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NO. Two 60W lamps will use 120W - which is more than 100W !!
Probably not. Sodium light fixtures generally have a "ballast".
Power is measured in Watts, power (Watts) = E (volts) x I (current - amps) current is determined by the internal resistance (R) of the lightbulb, the lower the resistance the more current will flow. 120v x 0.5a = 60W 120V x 0.83a = 100W the 100W lightbulb will draw more current We also have Ohm's law: E(volts) = I (amps) x R (ohms) Household voltage stays the same at 120v we have for a 100w lamp: 120v = I x R R = 120v/0.83 amps R = 144.6 ohms for a 60w lamp: 120v = I x R R = 120v/0.5 amps R = 240 ohms The higher watt lamp has lower resistance.
Depending on the brand, it will give about 18.000 Lumens and equals a 100W HPS lamp
Volts * Amps = Watts 12 V * ?Amps = 100 Watts OR ? Amps =100W/12V OR 8.33 Amps = 100W/12V Use a 10Amp fuse inline and you can get by with 16Ga wire.
NO. Two 60W lamps will use 120W - which is more than 100W !!
For general calculations VA is the same as Watts.
The formula you are looking for is E = W/I.
Probably not. Sodium light fixtures generally have a "ballast".
Power is measured in Watts, power (Watts) = E (volts) x I (current - amps) current is determined by the internal resistance (R) of the lightbulb, the lower the resistance the more current will flow. 120v x 0.5a = 60W 120V x 0.83a = 100W the 100W lightbulb will draw more current We also have Ohm's law: E(volts) = I (amps) x R (ohms) Household voltage stays the same at 120v we have for a 100w lamp: 120v = I x R R = 120v/0.83 amps R = 144.6 ohms for a 60w lamp: 120v = I x R R = 120v/0.5 amps R = 240 ohms The higher watt lamp has lower resistance.
As per the formula for power (Power (Watt) = Voltage (V) x Current (i) & Resistance (R) = V / i), 25w lamp bulb would have higher resistance compared to that of 5w lamp bulb.
Depending on the brand, it will give about 18.000 Lumens and equals a 100W HPS lamp
Watts measure power. Energy is measured in joules and is the product of power and time. 1 joule = 1 watt for 1 sec Energy used in the example = 100 x 1 = 100 joules
For a lamp to operate at its rated power, it must be subject to its rated voltage. The lower the voltage, the lower the resulting power. In fact, a small drop in voltage will cause substantial drop in power. Higher 'wattage' lamps have lower resistance values than lower 'wattage' lamps.So, if you put two lamps in series, the greater voltage drop will appear across the lamp with the greater resistance. In your example, that means the 100-W lamp will be subject to the greater voltage drop, and its loss of power will be less than that of the 200-W lamp. So the lower power lamp will be the brighter of the two.
Total supply voltage = 220vrated power of first lamp= 100 wattso current though it, I=p/vI=100/220=.45 ampsResistance offered by first lamp=220/.45= 488 ohms( i avoid fractions)..............................................................rated power of second lamp=60 wattsvoltage is same, so current through it = 60/220=.27 ampsresistance of second lamp = 220/.27=814 ohms( i avoid fractions).........................................................................power drawn by first lamp =I12 R=(.45)2 * 488 =98.82(=100)power drawn by second lamp =I22 R=(.27)2 * 814 =59(=60)...................................................................................SO 100 WATT BULB WILL GROW BRIGHTER AS IT HAS MORE POWER
Volts * Amps = Watts 12 V * ?Amps = 100 Watts OR ? Amps =100W/12V OR 8.33 Amps = 100W/12V Use a 10Amp fuse inline and you can get by with 16Ga wire.
Generally speaking, the rated power of any light fitting ('luminaire'), including a table lamp, is determined by the enclosure, shade, or reflector, rather than its socket. What it comes down to is (a) whether the rate of heat transfer away from the lamp is sufficient to prevent the luminaire from overheating, and (b) whether the reflector/shade can withstand the resulting temperature.