Volts/amps for an electric arc between points

In summary, if you want to create a continuous electric arc between two sharp metal contacts, you will need a voltage of around 30,000 volts and a current of around 1,000 amperes. If you want to create a smaller electric arc between two less sharp metal contacts, you will need a voltage of around 3,000 volts and a current of around 300 amperes.
  • #1
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Can someone help me find these calculations or give me a point in the right directions?

If I have a hollow insulating cylinder (has a diameter of 5 cm and a length of 14 cm) with two (conductive) sharp metal point contacts at each end (measuring 2cm each leaving 10 cm exactly between the points). Inside the cylinder is normal atmosspheric pressure of standard air.

How do I calculate how many volts and amps I require to have a continuous electric arc between the contacts?

Thanks for your help and all answers are welcome.

:smile:
 
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  • #2
For dry air, you can expect about 30kv/cm, but your sharp metal make me question that, as the air will see a higher potential gradient near the points, and therefore may create a path via ionization. Being sealed off may affect your equations, as ozone and nitrous oxides may accumulate.

As for current, that's determined by your voltage. My best bet, depending on your power supply, would be to gradually remove resistance until the arc is sustained.

Your current type (DC, AC, high frequency AC) will also greatly effect your question.
 
  • #3
Thanks for your reply, yes I had read that for STP air (standard temperature pressure) that is almost if not moisture free the voltage required is about 300,000 to 330,000 volts on average.

I had chosen the sharp contacts as I had read that they would also allow the electric to flow with less resistance than plate electrodes. Thanks for the pointer on the build up of nitrous oxide and ozone (forgot about that) I suppose I could incorporate a filtered gas exchanger of some description, the reason I had closed the cylinder is to regulate the moisture, pressure, temperature etc and I thought that the steady increase of the temperature inside the container would also lower the amount of voltages required to sustain the arc (Like the traveling arc on a Jacob’s ladder).

It would seem that a few EHV power supplies and trial and error are the best options (With extreme caution and safety at mind). I'll see if I can get my hands on a heavy duty volt meter and ammeter.

Thanks. :smile:
 
  • #4
Hmm ... If you want dielectric breakdown at a lower voltage and don't mind corrosion, you may actually want to leave the cylinder sealed, as the accumulated gases will actually help. They aren't good for you though. I found several other sources that give 3.0 × 10^6 V/m for the dielectric breakdown of air. see hypertextbook.com/facts/2000/AliceHong.shtml

It's not so much voltage that sustains the arc, especially since you're basically building a very small capacitor here. Current and frequency have a larger effect on the sustainability of the arc.

If you have a higher current power supply, you could make the electrode distance adjustable and slowly increase the distance. An established arc can sometimes be 'pulled' longer than the voltage can cause a breakdown.

I'm sorry I can't give you any specific equations for this. If expense is not so much an issue, certain gases will breakdown easier at certain pressures than air, such as the noble gases in low pressure. This will also change your capacitance though.

One more thing, your electrodes will corrode and melt, so, depending on current, they may melt quite fast if they're very sharp.
 
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  • #5
Thank you for your reply,

Yes, I see what you mean with the build up of gases, I should seal the container with electrodes spaced around 1 cm apart to start and as the ozone and nitrous build up and the temperature increases then stretch the arc. I can calculate the pace of stretching based on the rate of gaseous build up in the cylinder.

I may have to rethink how sharp the contacts will be to allow for the increased time of stretching and measure corrosion times and voltage drops.

I know of a good gas supplier so I may get trial and error readings with a noble gas at lower pressures, so I can postulate some calculations in the higher pressures and less conductive air.
 

What is an electric arc?

An electric arc is a type of electrical discharge that occurs between two electrodes when a high voltage is applied. It produces a bright, intense light and can reach extremely high temperatures, making it useful for welding, cutting, and other industrial applications.

What is the relationship between volts and amps in an electric arc?

The relationship between volts and amps in an electric arc is known as Ohm's Law. It states that the current (in amps) flowing through a conductor is directly proportional to the voltage (in volts) applied, and inversely proportional to the resistance of the conductor.

How do volts and amps affect the intensity and temperature of an electric arc?

The intensity and temperature of an electric arc are directly influenced by both the voltage and amperage. As the voltage increases, the intensity and temperature of the arc also increase. Similarly, as the amperage increases, the intensity and temperature of the arc increase as well.

What is the recommended voltage and amperage for a safe electric arc?

The recommended voltage and amperage for a safe electric arc can vary depending on the specific application and materials being used. However, it is generally recommended to keep the voltage below 100 volts and the amperage below 50 amps to prevent potential hazards such as electric shock or burns.

How can I calculate the required volts and amps for a specific electric arc?

The required volts and amps for a specific electric arc can be calculated by determining the resistance of the conductor and using Ohm's Law (V=IR). The resistance can be determined by the material and size of the conductor, as well as the distance between the two electrodes. It is also important to consider the specific requirements and limitations of the equipment being used.

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