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Published byAlejandro Jenkins Modified over 2 years ago

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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 melt down!!!! What if this coil were connected to 130 v ac???

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P T P T

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The actual current is 3.45 amperes AC. Input voltage is 130 V. 450 watts ???????? 130 v Why is the current so small? And why doesnt the coil even get warm? In fact, there are almost zero watts of heat !!!

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A little information to remember: R V V battery Zero volts 6 volts 6 v 0 volts Vbatt + Vresistor = 0 Gain 6 v + lose 6 v =0

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R V AC Power Source Power Voltage Current Current and Voltage in in PHASE; therefore, the Power curve is in phase with them both. Resistor Volt- meter Vapplied+Vinduced=0

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Induced voltage in coil Current in coil and resistor 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. V L = - t O I Prop

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Induced voltage Signal generator voltage Current in coil The voltages around a loop must add up to zero so the sum of V signal gen + V L = O Va V L

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A coil in series with an AC power source: AC V coil Voltmeter POWER SOURCE Vapplied Vinduced current Vapplied + Vinduced=0

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LECTENG/AC_POWER/AC.html

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Power = V battery x I battery

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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.

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(Vmax Cos 0 ) (I max Sin 0) = Power Flux building up Energy being stored and given back

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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 Theta is the angle between the two curves…in this case 90 degrees. Cos 90 degrees = zero = zero!

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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.

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The resistance of a coil to AC current is given by the equation: R coil = 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.

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Vbattery Icurrent VIcos0= Power (Asin0)(Acos0)= Power + _

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I V Power = VI (+) (-) 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!

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Vapplied Power Vinduced Current Time (s) Vapplied

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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. Rcoil Signal Gen VRVR VLVL gnd

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VRVR Gnd VLVL

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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.

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ac

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Power Capacitor

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Vapplied Power Vinduced Current Time (s) Vapplied Inductor

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