transformation of power.
Example 21-10 An average of 120 kW of electrical power is
sent to a small town from a power plant 10 km away. The
transmission lines have a total resistance of 0.40 . Calculate
the power loss if power is transmitted at (a) 240 V and (b)
24,000 V.
This problem does not use the transformer equation, but it
shows why transformers are useful.
P = IV
P
I=
V
PLOST = I2 R
2
PLOST
P
=  R
V
(a) at 240 V
PLOST
 120×10 W 
=

240
V


3
2
 0.4 Ω 
PLOST = 100×103 W = 100 kW .
(a) at 24000 V
PLOST
 120×10 W 
=

 24000 V 
3
2
 0.4 Ω 
PLOST = 10 W .
More than 80% of the power would be wasted if it were
transmitted at 240 V, but less than 0.01 % is wasted if the
power is transmitted at 24000 V.
http://www.howstuffworks.com/power.htm