There are no watts in a volt. A volt is a measure of electric intensity (the ability of a spark to jump a gap). An ampere (amp) is a measure of the amount of electricity (the number of electrons flowing past any specific point in the circuit). Watts are a measure of electric power (the ability of electricity to do real work) and are calculated by multiplying volts times amps (1 volt × 1 amp = 1 watt). So 270 volts doesn't really supply enough info without also stating the amperage too. 270 volts is more than enough for the body to feel but if the amperage is very low (say, one-thousandth of an amp or so) then the power level would be about one-quarter of a watt and any shock would be barely felt. This is how those concealed prank shockers work. A couple of D-cell batteries provides a small voltage which is boosted to 50,000 volts or so but at the expense of the current level which winds up somewhere in the area of millionths of an amp or so. This large voltage is enough to provide a strong tingle but nothing more since the current is almost nonexistent. But raise both the voltage and current at the same time and you start getting into dangerous territory. House wiring provides 110-120 volts at dozens of amps which is more than enough wattage to kill you if you should happen to grab a live wire. Word to the wise, don't try it.
2007-04-15 18:24:43
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answer #1
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answered by moodie1_ny 2
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Hey Chris, A watt is in units of Joules/Second, or basically a unit of power. Although the number varies according to different publications 1 horsepower is usually equal to about 746 watts just to give you an idea of the quantity you are working with here. A volt is a difference in electrical potential with a unit of Joules/Coulomb of charge. A common way to look at this is pressure in a pipe. This will affect how much "electrical potential" can "flow" through a circuit. So in reality these are not at all related so that is why you are having problems with finding a conversion factor. Hope this helps. What Amy said above is technically correct. If you have a circuit and you want to find the power (watts) lost over a given resistive load (a motor usually or just a resistor in the simplest case) the equation you use is: Power (watts) = Volts^2/Resistor value or Volts*Amps You can refer to her example if you think this is the correct idea for the question you originally asked. However, there is no conversion between just volts and watts. Good luck!
2016-05-21 01:30:24
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answer #2
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answered by susanna 3
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Wattage is determined by current times the voltage. The amount of current is determined by the resistance. So if you hooked up a 100 ohm load to 270 volts, the current would be 2.7 Amps. 2.7 amps * 270 volts would come out to 729 watts. If the resistance was doubled, the current would be cut in half so that wattage then would be 364.5 Watts
2007-04-15 18:02:23
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answer #3
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answered by J D 5
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270 Volts
2017-01-12 14:01:37
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answer #4
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answered by ? 3
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You're comparing apples and oranges. A watt is a unit of power that can be directly converted into horsepower. A volt is a unit of electrical pressure.
Electricity in a circuit behaves a lot like water in a pipe. Water pressure determines how far water will squirt out of a hose. Voltage determines, among other things, how far a spark will jump.
The basic equation regarding electricity is Ohm's Law: E=IR where E is electromotive force in volts, I (capital i) is a quantity of electricity (like a quantity of water). I is expressed in amperes (amps). R is resistance, measured in ohms. A skinny wire will offer more resistance to the flow of electricity than a fat wire, just like a skinny pipe will offer more resistance to the flow of water than a fat pipe.
You have three things, E=electromotive force in volts, I=current in amperes and R=resistance in ohms.
If you know two, you can calculate the other. By simple algebra we can rearrange Ohm's Law three different ways: E=IR or I=E/R or R=E/I.
Remember: Voltage is electrical pressure.
Current is an amount of electricity, usually
expressed in amperes per second, just like
gallons of water per second.
Resistance is opposition to the flow of
electricity expressed in ohms. Different
materials have different amounts of electrical
resistance, just as sand and clay offer different
amounts of resistance to the flow of water.
I haven't mentioned watts yet. They are not a measure of electricity. They are a measure of power. The word watt comes from the name James Watt, inventor of the first practical steam engine. It was used for pumping water out of mines and he never imagined any other use for it. He invented the term horsepower to compare his pump to the usual system of having a horse haul a big bucket of water up with a rope and pulley. It was a measure of how much water the pump could pump. A horse running around a racetrack is putting out a lot more than one horsepower.
Watts are calcualated by multiplying volts times amperes (amps) per second. One volt times one amp per second equals one watt. If you cut the resistance in the circuit in half, that one volt will now move two amps per second through the circuit and the power will be two watts.
I hope I have helped you.
2007-04-15 18:46:46
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answer #5
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answered by Anonymous
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All depends on the resistance or the amperage.
1 volt of potential difference is applied to a resistive load, and a current of 1 ampere flows, then 1 watt of power is dissipated.
1W= 1V * 1A or if you had only 1A draw then
270W=270V * 1A
2007-04-15 18:10:24
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answer #6
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answered by tequila_mike 3
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1 volt of potential difference and a current of 1 ampere flows, then 1 watt of power is dissipated.
1W= 1V * 1A or if you had only 1A draw then
We know the formula W=V*A
So,
270V * 1A =270W.
https://www.electrikals.com/
2015-10-12 19:48:11
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answer #7
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answered by Robert 4
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Completely different measurements. You may as well ask how many kilograms is there is a meter. You won't get an answer.
2007-04-15 19:12:02
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answer #8
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answered by Kasey C 7
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http://www.holroyd-components.co.uk/ohms_law.htm
2007-04-15 18:03:08
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answer #9
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answered by AuntTater 4
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how many amps?
2007-04-19 14:05:25
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answer #10
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answered by Anonymous
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