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The formulas you are looking for is I = E/R.
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The formula you are looking for is I = W/E.
Basically, Power = Current*Voltage Current = Power/Voltage Current = 15/120 Current = 0.125A or 125mA
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Since power is volts time amps, the current in a 60W lamp connected to 120V is 0.5A. Since a lamp is a resistive load, there is no need to consider power factor and phase angle, so that simplifies the explanation. ======================== Assuming this is an incandescent or halogen lamp (using a filament to make the light) there is a trick here: the resistance of a lamp filament varies with temperature and does not follow Ohm's law. The resistance will be much lower, thus the current will be much higher when the filament is cold, when the lamp is first connected. As the filament heats up, the resistance increases until it gets to a steady operating point of 0.5A. For a halogen lamp, the operating temperature is about 2800-3400K, so the R at room temperature is about 16 times lower than when hot... so when connected, the current is about 8A but drops rapidly. The current could be even higher if the lamp is in a cold environment. Non-halogen lamps operate at a lower temperature and would have a lower initial current--about 5A. And this all assumes the lamp is rated for 120V. If it is a 12V/60W lamp, the filament will probably break and create an arc, which may draw a very large current.
A ammeter will tell you how much current draw the load is drawing
The formulas you are looking for is I = E/R.
40W Bulb will spiol due to over current passing through its coilAnswerSince the 40-W lamp has a higher resistance than the 100-W lamp, the greater voltage drop will appear the 40 W lamp. As a result the 40 W lamp will be subjected to a voltage beyond its 100-V rating, and the 100-W lamp will be subjected to a voltage below its 100-V rating. Therefore, the 40-W lamp will burn much more brightly than the 100-W lamp.Incidentally, the symbols for the 'watt' and 'volt' are upper, not lower, case: W and V.
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5500Watts/220V=25 Amps
Either a short to ground, or too much current draw in that circuit.Either a short to ground, or too much current draw in that circuit.
How much what? Power? 102mW (0.102W). How much light? A filament bulb produces 1 to 3 percent of the power input as light if the proper voltage is applied. If the lamp is rated for a higher voltage, it won't get hot enough and the light out will be a lot less.
Without the battery hooked up, the current draw is too much for the alternator to keep fire on the plugs.
It would be pretty much undefined, since the filament of the halogen bulb would fail immediately then there would be an open circuit with no current draw. <<>> The formula for current is Amps = Watts/Volts. The lamp itself would draw 4.16 amps. Since the voltage of the lamp is 12 volts there is a internal transformer involved in the fixture itself. It doesn't matter what the input (primary) voltage to the transformer is, so long as it meets the manufacturer's specification as to the proper voltage to operate the fixture.
Yes you can. The battery supplies only as much current (amps) as the lamp draws when connected to 6 volts. The "12 amp" battery won't supply any more current when the lamp is shining than the "6 amp" battery did, but it'll last twice as long between charges.