# Basic electricity question

## Recommended Posts

A simple harmonic circuit with a capacitor and an inductor is made from superconducting materials.  At t = 0 current in the inductor is maximum and voltage across the capacitor is zero.  At t = 0 an external heat source collapses superconductivity but not conductivity.  The inductor now has high resistance and high current, so there should be a massive voltage spike across the inductor almost instantaneously.

Would it be the case that a similar huge voltage spike appears across the capacitor almost instantaneously and then climbs higher as the capacitor charges?

If all our components become plasma due to overheating, could we still generate potentials of billions of volts at least for a moment?  The idea being to create the most powerful sparkplug, or even better the most powerful hammer and anvil.

##### Share on other sites
17 minutes ago, farmerben said:

A simple harmonic circuit with a capacitor and an inductor is made from superconducting materials.

Is this even possible?

##### Share on other sites

This being a discussion about superconductivity I wouldn't call it basic.

That being said, what you are describing is called magnet quench, and is not uncommon when dealing with superconducting magnets.

##### Share on other sites

Thanks for the word "magnet quench".  I didn't find answers to my questions yet.  But I did find numerous articles about how to dump 50kV surges before the inductor explodes.  However, the quenches described begin at a single point and propagate as heat moves out.  This suggests that fast and complete quenching could greatly increase the voltage surge.

Something that could generate more than 2.5 MV would be really useful for igniting fusion.  On the other hand if there are other ways to ignite fusion, 50 kV is a good amount of potential for directing particles.

I'm not sure what happens when the inductor vaporizes.  It loses the number of loops, but my thinking is the magnetic flux is continuous and the plasma will have circular current.

##### Share on other sites

Essentially you're talking about a boost converter, using the inductor as both the energy storage and the switching element. Using magnetic quench instead of a transistor for the switch is interesting, but probably not terribly efficient. Generally a boost converter is switched at a very high rate to minimize losses, which wouldn't be practical with magnetic quench since the time to re-cool the superconductor would be very long.

##### Share on other sites

Sort of like this.  Once the switch is open the circuit on the right becomes a simple harmonic oscillator. The switch does not create voltage surges or sparks, the left side of the circuit can be removed completely.

What I want to do is disconnect the power supply completely and use the circuit on the right for a fusion fuel pellet.  When the quench happens there is a sudden voltage spike.  Is it as simple as dV/dt = I dR/dt ?  Superconductor quench makes dR/dt a huge positive number.

For a fusion fuel pellet we want the inductor to surround the capacitor.  So maximum electric and magnetic fields are centered at the same place.  We load the pellet with a dielectric which is also high explosive and mostly deuterium.  Detonate the chemical explosive, quench the entire superconductor fast, crush deuterium and tritium between the plates of the capacitor.

There is no oscillation or reset after the quench, the inductor and capacitor are vaporized.  I think the electric and magnetic fields persist into the plasma state.  It is a self contained sacrificial fuel nozzle.   If the superconductor is mostly lithium, it could be quite efficient as propellant.

Edited by farmerben

##### Share on other sites

By definition the voltage across the inductor equals the voltage across the capacitor.  Likewise, once the switch is open (and not arcing), the current across the inductor equals that of the capacitor.  Of course for sufficiently high frequencies, expect the "parts being equal" to effectively move at the speed of "light" (conduction inside the medium). I'd also expect that you need to model the resistances of the wires in the classic case as well.  I'd expect it to oscillate, but the conventions of schematic drawings don't always allow for the reality of HF effects.  Perhaps some HF engineer can help with how the HF world interprets schematics.[/Edit].

If the inductor remains conducting, the inductance shouldn't change (until it melts) and the circuit will act like a boring old classical circuit, only now with ESR.  ESR [Equivalent Series Resistance] is a spec commonly listed in capacitors indicating the amount of resistance you can expect across one.  I'd assume that inductors have a similar spec, but don't remember having to bother checking one.  I don't think I've used an inductor outside a switching power supply and don't really expect to.  So the oscillator simply decays in a boring old fashion (assuming it is sufficiently conductive to not explode).

In practice, I've heard that most superconductors are encased in copper (or similar conductive material), especially if the superconductor isn't naturally conductive.  In such a case, the inductor would be completely shorted (probably to ground) and the whole think would stop dead.  The whole "suddenly act classically" is dependent on being sufficiently conductive and superconductive in the same material (otherwise it would likely explode as I2R not only suddenly becomes high, but heating happens extremely *densely*, being an inductor after all).

Edited by wumpus
[/Edit]

##### Share on other sites

I believe wumpus is correct here. I can't say for sure, but I remember solving similar exercises as this one in my undergrad EE class, and nothing extreme ever resulted. I'd have to check my notebooks (I don't have them on hand) but I'd guess the max voltage you're going to get is a function of the capacitance and inductance, and the resistance of the wires if you're counting that (which at high voltage you probably should be).

The classical equations break down because 'derivative' and 'instantaneous' don't play nicely together, ever. Usually you end up with infinite voltage or current - practically, that means that some component burns out and opens the circuit/explodes/starts fires/other bad things. I've seen and LC loop (the RHS of your circuit there) used explicitly to prevent that component burnout in circuits with capacitors when the circuit is opened - the LC absorb the 'infinite' jump in voltage and dissipate the current because of their innate resistance. IDK how that would play out with superconductors though. I think that was with AC circuits though - its been a few years since I've done EE work.

##### Share on other sites

ive built a few dc/dc converters from scratch and this is over my head. this is well beyond basic electricity. il try though.

i know that in dc circuits inductors resist changes in current and capacitors resist changes in voltage. using the diagram in the post above it looks like you are going to need calculus to figure out what happens when you open the switch. when the power is on the cap will charge and the inductor will store energy in its magnetic field. the resistor will limit the current and will slow down the rate of charge of the cap and will limit the amount of energy stored in the magnetic field. when the switch is open the cap will try to dump its charge through the inductor since there is nowhere else for it to go. with no resistance limiting the current it should be pretty fast. but inductors also resist changes in current. also the inductor's magnetic field wants to collapse and it will push the voltage across it up to do this (dc-dc converters use this to boost voltage). you end up with the cap wanting to fight the inductor trying to dump its stored energy and reach a ground state. if the conduction is non-superconducting its just going to get dumped as waste heat oscilating back and fourth until the voltage reaches zero. superconducting will make this last longer essentially extending the time this energy remains in storage. quenching is just going to make it go non-superconducting and the conversion of energy to waste heat will continue. assuming you dont have any arcing or melting, this will be slowed down. if the goal is to dump this heat into a fusion pellet as fast as possible, i can think of better ways to do that. like just stick the pellet directly in the discharge arc of a capacitor bank.

Edited by Nuke

##### Share on other sites
1 hour ago, farmerben said:

What I want to do is disconnect the power supply completely and use the circuit on the right for a fusion fuel pellet.  When the quench happens there is a sudden voltage spike.  Is it as simple as dV/dt = I dR/dt ?  Superconductor quench makes dR/dt a huge positive number.

For a fusion fuel pellet we want the inductor to surround the capacitor.  So maximum electric and magnetic fields are centered at the same place.  We load the pellet with a dielectric which is also high explosive and mostly deuterium.  Detonate the chemical explosive, quench the entire superconductor fast, crush deuterium and tritium between the plates of the capacitor.

There is no oscillation or reset after the quench, the inductor and capacitor are vaporized.  I think the electric and magnetic fields persist into the plasma state.  It is a self contained sacrificial fuel nozzle.   If the superconductor is mostly lithium, it could be quite efficient as propellant.

In this case voltage, current, and resistance vary significantly with time and you also have to worry about impedance since we're dealing with an AC circuit with reactive components (see https://en.wikipedia.org/wiki/Ohm%27s_law#Reactive_circuits_with_time-varying_signals). I'm pretty sure the voltage will jump as high as it is going to get when you disconnect that switch. While it is superconducting you won't lose energy to resistance (at least through the inductor), but that will change when you quench (lots of energy dissipated as heat). That's assuming your capacitor doesn't burn out the moment you disconnect your supply.

You can get the conditions to power fusion using inductors and capacitors to boost voltage since Farnsworth-Hirsch fusors commonly use high voltage flyback transformers (with ordinary inductors) to drive them. Unfortunately high voltage fusors generally don't reach an energy positive point. They are useful as neutron sources and research, but not useful for doing net positive work.

The electric and magnetic fields of the plasma would change very quickly with time. This is one of the major hurdles of fusion, once you get it started the plasma generates its own electromagnetic fields that counter the beautifully orchestrated EM fields you used to get them to fuse in the first place.

##### Share on other sites

In the classical low power domain,  I think it works like this.  R battery >> R inductor.  With the switch closed the potential across the inductor and the capacitor is near zero.  The capacitor is near zero charge.  When the switch opened V inductor and V capacitor are zero, but current is maximum.  A quarter period later the voltage equals the previous voltage of the battery, but current is zero and there is no magnetic field.

One interesting thing I find on the Farnsworth fusor pages, and on pages about flyback diodes to handle accidental magnet quench, is that they all refer to exactly the same voltage as the spark plugs in my truck 50kV.   I think this number must be based on the maximum voltage surges that normal conductors can handle without vaporizing.

D + T fusion begins at 100 keV, but it's over a million times more likely at 2.5 MeV.  So I'm looking for a voltage surge about 50x larger than what people deal with every day.

If fuel pellets were tiny enough ~ 1mm.  They could be practically assisted by external beams and external magnetic nozzles.  Power and magnetic field add linearly, so the spacecraft only needs a fraction of nozzle capability and a fraction of the beam capability that would be required for inert fuel pellets.    While we add fusion mechanisms linearly the fusion rate goes up logarithmically, for a while.

Monoatomic layers of lithium and boron on graphene are superconducting in liquid deuterium.  They also have a lovely tendency to be disintegrated by neutrons into much higher energy alpha particles.

I'm not actually interested in a harmonic oscillator, it just happens to be the same in the classical low power case.

This is another concept for the capacitor/sparkplug.  It will fail in the narrow region first, but the electric field will accelerate all the positive charges in the same direction.  We want to achieve velocities with respect to each other of 2.5 MeV, which is more likely to occur in the pitted region.

Edited by farmerben

##### Share on other sites

Here is a more flushed out representation.  The power supply produces the blue current, when it is removed the red current begins, it has zero voltage initially.

At t = 1/12 period, we have a 30 degree phase angle where V = 1/2 Vmax, and B = sqrt(3)/2 Bmax.  We probably want to quench the superconductor at or before that point.  Let's say our initial power supply was 50 kV and our system is at 25 kV when we quench it.  The quench multiplies the voltage across the device by thousands of times.  Giving us 25 MV across the device and 2.5 MV across a hundred layers of capacitors.

Fusion begins in zone 1.  Hopefully leading to an even greater fusion rate in zone 2 which is enough to continue driving fusion in zones 3, 4, 5.

The molecule Li6 B10 H24 is a promising filler.  It is a transparent, dielectric salt, which is profoundly unstable.  Low energy photons cause it to explode into 6 radical ions with about 30eV of energy each.  Which means it can all be detonated virtually simultaneously before fusion energy reaches it.  That is good for increasing the number of possible collisions when fusion begins.

It really depends on being able to get the initial burst of fusion in zone 1 to be a million times greater than what can be achieved with a Farnsworth fusor or a neutristor.

Edited by farmerben

##### Share on other sites

fusors actually have pretty good fusion rates, but losses are really bad and prevent them from being breakeven. its why im a huge polywell proponent because you take the fusor, and replace the grids (which both contaminate plasma and provide conduction of energy away from the fusion reaction), with a virtual cathode (created by injecting electrons into a magnetic grid, or magrid) which should give you all the benefits of a fusor without the disadvantages. one thing they seem to be doing that nobody else seems to be doing is to design the machine entirely with computer models so that they can really optimize their demo machine before its built. but the next machine they will build will likely be a demo machine that can reach scientific breakeven. bussard just wanted to skip ahead and build a big reactor, and that was what he was trying to fund with his google talk before he died. park is taking a more conservative approach, but the plan is to still have a demo in the early 2020s.

Edited by Nuke

##### Share on other sites

The polywell stacks well with many other technologies.  In addition to electron beams and ultraviolet lasers it is possible to stack several other strategies.

With most types of particle accelerators the energy level is great, but the current is very low.  This can be alleviated in space, because with an infinite vacuum chamber we can focus as many beams as we want at the same point.  However, electrons will repel each other so only by having positive charge beams converging toward the same point at the same time, can bring the total energy to a high level.

A type of electron beam called a laser wakefield generator gives the most bang for the buck.

Good old fashioned cyclotrons give the highest current levels for ion beams.  A nice 1 meter cyclotron with a single magnet can get deuterium particles of about 500 MeV.  By itself it is an inefficient way to do fusion.  But as a well timed pulse combined with other systems it is extremely helpful.

Neutron beams from a fission reactor can scatter off hydrogen at fusion energies.  They also disintegrate lithium with extreme energy.  Neutrons would escape a polywell.  Krypton 78 absorbs neutrons to become Kr 79 which emits positrons with a halflife of 35 hours.   I'm not sure about the details, but using fission to drive fusion and antimatter, has positive feedback potential.

In the Van Allen belts there are plenty of high energy electrons and protons that can be scooped up with a magnet.  But I think protons are not what you want in a fusion reactor.

##### Share on other sites
2 hours ago, farmerben said:

The polywell stacks well with many other technologies.  In addition to electron beams and ultraviolet lasers it is possible to stack several other strategies.

With most types of particle accelerators the energy level is great, but the current is very low.  This can be alleviated in space, because with an infinite vacuum chamber we can focus as many beams as we want at the same point.  However, electrons will repel each other so only by having positive charge beams converging toward the same point at the same time, can bring the total energy to a high level.

A type of electron beam called a laser wakefield generator gives the most bang for the buck.

Good old fashioned cyclotrons give the highest current levels for ion beams.  A nice 1 meter cyclotron with a single magnet can get deuterium particles of about 500 MeV.  By itself it is an inefficient way to do fusion.  But as a well timed pulse combined with other systems it is extremely helpful.

Neutron beams from a fission reactor can scatter off hydrogen at fusion energies.  They also disintegrate lithium with extreme energy.  Neutrons would escape a polywell.  Krypton 78 absorbs neutrons to become Kr 79 which emits positrons with a halflife of 35 hours.   I'm not sure about the details, but using fission to drive fusion and antimatter, has positive feedback potential.

In the Van Allen belts there are plenty of high energy electrons and protons that can be scooped up with a magnet.  But I think protons are not what you want in a fusion reactor.

initially you are going to want to go with d-d. but moving to proton-boron you can just let the neutrons go as they are only a tenth of a percent of the fusion products anyway. the alpha particles can be direct converted to electricity, which for a spacecraft really reduces the radiator requirements and eliminates the need for a thermodynamic engine to recover the energy. direct conversion should also be more efficient than a brayton cycle. alternately you could leave a section of the direct conversion grids open to produce thrust from the high energy alpha particles while still recovering enough to sustain the reaction.

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.