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Old 03-10-2006, 08:30 PM   #11 (permalink)
 
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Re: ballast leaking electricity?

Quote:
Originally Posted by Electric_Sun
I don't want to split hairs, but if the part being touched is at ground there should be no voltage, thus no current, thus no shock, since it is at ground. If the part being touched is floating above ground (not solidly grounded, or a resistance to ground), then the current would generate a voltage and shock and the solution would be to correct the ground. yes no?
NO. Current (amperes) does NOT generate voltage (potential). Think as if it were like a water pipe. The water pressure is like voltage and the amount of water flowing is like current. Imagine for a moment that there is one small strand of a wire that is not touching ground but rather is close enough to ground to cause a slight and weak arcking. The voltage (potential) will be reduced as caused by the resistance (gap to ground). In addition, the current (amperes) will be reduced to a very small amount and not enough to trip the breaker. As in a water pipe, it's like a pin hole leak. Hardly any water pressure (voltage) and the amount of water (current) leaking is small but it's enough to make your finger wet.
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Old 03-10-2006, 08:41 PM   #12 (permalink)
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Re: ballast leaking electricity?

Did I have you for physics last semester? Still have that hot lab assistant? Talk about PHYSICS! Oh mama!! I would've loved to test a friction coefficient out with ... or was it biology? That assistant was hot too! I wanted to delve into her genus like nob ... or wait? Did I go to college?
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Old 03-10-2006, 08:58 PM   #13 (permalink)
 
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Re: ballast leaking electricity?

Yes, that was me and that was her.
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Old 03-10-2006, 09:00 PM   #14 (permalink)
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Re: ballast leaking electricity?

Quote:
Originally Posted by clubtan1
NO. Current (amperes) does NOT generate voltage (potential). Think as if it were like a water pipe. The water pressure is like voltage and the amount of water flowing is like current. Imagine for a moment that there is one small strand of a wire that is not touching ground but rather is close enough to ground to cause a slight and weak arcking. The voltage (potential) will be reduced as caused by the resistance (gap to ground). In addition, the current (amperes) will be reduced to a very small amount and not enough to trip the breaker. As in a water pipe, it's like a pin hole leak. Hardly any water pressure (voltage) and the amount of water (current) leaking is small but it's enough to make your finger wet.
You must be one of those left hand eddy current guys. Voltage across resistance generates current, current through resistance generates voltage,,, tomato, tomoto. I think we are talking about the same thing. What ever shocks you is not at ground potential (or is outside social mores).
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Old 03-10-2006, 09:01 PM   #15 (permalink)
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Re: ballast leaking electricity?

I sooooooooooo want to classify her butticular delecti. Such a rare specimen to be found in community college. I felt like Darwin, Austin Powers, and Forrest Gump during the dorm room scene all at once.
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Old 03-10-2006, 09:12 PM   #16 (permalink)
 
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Re: ballast leaking electricity?

Glad I could float your boat Belvey

Electric_Sun. Sorry, current (amperes) DOES NOT generate voltage through a resistive source. And YES, under certain conditions, a voltage CAN be applied to a grounded source WITHOUT current (amperes) increasing to infinity. In a NORMAL resistive state, voltage is inversely proportional to current. This is the case in a tanning bed when one turns the bed on. As current increases due to the resisitance of the bulbs, the voltage we decrease per OHMS LAW.


Ohm's Law

Ohm's Law defines the relationships between (P) power, (E) voltage, (I) current, and (R) resistance. One ohm is the resistance value through which one volt will maintain a current of one ampere.

( I ) Current is what flows on a wire or conductor like water flowing down a river. Current flows from points of high voltage to points of low voltage on the surface of a conductor. Current is measured in (A) amperes or amps.

( E ) Voltage is the difference in electrical potential between two points in a circuit. It's the push or pressure behind current flow through a circuit, and is measured in (V) volts.

( R ) Resistance determines how much current will flow through a component. Resistors are used to control voltage and current levels. A very high resistance allows a small amount of current to flow. A very low resistance allows a large amount of current to flow. Resistance is measured in ohms.

( P ) Power is the amount of current times the voltage level at a given point measured in wattage or watts.
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Old 03-10-2006, 09:14 PM   #17 (permalink)
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Re: ballast leaking electricity?

Ohm was a recluse. Let's get back to the lab assistants before one of you ends up in love up there. What was her name? Don't remember. But I do remember the way her sweaters reflected light! Talk about excited electrons!!
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Old 03-10-2006, 09:18 PM   #18 (permalink)
 
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Re: ballast leaking electricity?

Nice going. I think you insulted her. She's leaving the classroom now. SWELL.
Now what shall I do for a lab assistant?
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Old 03-10-2006, 09:19 PM   #19 (permalink)
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Re: ballast leaking electricity?

I don't know, but I want to help interview!
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Old 03-10-2006, 09:26 PM   #20 (permalink)
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Re: ballast leaking electricity?

[quote=clubtan1]Electric_Sun. Sorry, current (amperes) DOES NOT generate voltage through a resistive source. And YES, under certain conditions, a voltage CAN be applied to a grounded source WITHOUT current (amperes) increasing to infinity. In a NORMAL resistive state, voltage is inversely proportional to current. This is the case in a tanning bed when one turns the bed on. As current increases due to the resisitance of the bulbs, the voltage we decrease per OHMS LAW.
quote]

Wow, where did you learn that. We may be talking symantics, but current can generate voltage. A constant current source put throught a fixed resistance will generate a voltage and that voltage will vary if the resistance is varies, thus the current is generating the voltage. E=I*R

Voltage is NOT inversely proportional to current, it is directly proportional to current in a resistive circuit. Again E=I*R as voltage rises, current rises, as voltage drops, current drops.

A florescent tube can hardly be called a resistive circuit, especially when used with an inductive ballast.

Larger voltages are required to strike the arc (jump the electrode gap) in the lamp since the gases inside are not ionized. Once the gases ionize, lower voltage is required to maintain the arc.

....or so I've heard
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