it is 15000/440 based on basic power definition i.e., power =voltage *current
1.65KVA
The equation that you are looking for is I = E/R. Amps = Volts/Resistance.
The formula you are looking for is Watts = Amps x Volts.
I may be wrong, but I think you can calculate it by: Watts = Volts x Amps So: Amps = Watts / Volts Thus, if you're running them on a 240 volt circuit, it resolves to: A = 200 / 240 = 0.83 Amps
6240 watts if it's on a 240 volt circuit. A better answer is to just learn that amps X volts = watts.
Amps * Volts = Watts So, Watts / Volts = Amps 2000 / 240 = 8.333 Amps You should run the circuit on a two pole 15 Amp breaker, using 14 AWG, 2 conductor (plus ground) wire, just so you have a little safety factor in the circuit size.
Ohm's law: Volts = Amps * Ohms, or Amps = Volts / Ohms 12 volts / 0.5 ohms = 24 amps
The equation that you are looking for is I = E/R. Amps = Volts/Resistance.
15 amps at 80% = 12 amps continuous. Watts = Amps x Volts.
You want to know how many amps in that circuit. To do so, divide the Watts by the Volts. in your case it would be 60 watts / 120 volts = 0.5 Amps.
For a single phase circuit, the equation you are looking for is I = W/E. Amps = Watts/Volts.
The formula you are looking for is Watts = Amps x Volts.
The formula you are looking for is W = I x E, Watts = Amps x Volts.
That depends on circuit voltage. 1 watt is equal to 1 volt times 1 amp.
I may be wrong, but I think you can calculate it by: Watts = Volts x Amps So: Amps = Watts / Volts Thus, if you're running them on a 240 volt circuit, it resolves to: A = 200 / 240 = 0.83 Amps
Twenty amps is zero watts. You are missing one value. W = Amps x Volts. <<>> It depends on the resistance and the draw current in the electrical circuit.
6240 watts if it's on a 240 volt circuit. A better answer is to just learn that amps X volts = watts.
Amps * Volts = Watts So, Watts / Volts = Amps 2000 / 240 = 8.333 Amps You should run the circuit on a two pole 15 Amp breaker, using 14 AWG, 2 conductor (plus ground) wire, just so you have a little safety factor in the circuit size.