What size wire for a 240 volt 8800 watt stove?

Answer 1

8800w/240v = 36.6 Amps Since at some point (Thanksgiving, etc) you may use this range for more than 3 hours straight, your wire size must be calculated at 125 percent of that number: 36.6A * 1.25 = 45.75 (rounded to 46) Amps Using the NEC table 310.16, 75 degree column, the smallest copper wire that can carry 46 amps is #8 AWG. If you use aluminum the smallest is #6 AWG.

Answer 2

Technically it's a 20 amp load (20A x 240V = 4800W). However, you should not run a breaker at full load. The breaker is to protect the wiring from overheating and from short circuits. You should use 10 gauge wire and a 30 amp double pole breaker.

Answer 3

Amps X Volts = Watts

Amps X 220 = 4,000

Amps = 18.181818181818...

You CAN get by with 12 AWG, and a 20 amp circuit, but that's really pushing it. Nothing else could be on the circuit and you'd be getting real close to the edge. The breaker would no doubt run warm and you'd probably get quite a number of "false trips".

It isn't that much more expense to use a 10 AWG wire and a 30 amp circuit, and it will save you a LOT of trouble in the future.

Answer 4

First thing that has to be done is find the amperage as that is how the wire size is chosen. The formula you would use id I = W/E, Amps = Watts/Volts. 4500/220 = 20 amps.

A #10 copper conductor with an insulation factor of 75 or 90 degrees C are both rated at 30 amps. The heater conductor has to be de rated to 80% capacity so 30 x .8 = 24 amps. Your load amperage is only 20 amps so this conductor will be the right size.

Answer 5

This heater probably runs on 240 volts, draws 20 amps, would require minimum 12 gauge wire, or 10 Ga wire depending on the length of the wire

Answer 6

AWG # 10 with ground. Must be powered by a 30 amp breaker.

Answer 7

10 ga