Download PowerPoint: Self induction in a coil and Back

130 volts / 1.4 ohms
= 93 amperes!
What if this coil
were connected to
130 v ac???
VI = 12,000 watts !
It would melt
down!!!!
The Resistance is
1.4 ohms
Ohm meter
P
T
P
T
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?
130 v
In fact, there are
almost zero watts
A little information to remember:
6v
V
R
0 volts
6 volts
V battery
Zero volts
Vbatt + Vresistor = 0
Gain 6 v + lose 6 v =0
V
Voltmeter
AC
R
Resistor
Power Source
Vapplied+Vinduced=0
Power
Voltage
Current
Current and Voltage in in PHASE; therefore, the Power
curve is in phase with them both.
Current in
coil
and resistor
Induced voltage in coil
O Prop I
Flux in the coil follows
VL = -  
this curve also to we expect
t
maximum induced voltage to
be where the flux is changing at the highest rate.
Signal generator
voltage Va
Current in coil
Induced voltage
VL
The voltages around a loop must add up to zero so the
sum of Vsignal gen +
VL = O
A coil in series with an AC power source:
V
coil
AC POWER SOURCE
Vapplied
Voltmeter
Vapplied + Vinduced=0
Vapplied
current
Vinduced
http://ostc.physics.uiowa.edu/~wkchan/E
LECTENG/AC_POWER/AC.html
Power = V battery x I battery
Output voltage of signal generator
Current out of signal generator
Power output taken from the signal generator would be:
P=VI
If these are multiplied the puzzle is solved.
Signal generator
voltage Va
Current in coil
Induced voltage
VL
The voltages around a loop must add up to zero so the
sum of
Vsignal gen + VL = O
(Vmax Cos 0 ) (I max Sin 0) = Power
-
+
+
Flux building up
Energy being stored and given back.
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
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.
The resistance of a coil to AC current is given by the
F L where F is the frequency
equation: Rcoil = 2
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.
Vbattery
VIcos0= Power
(Asin0)(Acos0)= Power
Icurrent
+
_
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!
Power
Vapplied
Current
Time (s)
Vinduced
Vapplied
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
Signal Gen
gnd
VR
R
coil
VL
VR
Gnd
Current is min
voltage is max.
Current is max
voltage is min.
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.
ac
Power
Capacitor
Power
Vapplied
Current
Time (s)
Vinduced
Vapplied
Inductor