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Old 08-23-2003, 03:44 AM   #1
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Default Is this formula right ?

I is Amps
W is Watts
E is Volts
I = W/E


say i have a 50 watt globe and it runs off 1 volt then it means it draws 50 amps! is that right ? I got this formula out of a book

seems like this is more correct
I = E/W
0.02 = 1/50

help me out here guys
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Old 08-23-2003, 05:32 AM   #2
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The first formula is correct. Remember no one would run a 50 watt globe lamp on one volt. If run on 220 v it would draw only 0.23 A . At 110 v it would be 0.45 A.
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Old 08-23-2003, 07:32 AM   #3
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so how can that be right ??
if i did run a 50watt globe on 1 volt then it will draw 50 amps...
then if i ran a 50 watt globe on 50 volts it draws 1 amp

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Old 08-23-2003, 09:13 PM   #4
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A Watt is a derived unit of Volts * Amps

Thats how that works

V=IR
P=IV
P=I*I*V
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Old 08-24-2003, 02:21 AM   #5
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patcollins:
as per ur 1st 2 equations:
P=I x I x R
Or
P=V x V/R
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Old 08-24-2003, 03:07 AM   #6
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so this is the formula
I is Amps
W is Watts
E is Volts
I = W/E
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Old 08-24-2003, 03:50 AM   #7
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I think there is a bit of a confusion here.

A bulb rated at 50W will NOT draw 50 amp when connected across 1 V. The watt rating is wrt a fixed voltage.

A bulb can be discribed as a length of wire with known resistance. For the sake of simplification, assume this resistance to be invariant over temperature changes (fixed resistance)

The correct formula for power should be W=I X I X R

Given R= constant, W therefore is dependent on the value of I.

HOWEVER, V=IR, given fixed R, therefore I=V/R. (I is proportional to V)

Assume V=1V, R=1 Ohm, I=1amp therefore W=1W

Assume V=50V, R=1 Ohm, I=50amp thefore W=250W

Cheers
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Old 08-24-2003, 12:11 PM   #8
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Slight correction to my earlier post.
50W is the rated power output of the bulb in question, connecting too high a voltage across the bulb and it will increase the current flow across it, blowing the bulb prematurely.

Cheers
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Old 08-24-2003, 02:24 PM   #9
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Yes, the 50 watt bulb is only a 50 watt bulb at its rated current.
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Old 08-25-2003, 03:13 PM   #10
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SIMPLY:

Voltage times Amperage = Wattage

Wattage divided by amperage = voltage

Wattage divided by voltage = amperage.


Wattage is an expression of Total power displacement.




Imagine electricity as water, and wires as water pipe.

Higher voltage is like the water is moving faster.
Higher amperage is like having a larger diameter pipe.

Wattage is the total out put of the pipe over a given time period, say how long it takes to fill a bucket.....

A small pipe with water really spraying fast can fill it up in the same time as a really big pipe with the water flowing under less pressure.
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Old 08-30-2003, 04:09 AM   #11
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Default P=I^2 * R

I can't believe I remember it, but a physics professor once taught:

Twinkle Twinkle little star, P equals I squared R. If you're into physics that one just seems to stick...
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