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Current (amps) = power (watts) / voltage = 100/240 = 0.42 amps
The American system is 110 Volts and the UK (plus some other EU countries) are about 240 Volts. The formula is Volts X Amps = Watts. Don't panic, that's as hard as it gets. The problem is this, that for 240 Volts a current of .25 Amps is needed to light a 60 Watt bulb (240 X .25 = 60). At 110 Volts this would be about 110 X .5 = 60 giving over twice the Amps for the same Watts. The more Amps, the heavier the wires needed to carry the power. So, to be safe the fitting would have to be rewired. If in doubt contact an experienced electrician to rewire it for you. Better a few dollars than dialling 911 because the house is on fire!
Power(Watts) = I (Amps) x E(Voltage) PIE rule. so 1000 = I x 240. 1000/240 = 4.16667 amps.
Nobody has "240 watt mains". Perhaps you meant "240-volt mains".You would need to have a lighting or receptacle branch circuit, with over-current protection, in order to use any halogen bulb.However, if you have a 300 W bulb, it should work nicely in a 15-A or 20-A branch circuit.
No, you will need to get a step up transformer, it's probably more sensible to get a 12v sensor.
Please check it.It should be 240 X 12 if its is bulb angle. Manjush Navale
240 / 12 = 20
Current (amps) = power (watts) / voltage = 100/240 = 0.42 amps
The American system is 110 Volts and the UK (plus some other EU countries) are about 240 Volts. The formula is Volts X Amps = Watts. Don't panic, that's as hard as it gets. The problem is this, that for 240 Volts a current of .25 Amps is needed to light a 60 Watt bulb (240 X .25 = 60). At 110 Volts this would be about 110 X .5 = 60 giving over twice the Amps for the same Watts. The more Amps, the heavier the wires needed to carry the power. So, to be safe the fitting would have to be rewired. If in doubt contact an experienced electrician to rewire it for you. Better a few dollars than dialling 911 because the house is on fire!
The American system is 110 Volts and the UK (plus some other EU countries) are about 240 Volts. The formula is Volts X Amps = Watts. Don't panic, that's as hard as it gets. The problem is this, that for 240 Volts a current of .25 Amps is needed to light a 60 Watt bulb (240 X .25 = 60). At 110 Volts this would be about 110 X .5 = 60 giving over twice the Amps for the same Watts. The more Amps, the heavier the wires needed to carry the power. So, to be safe the fitting would have to be rewired. If in doubt contact an experienced electrician to rewire it for you. Better a few dollars than dialling 911 because the house is on fire!
A 240 v 14 w cfl bulb uses about 0.14 amps.
The resistance of a light bulb varies, depending on the type of bulb, the power rating, and the temperature. A typical incandescent 60 watt bulb, for instance has a cold resistance of about 30 ohms, and a hot resistance of about 240 ohms.
Power(Watts) = I (Amps) x E(Voltage) PIE rule. so 1000 = I x 240. 1000/240 = 4.16667 amps.
If you divide the watts of the bulb by the supply voltage, that is the current. For example a 60 w bulb on a 240 v supply gives a current of 60/240 which is ¼ amp.
That depends on what voltage it's designed to operate from. Power = (voltage)2 / R R = Voltage2 / power If it's a 117-volt bulb, R = (117)2 / 28 = 489 ohms. If it's a 240-volt bulb, R = (240)2 / 28 = 2,057 ohms.
Nobody has "240 watt mains". Perhaps you meant "240-volt mains".You would need to have a lighting or receptacle branch circuit, with over-current protection, in order to use any halogen bulb.However, if you have a 300 W bulb, it should work nicely in a 15-A or 20-A branch circuit.
You can use a lower wattage bulb on a higher wattage fixture provided the voltage rating is the same and the bulb base is the same.