1.44 kw
It depends on how many Amps (current) are applied to the voltage. Watt = Volts x Amps. e.g. 12 volts @ 5 amps = 60 watts
Yes it depends on many variables such as what the wire is made of the size of the wire whether the wire is a multi strand wire. What governs the amount of voltage a wire can carry is the insulation that is wrapped around the wire. Like wire with ratings of 300 volts, 600 volts and 1000 volts these are the highest allowable voltages that can be applied. A wire that is rated for 300 volts is good for 120 volts, 240 volts and 277 volts. At test research facilities, insulation is tested to destruction. The label that is given to the wire insulation as a result of the tests is the highest safest voltages that can be applied to that particular type.
The formula for this question is I=P/E Where I = Current P = Watts E = Volts Therefore applying this formula: I = 1500 / 120 = 12.5
3 Kv = 3,000 volts.
Volts and Kilowatts are different. no comparison can be drawn in respect to a straight conversion.
Amps can not give you a kilowatt with out a voltage being applied to the question. Watts = Amps x Volts. Amps = 1000/ Volts.
It depends on how many Amps (current) are applied to the voltage. Watt = Volts x Amps. e.g. 12 volts @ 5 amps = 60 watts
It depends on how many Amps (current) are applied to the voltage. Watt = Volts x Amps. e.g. 12 volts @ 5 amps = 60 watts
12.6 volts at full charge. During charging at least 13.8 volts are applied to the battery.
The formula you are looking for is I = E/R. Amps = Volts/Resistance.
144 watts <<>> The formula to find watts is W = A x V. Watts =Amps x Volts.
Yes it depends on many variables such as what the wire is made of the size of the wire whether the wire is a multi strand wire. What governs the amount of voltage a wire can carry is the insulation that is wrapped around the wire. Like wire with ratings of 300 volts, 600 volts and 1000 volts these are the highest allowable voltages that can be applied. A wire that is rated for 300 volts is good for 120 volts, 240 volts and 277 volts. At test research facilities, insulation is tested to destruction. The label that is given to the wire insulation as a result of the tests is the highest safest voltages that can be applied to that particular type.
Small bulbs come in many different voltages. It will always be marked on the bulb and will only work correctly with that proper voltage applied.
The formula for this question is I=P/E Where I = Current P = Watts E = Volts Therefore applying this formula: I = 1500 / 120 = 12.5
Some popular applied materials are difficult to pinpoint in this time for there are many applied materials that are being applied today. To better a search on the subject, to search would be beneficial when provided with the specific material that is being inquired.
45 volts
The first thing we have to do is clarify between the two items. 120 volt wire rating is an insulation rating of the wire. Like wire with ratings of 300 volts, 600 volts and 1000 volts these are the highest allowable voltages that can be applied. A wire that is rated for 300 volts is good for 120 volts, 240 volts and 277 volts. At test research facilities, equipment is tested to destruction. The label that is given to wire as a result of the tests is the highest safest voltages that can be applied to that particular wire. So when you see a wire that has a label stating that it is rated for 300 volts it means that any voltage under and up to 300 volts is safe to apply. The ability of a wire to carry current (amps) is related to the size of the wire. Now, watts is the product of amps x volts. To answer this question, assuming the supply voltage is 110 volts, an amperage needs to be stated as per the formula above.