[SOLVED] washer launcher with low voltage capacitors

A 2uf capacitor charged to 12v can release hundreds of amps in a small duration.

Wouldn't this be enough to propel some ferritic metal with an inductor but if it were made of super conducting filament to permit the entirety of the current?

A 2uf capacitor charged to 12v can release hundreds of amps in a small duration...

Click to expand...

That is the reason we do not bother with the current only; you must consider the total energy released.

A super conducting filament can support very high current (because there is no voltage drop) but very high magnetic field can quench the superconductor.

A circular ring made of Al or Cu or Ag can be made and the current produced by the capacitor discharge can throw the conducting ring into air.

But again you need to consider the total available energy and the magnetic coupling between the two.

But why ferritic material?

A simple experiment will be to hang a thin copper ring (or silver, if you have) by a thread. You discharge the capacitor via another similar coil hanging parallel to the first one.

But then what you are trying to prove? What is the purpose of this experiment?

Energy = 1/2 CV^2

With your values: 14 micro joules. Not very much energy.

Hi,

A super conducting filament can support very high current

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Even for super conducting filaments there is a current limit (A/ mm^2).
If you go beyond this thresold the filament becomes lossy, it loses it's super conductivity.

Klaus

c_mitra said:

That is the reason we do not bother with the current only; you must consider the total energy released.

A super conducting filament can support very high current (because there is no voltage drop) but very high magnetic field can quench the superconductor.

A circular ring made of Al or Cu or Ag can be made and the current produced by the capacitor discharge can throw the conducting ring into air.

But again you need to consider the total available energy and the magnetic coupling between the two.

But why ferritic material?

A simple experiment will be to hang a thin copper ring (or silver, if you have) by a thread. You discharge the capacitor via another similar coil hanging parallel to the first one.

But then what you are trying to prove? What is the purpose of this experiment?

Click to expand...

A spool of 32-45awg copper wire replaced by a spool of super conducting filament can make enough magnetism to propel a ferritic metal since its the type which is easier to propel.

The parasitic capacitance in a spool might prevent from metal being propelled.

KlausST said:

...If you go beyond this thresold the filament becomes lossy, it loses it's super conductivity...

Click to expand...

Considering a super conductor which doesn't saturate.

c_mitra said:

That is the reason we do not bother with the current only...

Click to expand...

Don't amp turns found in a magnetic element depend nearly entirely on current?

Consider conversation of energy law. Kinetic energy transfered to the metal object is detracted from the electrical circuit, causing the current to decay.

Don't amp turns found in a magnetic element depend nearly entirely on current?

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Yes and No.

In the steady state, the magnetic field is directly related to the amp-turns.

But creating the magnetic field needs energy. As you try to put more current, there is a back-emf produced and that voltage must be overcome to put more current.

In an inductor, the energy is stored into the magnetic field.

For a capacitor, the energy is stored into the electric field. You need to spend energy to increase the electric field from 0 to E.

In the same way, you need energy to create a magnetic field from 0 to H.

Energy is associated with the field, magnetic of electric. For a resistor, the energy is dissipative (joule heat; that cannot be recovered).

You can recover the energy from the magnetic field (or the electric field) when the field collapses.

You need to spend energy in a superconductor to get to a high current (see https://en.wikipedia.org/wiki/Type-I_superconductor for some more).

When the superconductor is quenched, all the energy is dissipated as heat.