0

5

10

15

20

25

1 3 5 7 9 11 13 15 17 19

Class Averages(3 exams w/bonus points

and homework quizzes)

average

70.4

+ 9590

- 85+ 80

75- 70+ 65

60- 55+ 50

45- 40

These cutoffswill not beraised…youcan use themto determineyour guaranteedlowest grade.

35 40 45 50 55 60 65 70 75 80 85 90 95 100

+_

1. up. 2. down. 3. left. 4. right.

5. into the screen. 6.out of the screen.

7. is zero.

The B-field everywhere within thiscurrent-carrying loop points

+ -

The SOLENOID pictured carriescurrent in the direction indicated:

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.

A positively charged sphere is set spinning as shown at left.

N

SS

N

1s2s2p3s3p4s3d4p5s4d

Oxygen O8

IronFe26

Fe3O4 [Magnetite]

Or the simple cubic IRON crystal

The slightly magnetic iron and oxygen atoms are locked into a crystal that fixes their orientation.

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 microscopicregions 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 domainsto grow.

In the presence of a strong external magnetic field

magnetic flipping can help some domainsto grow

and a sample of iron becomes magnetized!

An “induced”magnetic

North pole.

What’s the resistance of a 60-Watt bulb (designed to be operated with 120 volt )?

Ohm’s Law Power

V = IR P = IVP = I2RP = V2/R

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 = 602120 = 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 toasterI = 4.5 AR = 26.6 ohm

1500 Watt hair dryerI = 12.5 AR = 9.6 ohm

Household wiring12 AWG copper wire(0.08in diameter)0.0052 /meter

Power cords18 AWG copper(0.04in diameter)0.0210 /meter

Appliance wiring“telephone wire”

24 AWG copper(0.02in diameter)0.0842 /meter

Utility linespower transmission

6 AWG copper(0.162in diameter)0.00130 /meter

Unlike electrical components (resistors, inductors, transformers, motors)

we can think of the short copper wiresand 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 voltage across 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

Two identical bulbs would share the energy,while doubling the resistance in the circuit.

120 V

“60 Watt”

240

“60 Watt”

240

IRV R

VI

480

120volts

= 0.25 A

total

Notice: each would really be burning only

)240()25.0( 22 ARIPA

= 15 Watts

total

Each bulb only sees a fraction of the totalavailable voltage across its ends:

+120 V240 240

AAIRV

= 60 V

We say the voltage “drops” across each circuit element that uses energy.

0 V

+60 V0 V

)240)(25.0( A

BBIRV

= 60 V)240)(25.0( A

Schematics include test point voltages fordiagnosing parts of complicated circuits.

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?

How much power would the bulbreally consume?

How big would the voltage drop be across the extension cord?

How much power is wasted in the cord?

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?

RVI /)64.024064.0/(120 volts

)28.241/(120 volts = 0.49734748 Ainstead of 0.50 A

How much power would the bulbreally consume?

RIP 2 )28.241()497374748.0( 2 A= 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.

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.

I = 0.49734748 A

Voltage drop across the extension cord?

59.365 Watts

IRV )28.1(48734748.0 A= 0.6238 volts

How much power is wasted in the cord?

IVP VA 6238.048734748.0 = 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.

What about running a 1500 Watt (9.6 ) space heater with that extension cord?

0.64

120-V 9.6 0.64

How much current would be drawn?RVI / )6.928.1/(120 volts

)88.10/(120 volts = 11.294 AVoltage drop across the extension cord?

)28.1(294.11 A= 14.45632 volts

How much power is wasted in the cord?VA 45632.14294.11

= 163.270 Wattsclose to 12% of the energy

IRV

IVP

QUESTION 2

5. into the screen. QUESTION 1

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!

D. R = 1202/60 = 240 QUESTION 3

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.

C. 144 QUESTION 4

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 toAs 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.

3. greater thanQUESTION 6

The brightness is evidence that B uses more energy!

3. greater thanQUESTION 7

P=IV. If both have the same size current passing through them, the brighter (greater P) must have the larger V.