Voltage difference
How can I efficiently deliver electrical power
to your house?
What do magnets have to do with electricity?
Announcements
• Power Transmission (today and Thursday)
– Power loss to wires
– Delivering Power (role of transformers)
– Creating electrical current (role of generators)
• Big Picture
– Electronics: all about manipulating charges,
transmitting power
Lab #2 (Electromagnets and Transformers) this week.
time
Day 5::
Questions?
Read Power transmission Blm 11.2
and Transformers Blm 11.3
Reminders/Updates:
Labs Today! Build your own
Electromagnet & Transformer
Homework #2 due this Friday
Make sure to do pre-lab on separate sheet ahead of time.
Homework #2 on web page due on Friday at 5 pm.
Problem Solving Sessions Thursday 2-4pm, Friday 1-2pm.
* Sessions in the same room as labs!
Must be in a group for clickers by end of week!
Power distribution and generation
Distributing Power Across the Country
• Long wires from power generating plants to
homes (100’s of miles)
Why use AC power?
1. Energy loss in wires
2. Virtues of high voltage
3. Transformers and how they work
Power system and how transformers work in it.
• How much power will we lose in the wires?
Where does the “lost” power go?
Power generation (importance of Niagara Falls)
Big ideas:
• How can we minimize power lost in the wires?
• Changing magnetic fields produce
Voltages and currents
• Currents produce magnetic fields.
 Explore what affects power loss.
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Electron man!
Compare the brightness of bulb #1 in both cases…
1) Lose energy by bumping into stuff in atoms.
2) Lose energy everywhere but in some parts more than others.
3) Voltage change across something represents amount of energy lost
across it.
A) Brighter in case i
case i
B) Brighter in case ii
C) Same Brightness
+
1.5 V
1
lots of energy
at start.
Lots of volts
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Wires: glide down pretty easily, just a few bumps.
Lose a little bit of energy.
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1
case ii
2
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Usually
small
Rtotal = Rlight + Rwires
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Exhausted! energy used up getting
through course.
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Light Bulb…
high resistance like
trudging through
three feet of mud!
takes lots of energy
to get through.
B) Brighter in case ii
Two bulbs have more resistance than
one  RTotal = R1+R2
so less current flows by V = I R.
Then Power = P = IV = I2R
1.5 V
1
“60 Watt” Bulb
Light Bulb
Invented by Thomas Edison
(incandescent type)
Tungsten filament with resistance R,
gets hot and emits light
This means when hooked up to 120
Volts (typical outlet), you get
60 Watts of Power [energy/time].
What is resistance of this filament?
A) 60 Ohms
B) 120 Ohms
C) 240 Ohms D) 1000 Ohms
Power = V I  60 = 120 I  I = 0.5 Amps
V = I R  120 = 0.5 R  R = 240 Ohms
“60 Watt” Bulb
120
Volts
Wall Outlet
120 Volts AC
No. So the name is given for 120 Volts…
240
Volts
For a light bulb labeled “125
Watts” is the resistance of the
filament higher or lower than for
a “60 Watt” bulb?
A) Higher
B) Lower
C) Same
Power = V I = I^2 R
“125 Watt” Bulb
120
Volts
Thinking like an electron….
Lights and Power
No matter what the path, electrons lose all Voltage
(energy per electron) by the time they return to outlet.
Wiring inside home
Light
125 Watts
Light
Vpath1 = Vpath2 = Vbattery (approx. for ideal R=0 wires)
60 Watts
+
“Power Source”
(We can vary
voltage output)
If you hooked up this “60 Watt” Bulb
to a 240 Volt Battery (instead of
120 V), would you still get 60 Watts
of power?
What is different between a 60 Watt and a 125 Watt bulb?
a. the Voltage difference (amount of energy lost by each electron)
across the bulb
b. the number of electrons passing through the filament each second
c. Choice a and the resistance of the filament
d. Choice b and the resistance of the filament
e. Choice a and Choice b and the resistance of the filament
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V1 = Vstart
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V2 = Vstart
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Path 1
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Path 2
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No energy left
Thought process:
Same current must pass through both bulbs.
Can figure out current from total resistance:
Rtotal = (R 60 W bulb + R 120 W bulb + Rwires)
current i = V / Rtotal
Lights and Power
Wall Outlet
120 Volts AC
Wiring inside home
“Power Source”
(We can vary
Voltage output)
Assume wires don’t waste energy
lots of energy
at start. (V)
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Wiring inside home
+
Light
125 Watts
60W
Light
120W
60 Watts
What if we hook a 60 Watt and a 125 Watt light bulb in
series, which will be brighter?
(Share reasoning before experiment)
A) 125 Watt bulb will be brighter
B) 60 Watt bulb will be brighter
C) Same brightness for both
lots of energy
at start.
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60 Watt,
Higher Resistance
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Power = i2 R
125 Watt bulb
So which one is brighter? Lower Resistance
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120 V
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126
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glide down easily, just a few bumps. Hardly any energy.
Home Wiring
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!?#%, bridge
out, stuck.
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energy used up getting
through course. Vigor (V)
Wiring inside home
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lots of energy
at start.
Wall Outlet
120 Volts AC
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Light
HEATER
60 Watts
\/\/\/\/\/
Wiring inside home
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lots of energy
at start.
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deep
mud!
(bulb)
energy used up getting
through course. Vigor (V)
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pretty
easy
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e’s piled up down both routes, so still
divide up and go down both, just end
up faster on bridge route
What happens when bridge gets
fixed so have another route?
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Power into bulb = current x voltage drop in bulb
= 0.5 Amps x 119.95 Volts = 59.975 Watts
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Total resistance of wires = 0.1 Ohms
Resistance of lightbulb = 240 Ohms
Total resistance = Rwires + R bulbs = 240.1 Ohms
e’s piled up down both routes, so
still divide up and go down both,
just end up faster on bridge route
What happens when bridge gets
fixed so have another route?
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Voltage drop in wires = Current x resistance = 0.5 Amps x 0.1 Ohms = 0.05 Volts
Power into wires = Current x Voltage drop in wires
= 0.5 Amps x 0.05 Volts = 0.025 Watts
Light bulb dims. Why?
Discuss with your group and come up with possible reasons.
Share with class.
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How much power and where is it going? Work through together.
Plug voltage = 120 Volts
Current = Plug Voltage / Total R = 120 Volts / 240.1 Ohms = 0.5 Amps
Turn on Heater (close switch), observe light bulb.
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exhausted!
1500 Watts
120 V
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“Power Source”
(We can vary
voltage output)
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deep
mud!
(bulb)
energy used up getting
through course. Vigor (V)
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exhausted!
e’s piled up down both routes,
so still divide up and go down
both, just end up faster on
bridge route
What happens when bridge gets
fixed so have another route?
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lots of energy
at start.
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deep
mud!
(bulb)
energy used up getting
through course. Vigor (V)
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pretty
easy
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HINT: this is tricky – REMEMBER… the wires have small resistance!
I. Current through upper wire is now larger
II. Voltage drop across upper wire is now larger
III. Voltage drop around whole circuit is now larger
IV. Current through bulb is smaller
V. Voltage drop across bulb is smaller
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b. I and II
c. I, II, V
d. I, II, IV, V
e. I, II, III, IV, V
What will make bulb even dimmer?
I. Shorter wires
II. Longer wires
III. Adding another heater
IV. Thinner wires
V. Fatter wires
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What changes compared with bridge out (heater off) ?
a. I and III
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exhausted!
What changes compared with bridge out (heater off) ?
A)Current through upper wire is now larger
B) Current through the upper wire is the same
C) Current through the upper wire is smaller
D)It depends on the kind of wire
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pretty
easy
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lots of energy
at start.
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exhausted!
a. I and V
d. II, III, V.
b. I, II, and III
e. III only.
c. II, III, and IV
a. Can go easy path across bridge. Moves faster through (higher current).
b. Have to go down entry and exit road much faster. Lose more energy in wires
hitting bumps at high speed than when no heater.
c. Ones that pick mud path have a little less energy to
get through it than they did without bridge path, so get through it slower.
Answer is C.
Anything that either increase resistance of wires (e.g. thinner or
longer) OR increases current through the wires (e.g. adding
another heater).
Current through mud, I = (V - Vroad)/R, but going faster so Vroad bigger than before,
go through mud slower than if bridge out. I smaller. (Vroad = I Rroad)
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Power distribution and generation
Why use AC (Alternating Current) power?
1. Loss in wires
2. Virtues of high voltage
3. Transformers and how they work
Power system and how transformers work in it...
•Changing magnetic fields produce
Voltages and currents
•Currents (moving charges) produce
magnetic fields
Why is High Voltage Good
Power
Plant
want thick wires, no longer than
necessary, still have some small
Example: Deliver 100 Watts to house, P = IV
V = 10 V, I has to be 10 A.
V= 100 V, I has to be 1 A
V= 1000 V, I has to be? 0.1A
What is power loss in wires for each Voltage if R wire = 1 Ohm?
P = I Vdrop in wire (note: this is not Vcircuit)
P = I (IR)= I2R = (I2) 1 Ohm
10A  P= 100 Watt = same as power to house
1A  P = 1 W
0.1A  P = 0.01 W = 0.0001 x power to house.
so tremendous advantage to transmitting power at high V!
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