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.

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?


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=

€

=(0.497374748A)2(240Ω)= 59.372 Watts instead of 60 W




18 copper wire) to a 60 Watt bulb.

I = 0.49734748 A

Voltage drop across the extension cord?

59.372 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


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 =

U.S. Dept of Energy estimates of power use by typical household

appliances:

refrigerator: 725 Watts PC w/monitor : 270 watts coffee maker: 1000 Watts dishwasher: 1200-2400 washer: 350-500 dryer: 1800-5000 hair dryer: 1200-1875 microwave oven: 750-1100 vacuum cleaner: 1000-1440 water heater: 4500-5500

http://www.eere.energy.gov/consumer/your_home/appliances/index.cfm/mytopic=10040

Just to run the fridge continuously a single household will draw 7-8 A.

http://www.eere.energy.gov/consumer/your_home/




The greater the current flow, • the greater the fraction of power wasted!• P = I2R…depends on current SQUARED!

The greater the distance power needs to be transmitted:

• more wire needed more total resistance• the greater the fraction of power wasted!

Supplying household currents by centrally located batteries or DC generators

is simply not practical!

Huge loss in voltage (dropping over the power lines)

Huge loss of wasted power(especially for remote distribution)

May heat copper to the point it sags under expansion, becomes a fire hazard or even melts.

Current actually flows in surges in and outof the outlet, reversing 120 times/second

(repeating itself 60 times/second).

Neutral(0 volts)

Ground(0 volts)

HOT+/-V

Locally groundedat your building

Vtime

How can an average of zero currentand zero voltage do a darn thing?

+Vmax

The average voltage is:

Vmax

A. Vmax. B. zero. C. Vmax 2

The average current this pushes through a resistance, R is:

A. Imax= B. zero. C. Imax 2

Vmax R

P = I2R = V2/RThe POWER delivered

V

average square voltage

The average (or “mean”) square voltage is

(Vmax)212

A glowing bulb filament orheating coil doesn’t care the direction current moves!

The average (or “mean”) square voltage is

(Vmax)212

The square root of the “mean” square voltage is

(Vmax)212

Vmax12Vrms=

is called the root mean square voltageand is the appropriate average to use

for alternating current (AC).

Imax12Irms=similarly:

Vmax12Vrms=

The standard 120 volts we’re provided by the power company is the rms value!

The peak voltage it oscillates between is

C. volts 852

120 ±=±

volts 1702120 ±=×±D.

volts 151202 ±=×±A.

volts 2402120 ±=×±E.

B. volts 602

120 ±=±

With the rms values giving the effectiveaverage for current and voltage:

all the equations we developed for DC circuits

still apply to AC circuits!

Ohm’s Law: RIVrmsrms

=

Power:rmsrms

VIP =

RIPrms2=

RV

P rms2

=

What advantage does THIS offer over DC current?

P = I2R

We’ve argued enormous heat lossesand voltage drops result when transporting

huge DC currents over large distances.

Ohmic heat losses:

need to keep current low

as well asresistance

Low currents would mean small voltage drops (V=IR) across the power lines and less heat loss.

We have seen: current produces a B-field.

We should expect:

A. a permanent magnet has an electric field surrounding it.B. a moving magnet to produce an electric field.C. a strong magnetic field produces current in nearby conductors.

S N

S N

Moving a magnet closer to a conducting loop

increases the strength of the magnetic field near the conducting loop

effectively increasing the flux of field lines that run through the open area of the loop.

But which way does current appear in response?

Note: the current induced would turn the loop itself into an electromagnet!

Will its field point in the same direction as theexternal permanent magnet? Or opposite it?

S N

Imagine nudging a magnet toward a conducting loop

If the induced current flowed in a direction thatcreated a new B-field parallel to the magnet’sit would strengthen the field near the loop even more!

But an increase in the B-field strength is what caused the generation of current!

Which would only produce MORE current!

Furthermore: notice the poles of the electromagnet formed! It will PULL the magnet in toward it!This would intensify the fields further and

generate more and more and more current!

Clearly violating conservation of energy!

S N

Imagine nudging a magnet toward a conducting loop

but the induced current creates a B-field opposite to the magnet’s

diminishing the field’s increase near the loop!

Furthermore: the electromagnet’s poles formin a direction that repels the oncoming magnet.

WORK is required to move the magnet.

Where does this work go?Into a voltage! And the production of current.

Conservation of energy!

Lenz’s Law: B-fields form toproduce a frictional resistance against

the changes that create them.

Generated current creates B-fields which OPPOSE the forces that create the current!

Inertia for electric and magnetic fields!

A conducting loop is pulled away from the South pole of a permanent magnet. As viewed

by the observer shown, the loop develops

A. a clockwise current. B. no current.C. a counterclockwise current.

S N

Moving a magnet closer to a conducting loop

which lies flat (as pictured), its open loopNOT facing the moving magnet.

Although the magnet field near the loop, is in fact increasing,

none of the increasingly dense field lines(the sign of increasing field strength)

passes through the loop’s enclosed area.

Is there ANY direction an induced currentin the loop could build its own magnetic

field to reduce the increasing field?

QUESTION 1

QUESTION 2

QUESTION 3

QUESTION 4

Just imagine trying to average a long list of numbers that are

evenly distributed either side of zero: as many positive as negative, and, in fact, for every positive number you need to add, its negative also appears in the list.

Of course you SAW this with your own eyes: a strong B-field by itself has no effect on nearby circuits…unless it is changing!The philosophical argument goes: moving CHARGE (current) produces the B-field. Charge is the source of electric fields.SO…moving MAGNETS (the source of B-fields) will produce an E-field.

B. Zero.

Current also swings smoothly and symmetrically either side

of zero. So same as above. But you can also reason this from:if the average voltage is zero, and I = V/R, the average currentwill have to be zero.

B. Zero.

Alternating current means it surges out, then (returns) in from the receptacle in the wall. No net charge ever passes completely down the wire, and I = Q/t.

volts 1702120 ±=×±D. Looking at the definition of rms I can see the rms “average” must be a little lower than the peak values: .

2V

V maxrms

=

Or, actually doing the math : .rmsmaxVV ×= 2

B. a moving magnet to produce an electric field.

the resistance of a 100 ft ( 30.48 m ) of gauge 18 copper wire is about 0.64

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