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.
You need to know the amps the bulb draws and to find this you need to find the resistance of the bulb. The formula you need is Volts = Sq root of Watts x Ohms.
Answer
The rated voltage of a lamp is displayed, together with its rated power, on its glass envelope.
The problem with the solution suggested in the original answer, is how do you find out the resistance of the lamp filament. The problem being that there is a very large difference between its 'cold' and 'hot' (operating) resistance, so if you used, say, an ohmmeter to measure its 'cold' resistance, you would have no means of knowing what its 'hot' temperature would be!!
The maximum amount of wattage that is allowed to be used in any lighting fixture is on the label of the fixture. The manufacturer is bound by rules that they have to state on each fixture what the maximum amount of watts that the fixture can take before damage is done to the fixture or any surrounding objects. The damage from a lamp is in the form of heat, the higher the wattage of a lamp results in a higher heat from the lamp. Fixtures are allowed any wattage up to that recommended nameplate label wattage.
The wattage is stamped on the bulb. The rating might be hard to read, because of the heat dimming the stamping over time.
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.
Energy use is measured in watts. The watts used by a light bulb is on the package or the bulb. Find bulbs with the same wattage but different physical sizes if you want to prove this.
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.
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.
You generally need the same number of volts for a given amount of light (lumens), regardless of how many hours you use it. They typically measure the amount of energy used by a bulb in "watts", not volts, and you can find a wide range of wattage ratings from milliwatt LEDs to 1000-watt floodlights and on up.
Ge894 should be the bulb
which light bulb are you talking about?