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The Resistance is 1.4 ohms
Ohm meter
130 volts / 1.4 ohms= 93 amperes!
VI = 12,000 watts !
It would meltdown!!!!
What if this coilwere connected to 130 v ac???
P
T
P
T
The actual current is3.45 amperes AC.Input voltage is 130 V. 450 watts ????????
130 v
Why is the currentso small? And whydoesn’t the coileven get warm?
In fact, there arealmost zero wattsof heat !!!
A little information to remember:
R VV battery
Zero volts
6 volts6 v
0 volts
Vbatt + Vresistor = 0Gain 6 v + lose 6 v =0
RV AC Power Source
Power
Voltage
Current
Current and Voltage in in PHASE; therefore, the Powercurve is in phase with them both.
ResistorVolt-meter
Vapplied+Vinduced=0
Induced voltage in coil
Current in coil and resistor
Flux in the coil follows this curve also to we expectmaximum induced voltage to be where the flux is changing at the highest rate.
VL = - t
O IProp
Induced voltage
Signal generator voltage
Current in coil
The voltages around a loop must add up to zero so the
sum of Vsignal gen + VL = O
Va
VL
A coil in series with an AC power source:
ACV coil
Voltmeter
POWER SOURCEVapplied
Vapplied
Vinduced
current
Vapplied + Vinduced=0
http://ostc.physics.uiowa.edu/~wkchan/ELECTENG/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 = V I If these are multiplied the puzzle is solved.
(Vmax Cos 0 ) (I max Sin 0) = Power
Flux building upEnergy being stored and given back.
+
-
+
-
Induced voltage
Signal generator voltage
Current in coil
The voltages around a loop must add up to zero so the
sum of Vsignal gen + VL = O
Va
VL
Why is there no heat being generatedin the coil?…what energy is taken from the batteryis given back so there is NO heating ofthe coil. Power = (Va )( I) cos O
Theta is the angle between the twocurves…in this case 90 degrees.Cos 90 degrees = zero
= zero!
Now, why is the current so much lowerthan expected? If we use I=V/R it mustbe because somehow the resistance of thecoil is larger to AC than to DC.
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 hzR then = 38 ohms. I= V/R = 130 v/38 ohms = 3.4 amps.
The resistance of the coil if proportionalto both the frequency and L. As f increases the resistance gets larger….Choking off the current.
Vbattery
Icurrent
VIcos0= Power(Asin0)(Acos0)= Power
+
_
I
V
Power = VI
(+) (+)
(-)
The power taken from the signal generator (+) balances the power given back (-) and therefore no heat is produced in thecoil even though the current is 3.4 amps!
Vapplied
Power
Vinduced
Current
Time (s)
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.
R coil Signal Gen
VR
VL
gnd
VR
Gnd
VL
Current is maxvoltage is min.
Current is minvoltage is max.
The current in coil and resistorare the same. Thevoltage across theresistor is in phase with the current so we cansee the phase relationshipbetween currentand voltage of thecoil.
ac
Power
Capacitor
Vapplied
Power
Vinduced
Current
Time (s)
Vapplied
Inductor
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