On the end of any light bulb except for the fluorescent type there are numbers on the end of light bulb. The number will be followed by the letter w I.e.(65 w). W represents watts-amount of power used. On fluorescent light bulbs the curly cue ones it will be around the base where it screws into the socket.
Look on the light bulb for the voltage and the power in watts. Then divide the watts by the voltage and that gives the amps. Some CFL bulbs also state the current as well as the voltage and power, which is because they can have a poor power factor.
Manipulate the following equation, to make I the subject: P = I2R, where P = power, I =current, and R = resistance.
A burned out light bulb has high resistance - it is open - so, in a series circuit, it will have full supply voltage across it while the other bulbs in the circuit have zero volts. In a parallel circuit, just look and see which bulb is not lit.If you are talking about Christmas tree lights, however, they are generally designed to short out when they burn out, so that bulb goes dark while the others stay lit, even in a series circuit. The down side of that design is that the remaining bulbs will get brighter and hotter, and they will tend to burn out faster.
you can find simple machines everywhere,theirs a door knob,stairs,can opener,the faucet,clock,hammer,knife,light bulb and a spray bottle there's a lot more but i cant name them all;)
A zener diode with a rating of 500 mW will pass 50 mA at 10 V. (Power = voltage times current)Note: The question appears mis stated, in that it states a rating of 500 MW, not 500 mW. To my knowledge, there is no zener with a rating of 500 MW.
Find the wattage rating on the dimmer controller. This is the maximum allowable wattage that the lamp in the fixture should be. Lamp wattage ratings under the rated dimmer wattage is fine but do not install a larger wattage lamp that is over the dimmer control's rating.
Find out what the wattage of the bulb in the fixture is and you will have your answer.
You can prove it by using an energy monitor that measures the actual energy consumption of the light bulb. The wattage directly correlates to the amount of energy used by the bulb, regardless of its physical size. A higher wattage bulb will consume more energy than a lower wattage bulb, even if they are of the same physical size.
A watt is a measurement of electricity, usually pertaining to light. Light bulbs luminosity is graded by wattage, such as a 60 watt bulb, 100 watt bulb and so on. You would find a watt in a light bulb, to start.
The answer depends entirely on the wattage of bulb used! You can find the amperage of your light by using the Power Law which states that amperage = wattage divided by voltage. Thus a 60 watt bulb on a 120 volt system would draw .5 or 1/2 an amp.
Look on the light bulb for the voltage and the power in watts. Then divide the watts by the voltage and that gives the amps. Some CFL bulbs also state the current as well as the voltage and power, which is because they can have a poor power factor.
The surface of a 12-volt light bulb can reach temperatures of around 150-200 degrees Fahrenheit during operation. This temperature can vary depending on the wattage and type of bulb being used.
The best way to find a replacement for a 25 watt type G bulb is to check the wattage and size specifications on the current bulb, then search for a similar bulb with the same wattage and type G fitting at a hardware store or online retailer.
Get your bulb, it should be a very small wattage (weak) one, get your battery and wires, connect one wire to plus side of abttery and other wire to minus side of battery, then connect to the bulb, you may have to play around a bit to find a connection that produces light.
No, the wattage is determined by the resistance of the filament in the light bulb. The formula to determine the wattage is Watts = Voltage (squared)/Resistance in Ohms. To find the resistance of a 120 volt light bulb use the formula, Resistance in Ohms = Voltage (squared)/Watts. So for a 100 watt bulb at 120 volts the resistance is 120 volts x 120 volts = 14400/100 = 144 ohms. For a 60 watt bulb at 120 volts the resistance is 120 volts x 120 volts = 14400/60 = 240 ohms. As you can see this holds true to Ohm's law, current is inversely proportional to the resistance of the circuit. The higher the resistance of a load, the harder it is for the current to flow. In this case less current results in less light being emitted from the filament in the light bulb.
The average deep freezer or chest freezer uses about 130 watts. There are some energy saver models that might use less wattage.
Ge894 should be the bulb