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.
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.
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.
If you are using typical residential voltages ( 120v/240v) a 40 amp breaker is plenty big enough. You could even use a twenty amp breaker.
To calculate the breaker size for a 1500-watt 120-volt heater, use the formula: Breaker size = Power (Watts) / Volts. In this case, 1500 watts / 120 volts = 12.5 amps. Therefore, you would need a 15-amp breaker for a 1500-watt 120-volt heater to allow for a safety margin.
AWG # 10 wire on 30 amp circuit.
If the voltage supplied to the lamps is its operating voltage both lamps will have relatively the same output in brightness. If the 60 watt 110 volt lamp is used on a 220 volt supply, it will glow very brightly and then the lamp's filament will burn open. If the 60 watt 220 volt lamp is used on a 110 volt supply, the lamp will glow at half brightness, but it will last for a very long time before the filament burns open.
A volt is the unit of electric potential, an amp is the unit of electric current, and a watt is the unit of power. The relationship between them is described by the equation: Power (in watts) = Voltage (in volts) x Current (in amps).
If you are using typical residential voltages ( 120v/240v) a 40 amp breaker is plenty big enough. You could even use a twenty amp breaker.
To calculate the breaker size for a 1500-watt 120-volt heater, use the formula: Breaker size = Power (Watts) / Volts. In this case, 1500 watts / 120 volts = 12.5 amps. Therefore, you would need a 15-amp breaker for a 1500-watt 120-volt heater to allow for a safety margin.
20 amp
A watt is the product of amperage times voltage.
AWG # 10 wire on 30 amp circuit.
Ampire. watt/volt
I recommend 120 volt 20 watt halogen if you are on a budget. If you can afford them 120 volt LED is even better and will last for years.
Probably about 180 watts, assuming 90% efficiency.
no , it will burn out
yes the bulb will actually last longer
the 220 volt bulp in 220 volt ac current
A zero watt bulb does not consume any power, so it does not have a voltage rating. Voltage is a measure of electrical potential difference, typically associated with the amount of energy transferred per unit charge. In the case of a zero watt bulb, there is no power being consumed, so there is no voltage required to power it.