Yes, and it's proportional.
V * I = P
Where V is the voltage in Volts, I is the current in Amperes, and P is the power in Watts.
So we get:
I = P / V
For example, with a 240 Volt supply, a 12 Watt lamp would draw:
12/240 0.05 Amperes, or 50 milliamps
If it is a 120volt light, then it is watts / volts. 32 watts / 120 = .2667 amps. <<>> fluorescent lights usually have a power factor around 0.6 so a 32 watt bulb would take around 32/(120 x 0.6) amps or 0.44 amps.
Depend on watt and voltage use of light bulbs. You can use this ohm's law formular to calculate the current draw on light bulbs. I (current in amp) = P (watt)/ E (voltage) If 25W light bulb use in 115V AC (resident home) then current draw will be: 25/115 = 0.22A or 22 miliamperes. Hope this help.
no there is not thousands of light bulbs on a computer screen. Instead there is a projector built in the computer projecting whatever for example google and then it appears on your screen. It is very clever i think personally
You just have to divide the watts by the voltage to find the amps. For example 60 watts on a 120 v system would take ½ amp.
amps equals watts divided by volts.
The amperage that a chandelier draws is based on the amount of bulbs and the wattage of the bulbs used in the fixture. Count the amount of bulbs and multiply that number by the wattage of the bulbs. Take this total wattage and use this formula. I = W/E. Amps = Watts/ Volts.
Yes, wattage is wattage, is wattage, is wattage. "Power" is calculated in wattage. It equals the voltage times the current in amps. In a light bulb, the resistive filament will cause a certain amount if current to flow making the filament hot and producing light.
I would use no more 14 100watt bulbs on a 15 amp circuit or 19 bulbs on a 20 amp circuit. You can calculate this by taking 80% of circuit amperage I.E 12 amps or 16 amps and then multiply by circuit voltage(120) to get 1440 watts and 1920 watts. Then take these values divided by bulb wattage(100 watts) to get 14.4 bulbs and 19.2 bulbs. Then round down for partial bulbs to get 14 bulbs and 19 bulbs.
Have to know the wattage or resistance of the lamp to answer this question.
If we assume that you are using a common 15 Amp lighting circuit and switch and using 120 volts to power the bulbs then you need to keep the wattage at 80% of 15 amp worse case or 12 amps. Watts = amps x volts for standard incandescent bulbs. 12 x 120 = 1440 watts.
Between 0.5 - 0.9 amps (not including the energy for the light bulbs) depending on the make and model.
The number of CFL bulbs that can be used on a single pole dimmer depends on the wattage of the bulbs and the maximum load capacity of the dimmer. Each dimmer has a specified maximum load capacity, usually measured in watts, which indicates the total wattage it can handle. To determine how many CFL bulbs can be used, divide the dimmer's maximum load capacity by the wattage of the CFL bulbs being used. Make sure not to exceed the dimmer's maximum load capacity to avoid potential overheating and damage.
The heater should have a wattage rating (very few list amps). Calculate the amps using the wattage and voltage. Amps = Watts/Volts(480).
Two thoughts here, one the fixture should be rated at the maximum wattage allowed for the socket the bulb screws into. A label should state "maximum wattage allowed". To do a calculation Watts = Amps x Volts. Amps = Watts/Volts. 5 x 60 watts = 300 watts. 300/120 = 2.5 amps. The electrical code only rates down to #14 wire which is rated at 15 amps. From here you have to make the decision.
If you can tell me the resistance across the bulb, I can tell you the total wattage. Watt = amps x volts. I (amps) = E (volts 2.5)/ R (resistance in ohms). Calculate for amps draw on one lamp and then times 300 will give you total amperage. Take this value times total voltage and it will give you total wattage.
To answer this question the total wattage of the fixture is needed. To find this you need to find out the wattage of one bulb. If all of the bulbs are the same multiply the wattage of the bulb times 19. Use this total wattage in the following equation. I = W/E, Amps = Watts/Volts. In North America a #14 conductor is rated at 15 amps but has to be de-rated for a continuous load. This allows the conductor to carry 12 amps legally.A #12 conductor is rated at 20 amps but has to be de-rated for a continuous load. This allows the conductor to carry 16 amps legally.A #10 conductor is rated at 30 amps but has to be de-rated for a continuous load. This allows the conductor to carry 24 amps legally.Hopefully the chandelier falls within this amperage range
No, you are charged by the wattage that you use. Watts are the product of amps times volts. Say you have a device that draws 5000 watts and you are using a voltage of 480 volts then the amperage will be A = W/E = 5000/480 = 10.4 amps. Now using that same wattage at 208 volts the amperage will be A = W/E = 5000/208 = 24 amps. As you can see the only thing that changes is the current (amps) on a consistent wattage with different voltages. One big advantage on using higher voltages over smaller voltages is, with the amps being lower, a smaller wire size can be used for the same wattage load. In the end though, you are still billed on the wattage used.