'Voltage' and 'electric potential difference' are the same thing.
None of the others is like anything else, and they don't belong.
Your question should really be the other way around! Technically, the correct term is 'potential difference'. Since this is measured in volts, over time it has also become known as 'voltage'. A similar thing has happened, but to a lesser extent, with 'power' which, being measured in watts, is often referred to as 'wattage'.So, 'voltage' is simply another word for 'potential difference'. Originally, it meant 'potential difference expressed in volts' but, these days, it applies to microvolts, millivolts, volts, kilovolts, or megavolts!Take care, however, not to confuse 'voltage' and 'potential'. These are different, and voltage should never be used when you mean potential.
Wire size is based of the amperage of the device. To answer this question the amperage is needed or the wattage and voltage of the device.
Voltage does not affect the cost to run. You pay for the wattage, the higher the wattage the more it will cost to run.
Electric lamps have a wattage rating, but also have voltage ratings. Additionally, there are different socket styles for lamps, too.
That depends on the voltage V. Wattage P = amperage A times voltage V.
2800 watts at 240 volts would give you the 80 percent derated ampacity. <<>> Without a voltage voltage being stated, an answer can not be given. Electric heaters are sized by wattage and are used on many different voltages. Wattage is the product of amps times volts. So as you can see without the voltage the question can not be answered.
It depends on the voltage it runs on. The answer would be the wattage 15,000 divided by the voltage. Example at 240 volts it would run on 62.5 amps.
Wattage= current*voltage*power factor. Wattage=VI Cos(@)
wattage is nothing but power product of current and voltage p=v*i
Wattage, you mean power. Power = V * I. V - the voltage and I the current.
If running at 110V, that is 10Amps. Wattage = Voltage x Current Current = Wattage / Voltage - Neeraj Sharma
Wattage = Outage Voltage + Outage Wattage