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Take the wattage of the bulb and divide that by the voltage of the bulb. This will give the current the bulb draws. Amps are a measure of charge (electron) at an instant of time through a conductor. In an incandescent bulb the filament is heated by the current and the characteristics of the filament, usually tungsten, is that it gives off light when heated.
To find the resistance necessary, one would need to know how much current the bulb draws. If one knows the current the bulb draws, then one would subtract the 14 volts from 120 volts then divide that by the current the bulb draws and one will find the resistance needed. Once this has been done, one would need to multiply the current drawn by the voltage drop to get the wattage rating necessary. Another important detail to note is that the power dissipated by the resistor will be much greater than the power consumed by the bulb itself. Finally if the bulb burns out the voltage across the contacts will be 120V. I would not recommend using this method to drop the voltage for the bulb.
No, they overheat and chip out a draws too much current.
Yes. It draws less current.
No way of knowing w/o knowing the voltage.
Take the wattage of the bulb and divide that by the voltage of the bulb. This will give the current the bulb draws. Amps are a measure of charge (electron) at an instant of time through a conductor. In an incandescent bulb the filament is heated by the current and the characteristics of the filament, usually tungsten, is that it gives off light when heated.
1.7amp
To find the resistance necessary, one would need to know how much current the bulb draws. If one knows the current the bulb draws, then one would subtract the 14 volts from 120 volts then divide that by the current the bulb draws and one will find the resistance needed. Once this has been done, one would need to multiply the current drawn by the voltage drop to get the wattage rating necessary. Another important detail to note is that the power dissipated by the resistor will be much greater than the power consumed by the bulb itself. Finally if the bulb burns out the voltage across the contacts will be 120V. I would not recommend using this method to drop the voltage for the bulb.
100/220 = .45A or 450 mA
No, they overheat and chip out a draws too much current.
I have no idea
The voltage of a battery goes as the current times the resistance (V=IR). Because the voltage is being held constant, the resistor that draws the most current will have the lower resistance.
In most of North America, the mains supply to the home is 120 volts. Thecurrent through a device dissipating 40 watts is 40/120 = 1/3 Ampere.In most of the rest of the world, the mains supply to the home is 220 volts.The current through a device dissipating 40 watts is 40/220 = 0.18 Ampere.
Yes. It draws less current.
very low current
P = I x V Power = volts x amps 240 x 4 = 560 watts. This is how much power will be consumed by the motor. The actual power produced at the shaft, will vary on type and efficiency of the motor.
A light doesn't output current, it "draws" current based on voltage and its resistance. Voltage = Current x Resistance or Current = Voltage / Resistance. (Ohm's Law)