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The amazing world of self induction !. 130 volts / 1.4 ohms = 93 amperes! VI = 12,000 watts !. What if this coil were connected to 130 v ac???. It would melt down!!!!. The Resistance is 1.4 ohms. Ohm meter. Now follows proof that the power converted into heat is LESS

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Presentation Transcript
slide1

The amazing

world of self

induction !

slide2

130 volts / 1.4 ohms

= 93 amperes!

VI = 12,000 watts !

What if this coil

were connected to

130 v ac???

It would melt

down!!!!

The Resistance is 1.4 ohms

Ohm meter

slide3

Now follows proof that the

power converted

into heat is LESS

than expected.

slide4

A Square Wave

input into a coil.

slide7

P

T

P

T

slide8

The actual current is

3.45 amperes AC.

Input voltage is 130 V.

450 watts ????????

Why is the current

so small? And why

doesn’t the coil

even get warm?

In fact, there are

almost zero watts

of heat !!!

130 v

slide9

Conservation of

Energy around a

loop.....in all

circuits!

slide10

A little information to remember:

6 v

6 volts

V

R

battery

V

Zero volts

0 volts

Vbatt + Vresistor = 0

Gain 6 v + lose 6 v =0

slide11

V

R

AC

Power Source

Volt-

meter

Vapplied+Vinduced=0

Resistor

Power

Voltage

Current

Current and Voltage in in PHASE; therefore, the Power

curve is in phase with them both.

slide12

The voltage

supplied always

equals the voltage consumed.

slide13

A sin wave input

into a coil

slide14

Current in coil

and resistor

Induced voltage in coil

O

I

Prop

Flux in the coil follows

this curve also to we expect

maximum induced voltage to

be where the flux is changing at the highest rate.

VL = -

t

slide15

Signal generator

voltage

Va

Current in coil

V

Induced voltage

L

The voltages around a loop must add up to zero so the

sum of Vsignal gen + VL = O

slide16

A coil in series with an AC power source:

V

AC

coil

POWER SOURCE

Vapplied

Voltmeter

Vapplied + Vinduced=0

Vapplied

current

Vinduced

slide17

http://ostc.physics.uiowa.edu/~wkchan/ELECTENG/AC_POWER/AC.htmlhttp://ostc.physics.uiowa.edu/~wkchan/ELECTENG/AC_POWER/AC.html

slide18

Look at the

POWER CURVE!

Power = V battery x I battery

slide19

Output voltage of signal generator

Current out of signal generator

Power output taken from the signal generator would be:

P = V I If these are multiplied the puzzle is solved.

slide20

= Power

(Vmax Cos 0 ) (I max Sin 0)

-

-

+

+

Flux building up

Energy being stored and given back.

slide21

Why is there no heat being generated

in the coil?

…what energy is taken from the battery

is given back so there is NO heating of

the coil.

Power = (Va )( I) cos O

= zero!

Theta is the angle between the two

curves…in this case 90 degrees.

Cos 90 degrees = zero

slide22

Now, why is the current so much lower

than expected? If we use I=V/R it must

be because somehow the resistance of the

coil is larger to AC than to DC.

slide23

The resistance of a coil to AC current is given by the

equation: Rcoil = 2

F L where F is the frequency

and L the inductance of coil.

L for the coil is 0.10 henrys. F = 60 hz

R then = 38 ohms. I= V/R = 130 v/38 ohms = 3.4 amps.

The resistance of the coil if proportional

to both the frequency and L. As f

increases the resistance gets larger….

Choking off the current.

slide24

Vbattery

VIcos0= Power

(Asin0)(Acos0)= Power

Icurrent

+

_

slide25

Power = VI

(+)

(+)

I

(-)

V

The power taken from the signal generator (+) balances the

power given back (-) and therefore no heat is produced in the

coil even though the current is 3.4 amps!

slide26

Power

Vapplied

Current

Time (s)

Vinduced

Vapplied

slide27

An arrangement for viewing the phase relationship between current in and voltage across a coil. A signal generator is connected in series with a coil and a resistor. An oscilloscope (for viewing how voltage changes in time) is connected (via computer) to both the coil and resistor.

VL

VR

gnd

R

coil

Signal Gen

slide28

VL

VR

Gnd

slide29

Current is max

voltage is min.

Current is min

voltage is max.

The current in

coil and resistor

are the same. The

voltage across the

resistor is in

phase with the

current so we can

see the phase

relationship

between current

and voltage of the

coil.

slide30

A capacitor and

AC voltage:

ac

slide31

Power

Capacitor

slide32

Power

Vapplied

Current

Time (s)

Vinduced

Vapplied

Inductor