The potato-battery experiment only generates about 1.5 volts a few milliamps. For a 40W incadescant bulb you typically need 120V and about 330 mas. If you could control all your losses in circuitry and somehow group the potatoes in series you would need about 160 potatoes. Wire 80 in series and parallel that with another 80 in series. This will provide DC vice AC. You need the proper anode and cathode and proper gauge wire. 40W isn't practical application. Try smaller bulb for starters and work your way up.
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.
This varies widely between manufacturers, whether it's a "clear" bulb or a "frosted" bulb, and what type bulb as in incandescent, florescent , quartz, etc etc
There is no direct conversion for this, because these are two separate things. Watts is actual power usage. "Lumens" is how much light is provided from a specified amount of power. The typical light bulb converts some of the electricity driven through it into light, and the rest into heat. The more efficient the light bulb, the less power per lumen will be required.That is why newer bulbs can be purchased that are 13 watts and advertised as equivalent to 60 watt incandescent bulbs - the light output is roughly equivalent (lumens), but the power usage is substantially different.
in a compact flourescent about 8mg.
6 yrs
You wouldn't use a potato to screw in a light bulb... if the glass in the light bulb breaks as you're removing it, you can use a potato to take the light bulb out.
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Onions don't have volts.
Yes, even a potato can light a light bulb. Yes. If the batteries match the voltage of the bulb, they can light it. Flashlights have bulbs and batteries that power them. If you mean a household light bulb, then you'd need many batteries in series (80 of the 1.5 volt batteries).
their are 8 parts to the light bulb
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.
It depends how big the light bulb is to be honest
Define "light bulb"
It would take a lot of apples to generate enough electricity to light a light bulb. Apples can produce a small amount of electrical current through a process called oxidation, but it would likely require hundreds or even thousands of apples to power a light bulb.
A flashlight bulb typically has one cell, which is a single unit that provides power to the bulb. The cell can be a battery or a rechargeable battery, depending on the type of flashlight.
The number of batteries needed to burn out a light bulb depends on the type of battery and the power rating of the light bulb. Typically, for a standard household light bulb (60-100 watts), one high-power battery or two to three regular batteries might be enough to burn it out due to overload. However, attempting to intentionally burn out a light bulb can be dangerous and is not recommended.
A 60 watt light bulb typically draws 0.5 amps from a 120-volt power source. This is calculated by dividing the wattage (60 watts) by the voltage (120 volts).