mine shack cave
Like The Light BULB :P
Only 1 :p :P :
To find the intensity of a bulb, you can use the formula I = P / A, where I is the intensity, P is the power of the bulb in watts, and A is the area over which the light is spread. Alternatively, you can use a light meter to measure the intensity directly in illuminance units such as lux or foot-candles.
I think there is no 3 way LED light bulb :) :P :)
The power of a light bulb or appliance depends on its voltage (V) and current (I). The power (P) can be calculated using the formula P = V x I.
Since the light bulb is purely resistive (has very little reactance), you can just measure the RMS voltage across the light bulb (usually 120 V) and the RMS current going through the light bulb. Power (P) is:P = VRMS x IRMSwatts
To calculate the resistance of a 5 watt light bulb, you can use the formula P = V^2 / R, where P is power (5 watts) and V is voltage (typically around 120V for a household light bulb). Rearranging the formula to solve for resistance, you get R = V^2 / P. Plugging in the values, R = (120V)^2 / 5W = 2880 ohms. So, a 5 watt light bulb at 120V would have a resistance of 2880 ohms.
That doesn't matter. What matters is that he didn't give up and he was able to create the light bulb. :P
http://www.clubxterra.org/forums/showthread.php?p=18380
No, a higher wattage INCANDESCENT light bulb uses more current than a lower wattage INCANDESCENT light bulb. Some CF and LED bulbs are rated by the amount of light that an incandescent bulb would produce, but they are also rated by the wattage that they use.
It is p3.
To find the power output of a light bulb, you can use the formula ( P = V \times I ), where ( P ) is the power in watts, ( V ) is the voltage across the bulb in volts, and ( I ) is the current flowing through it in amperes. You can measure the voltage using a voltmeter and the current with an ammeter. Alternatively, if the bulb has a specified wattage rating (e.g., 60W), that value represents its power output directly.