2007-07-20 19:16:15 · 15 answers · asked by daisy_hilo 1 in Science & Mathematics ➔ Physics
If both, do we use the formula Power = voltage^2 / R or Power = Current^2 x resistance to determine the brightness?
2007-07-20 19:24:37 · update #1
The watts determine the brightness. Watt is the unit of Power. And Power is equal to Voltage x Amps (current). So both the current and voltage determine the brightness.
2007-07-20 19:34:14 · answer #1 · answered by jnice160 2 · 6⤊ 4⤋
Light Bulb Brightness
2016-10-01 05:14:54 · answer #2 · answered by Anonymous · 0⤊ 1⤋
Bulb Brightness
2017-01-05 09:18:09 · answer #3 · answered by ? 4 · 0⤊ 1⤋
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RE:
What determines the brightness of light bulb? Is it current, voltage or both?
2015-08-13 08:37:52 · answer #4 · answered by Max 1 · 0⤊ 0⤋
The unit of brightness is lumen. So the higher the lumen, the brighter it is. For the same wattage, LED light bulb produces much higher lumen than the older type light bulb. Yes, wattage (voltage x current) does matter but only to specific type of light bulb. Ultimately it is the amount of lumen that a light bulb can produce that determine the brightness.
2015-07-20 14:02:11 · answer #5 · answered by waysc 1 · 2⤊ 0⤋
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Ok, I see your confuzzlement. LOL When you connect two 50-Watt lamps in series, they will NOT each get 50 Watts - They will only get 1/4 of the power, or 12.5 Watts apiece! When you connect two bulbs in series, they will NOT draw the same power - they will have only half the Voltage, and will draw only 1/4 of the power, thus will have only 1/4 the light! Example, where P = Power in Watts E = Electromotive Force in Volts I = Current in Amperes R = Resistance in Ohms So using Ohms law for one lamp in your example: P = 50 Watts E = 10 Volts I = 5 Amps R = 2 Ohms But when you connect two bulbs in series, look what happens for ONE bulb: P = 12.5 Watts - only 1/4 the POWER thus 1/4 the light! E = 5 (Half Voltage) I = 2.5 (Half the Amps) R = 2 (Same resistance) And the total of the two bulbs in series is: P = 25 Watts - 1/2 of what 1 bulb alone dissipates (turns to light & heat). E = 10 Volts (Same TOTAL Voltage - Half for each lamp.) I = 2.5 Amps - Half the current. R = 4 Total resistance. I hope I have cleared this up for you. Good luck!
2016-04-05 02:40:13 · answer #6 · answered by Ethel 4 · 0⤊ 1⤋
Since the supply voltage is constant for all light bulbs, the resistance of the bulb and hence the current flowing through determines the wattage.
Power = V x I where I is the current. Since I = V/R, we can write
W = V^2 / R
The lesser the R, the higher the power.
2007-07-20 19:27:58 · answer #7 · answered by Swamy 7 · 5⤊ 3⤋
Wattage is what determines the brightness of the light. For example, a 100 watt bulb is brighter than a 60 watt bulb. Lumens is the unit of measurement for that brightness. For example, I believe that a 60 watt bulb gives 1300 Lumens of light. As where 100 watt bulb gives 1450 lumens. But, it is the wattage that determines the lumens that are given.
2007-07-20 19:22:24 · answer #8 · answered by river85715 3 · 2⤊ 2⤋
Watts and efficiency.
The higher the watts, the brighter the bulb. But when you buy a bulb today, they have 15 watt fluorescent bulbs that are as bright as 75 watt incandescent because they are more efficient.
2007-07-20 19:21:09 · answer #9 · answered by hebb 6 · 1⤊ 0⤋
The brightnes of a light bulb is depending upon its power.
The formulae to calculate the electric power are :
Power = Voltage X current
or.........= (current)² x Resistance
or.........= (Voltage)² / Resistance
As long as the voltage is constant, the power can be increased either by increasing the current or the resistance.
When the resistance is constant, by voltage or current.
When the current is constant, by voltage or resistnce.
But by increasing the resistance the current will be reduced. So it is not advisable.
Effective method is either increase the voltage or the current.
2007-07-20 19:59:55 · answer #10 · answered by Joymash 6 · 2⤊ 3⤋