Transcript Lecture33_DCpower

average
Class Averages
70.4
(3 exams w/bonus points
and
homework quizzes)
7
9
11
13
15
17
19
35 40 45 50 55 60 65 70 75 80 85 90 95 100
+
These cutoffs
will not be
raised…you
can use them
to determine
your guaranteed
lowest grade.
+
+
+
-
95
90
85
80
75
70
65
60
55
50
45
40
The B-field everywhere within this
current-carrying
loop points
_
+
1. up.
3. left.
2. down.
4. right.
5. into the screen.
6.out of the screen.
7. is zero.
+
-
The SOLENOID pictured carries
current in the direction indicated:
A positively charged
sphere is set spinning
as shown at left.
1)A magnetic field is set up with a
North Pole at the sphere’s top.
2)A magnetic field is set up with a
South Pole at the sphere’s top.
3)The magnetic field lines form
closed concentric circles
centered on the spin axis.
4)Since the charge stays centered
on a fixed position, no magnetic
field is generated.
Oxygen
O8
Iron
Fe26
1s
2s
2p
3s
3p
4s
3d
4p
5s
4d
Fe3O4 [Magnetite]
The slightly magnetic iron and oxygen atoms
are locked into a crystal that fixes their orientation.
Or the simple cubic IRON crystal
the magnetic field of any atom may
point in any of six different directions.
As molten iron solidifies, the magnetic
orientation of any atom is totally random.
So generally, any sample of iron is
overall (on average) UNMAGNETIZED.
By chance, neighboring atoms may share
the same orientation,
making microscopic, highly localized magnets.
The iron sample settles into microscopic
regions that are locally magnetized.
But overall, with no single dominant
direction, the iron remains UNMAGNETIZED.
These microscopic neighborhoods
are called DOMAINS
In the presence of a strong external
magnetic field
magnetic flipping can help some domains
to grow.
In the presence of a strong external
magnetic field
magnetic flipping can help some domains
to grow
and a sample of iron becomes magnetized!
An “induced”
magnetic
North pole.
Ohm’s Law
Power
V = IR
P = IV
P = I2R
P = V2/R
What’s the resistance of a 60-Watt bulb
(designed to be operated with 120 volt )?
A. R = 60/120 = 0.50 
B. R = 120/60 = 2.00 
C. R = 602/120 = 30 
D. R = 1202/60 = 240 
E. R = 60 120 = 720 
F. R = 602120 = 432,000 
What’s the resistance of a 100-Watt bulb
(designed to be operated with 120 volt )?
A. 1.20 
B. 83 
C. 144 
D. 360 
Nichrome (nickel/chromium alloy) elements
high resistance wire that can generate
enough heat to glow red hot:
541 Watt toaster
I = 4.5 A
R = 26.6 ohm
1500 Watt hair dryer
I = 12.5 A
R = 9.6 ohm
Household wiring
12 AWG copper wire
(0.08in diameter)
0.0052 /meter
Power cords
18 AWG copper
(0.04in diameter)
0.0210 /meter
Appliance wiring
“telephone wire”
24 AWG copper
(0.02in diameter)
0.0842 /meter
6 AWG copper
power transmission (0.162in diameter)
0.00130 /meter
Utility lines
Unlike electrical components
(resistors, inductors, transformers, motors)
we can think of the short copper wires
and tiny gold-plated copper traces on
a circuit board as being resistance-less.
Two bulbs are connected “in series”
with a 120-volt backup regulator.
Bulb B burns much brighter than Bulb A.
The current through Bulb B is ___ Bulb A.
1. less than
2. equal to 3. greater than
The power used in Bulb B is ___ Bulb A.
1. less than
2. equal to 3. greater than
The voltage across Bulb B is ___ Bulb A.
1. less than
2. equal to 3. greater than
120 V
“60 Watt”
“60 Watt”
240 
240 
Two identical bulbs would share the energy,
while doubling the resistance in the circuit.
Vtotal IR
Vtotal 120volts
I 
480
R
= 0.25 A
Notice: each would really be burning only
2
2
PA  I R  (0.25 A) ( 240)
= 15 Watts
Each bulb only sees a fraction of the total
available voltage across its ends:
VA  IR A  (0.25A)(240)
= 60 V
VB  IRB  (0.25A)(240)
= 60 V
+120 V
0V
240 
240 
+60 V
0V
We say the voltage “drops” across
each circuit element that uses energy.
Schematics include test point voltages for
diagnosing parts of complicated circuits.
A 120 volt generator supplies DC current
through a 100 foot extension cord (gauge
18 copper wire) to a 60 Watt bulb.
How much current would be drawn?
How much power would the bulb
really consume?
How big would the voltage drop be
across the extension cord?
How much power is wasted in the cord?
The resistance of a 100 ft (30.48 m)
of gauge 18 copper wire is about 0.64.
A 120 volt generator supplies DC current
through a 100 foot extension cord (gauge
18 copper wire) to a 60 Watt bulb.
How much current would be drawn?
I V / R
 120volts /(0.64  240  0.64)
 120volts /(241.28) = 0.49734748 A
instead of 0.50 A
How much power would the bulb
really consume?
P  I R  (0.497374748A) (241.28)
2
2
= 59.365 Watts instead of 60 W
The resistance of a 100 ft (30.48 m)
of gauge 18 copper wire is about 0.64.
59.365 Watts
I = 0.49734748 A
A 120 volt generator supplies DC current
through a 100 foot extension cord (gauge
18 copper wire) to a 60 Watt bulb.
Voltage drop across the extension cord?
V  IR  0.48734748A  (1.28)
= 0.6238 volts
How much power is wasted in the cord?
P  IV
 0.48734748A  0.6238V
= 0.304 Watts
which is 0.5% of the total energy supplied
The resistance of a 100 ft (30.48 m)
of gauge 18 copper wire is about 0.64.
9.6 
120-V
0.64 
0.64 
What about running a 1500 Watt (9.6 )
space heater with that extension cord?
How much current would be drawn?
I  V / R  120volts /(1.28  9.6)
 120volts /(10.88) = 11.294 A
Voltage drop across the extension cord?
V  IR  11.294 A  (1.28)
= 14.45632 volts
How much power is wasted in the cord?
P  IV
 11.294 A  14.45632V
= 163.270 Watts
close to 12% of the energy
The resistance of a 100 ft (30.48 m)
of gauge 18 copper wire is about 0.64.
QUESTION 1
5. into the screen.
QUESTION 2
2) South Pole at top.
Curl the fingers of your right hand in the direction the sphere
spins. Since it is a charged sphere, current circulates that way.
Your thumb reveals the North pole of this electromagnet!
QUESTION 3
D. R = 1202/60 = 240 
Given Power and Voltage, while asked for R, clearly the most
direct approach is to simply use P=V2/R. Solving for R gives:
R = V2/P.
QUESTION 4
C. 144 
The only difference now is P = 100 Watts. So R = V2/P means
(120 volts)2/100 Watts = 14400/100 = 144.
QUESTION 5
2. equal to
As much current enters as leaves this circuit (returning to the
battery). Since this is a series circuit, exactly as much current
flows through bulb B as A.
QUESTION 6
3. greater than
The brightness is evidence that B uses more energy!
QUESTION 7
3. greater than
P=IV. If both have the same size current passing through
them, the brighter (greater P) must have the larger V.