How long would inductor hold energy disconnected from battery but current flowed in alternate path

[Robert Clark]

Usually this question is asked in regards to just disconnecting the battery, like in the attached image. In that case the answer is just milliseconds. But in the analogous scenario for a capacitor, it could hold the charge for significant time because the charge would be held on the parallel plates as static electricity.

But it seemed to me there should be an analogous scenario for an inductor where the magnetic field persists. But for the inductor you need the current to continue to flow to maintain the magnetic field. So what about the scenario where the battery is disconnected by a switch but the switch then automatically closes an alternative pathway that allows current to flow, sans battery?

Bob Clark

 

[TugboatEng]

Inductors and capacitors behave differently. A capacitor pushes current to counter changes in voltage. An inductor pushes voltage to counter changes in current. An inductor is not analogous to a capacitor.

The problem with an inductor is that they can generate flyback voltages that will damage some state electronics. A flyback diode is the standard means to discharge energy from an indicator when de-energized.

Again, capacitors are not the same. They can draw very high currents when energized which means they may require some charging control. When disconnected they simply maintain their voltage.

 

[IRstuff]

The short answer is no; that's because the analogy is not perfect, a capacitor can have negligible losses, since its dielectric can be pretty near a perfect insulator, there is no equivalent in an inductor, since always has losses.

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[Robert Clark]

Thanks for the response. You want the switch so rapid that the field doesn’t collapse. There are also “make-before-break” switches that can switch to the alternative current flow pathway before the battery is fully disconnected:

Transfer switch 101.
Closed transition
A closed transition is a “make before break” transfer, in that the transfer switch makes a connection to the new power source before breaking its connection to the old one. As there’s no gap between disconnection and connection, downstream loads receive continuous power throughout the transfer process.

https://www.eaton.com/content/dam/e...tribution-controls-systems/ats/wp140001en.pdf

I’m informed that Buck-boost converters conform to such a configuration. Buck-boost converters are so common, the equations for such a scenario must have already been developed.

​

https://en.m.wikipedia.org/wiki/Buck–boost_converter

Note all versions of the buck-booster converter shown include diodes.

 

[3DDave]

As soon as current is interrupted there is no longer an electromagnetic force to maintain the field, which instantly, for all practical purposes, forces the current to continue to flow. That is the function in buck/boost converters. Because it is forcing a current flow, it will generate whatever voltage is associated with that flow. Another practical use is generating the thousands of volts for spark ignition. The trade is the time - lower resistance = lower voltage and higher current.

Inductors are electromagnetic dynamic
Capacitors are electrostatic and passively store charge

The main exception is using superconductors in a closed loop; in that case there is zero resistance and zero voltage. The cost and technical difficulty make that impractical as an energy storage device.

An analog to a capacitor is a water reservoir - it stores water against gravitation and the higher the pressure at the bottom, the more water it stores.

An analog to an inductor is a mass of water flowing in a river. It isn't possible to cut a section out of a river and still have that removed water continue to flow. The superconducting analog is freezing water being directed in a circle to make a flywheel from the ice.

 

[zeusfaber]

If you can introduce the alternative path without interrupting the current, and what you're left with is purely inductive and resistive (introducing the forward voltage drop of a diode, for instance, spoils the maths), then the current decays exponentially - the time constant being the inductance divided by the circuit resistance.

A.

 

[waross]

The time constant in seconds of an inductive-resistive circuit is calculated by multiplying[sup]Correction:[/sup]dividing the Inductance in Henries times the resistance in Ohm,
The time to reach 0% or 100% is accepted as 5 time constants.
This is commonly encountered in large DC fields where a diode is connected across the field to allow stored energy (both Volts and Amps) to decay to zero.

How long would inductor hold energy disconnected from battery but current flowed in alternate path

Answer:- Five Time Constants.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[FacEngrPE]

Inductors do store energy.
Lifting magnets can be particularly hazardous. Unplugging one of these without first turning the magnet power off will result in an arc flash across the plug connected to the magnet.

Mitigation is provided by a free wheel diode (resistance losses in the magnet eventually consume the energy).
If a resistance is added in the controller the energy will decay faster.

Hubble sells the line of lifting magnet controllers formerly sold by Euclid.

https://www.hubbell.com/hubbellindu...fting-Magnet-Controllers/HC91814202/p/1540621

Selecting fuses for lifting magnet circuits is particularly challenging. The Euclid controllers include an energy dissipating device arranged so it can never be disconnected from the magnet. It is arrange to protect the fuses from the stored energy.

 

[waross]

If a resistance is added in the controller the energy will decay faster.

Yes, but this will also raise the voltage across the external resistor.
For example, if the external discharge resistor is 10 times the resistance of the coil resistance the initial voltage across the external resistor will be 10 times the applied voltage.

When a switch is opened, the resistance of the arc across the switch points rises very quickly.
Thus the "Kick Back" Voltage rises to a high value very quickly.
The faster the switch action, the higher the voltage.

Lifting magnets are a special case.
It is not enough to deenergize the magnet.
Small parts such as nuts and bolts are often magnetized by the lifting magnet and will not drop free.
A scrap lifting magnet is often energized with reverse polarity for a short time to "kick off" small objects that have become magnetized.

In the circuit shown, there is not the level of interlocking that most designers will use.
In the absence of complete interlocks, the Drop fuses are there to clear the circuit in the event that both the Lift and Drop contactors are energized simultaneously.

The circuit is an "H" circuit.
When lifting, the Magnet Coil, M is energized with M1 Positive and M2 Negative through contactors 1L and 2L.
When dropping the Magnet Coil, M is reverse energized with M1 Negative and M2 Positive through contactor D

The Discharge Sensor Module, DSM, monitors the decaying voltage across the Magnet Coil and energizes the Drop Relay and Drop Contactor when safe to do so.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[electricpete]

> but it seemed to me there should be an analogous scenario for an inductor where the magnetic field persistb

Your starting point seems to be pretty basic, so I'm going to go back more toward the basics.

Linear inductors and capacitors act in an analogous way. It is also called duality. The energy storage variables are V for capacitor and I for inductor so they play similar roles. The natural response to both with initial conditions at t= 0 is as follows:

* simple loop with capacitor and resistor:
* Vc(t) = Vc(0) * exp(-t/tau) where tau = R*C

* simple loop with inductor and resistor:
* IL(t) = IL(0) * exp(-t/tau) where tau = L / R


high tau means slow decay.

(Aside: Actually for duality a series connection of an inductor corresponds to a parallel connection of a capacitor, but if there is only one external element then it forms a simple loop regardless of whether you consider the external element to be in series or in parallel)

Why did I say simple loop when you asked about the scenario of open circuited cap? As IRstuff observed, even for a capacitor which is not obviously connected to any circuit, there will still be leakage which can be modeled as a leakage resistance connected across the capacitor terminal (although the leakage resistance may be quite high so capacitor can retain charge for a long time with a time constant of up to hours or days). Likewise for an inductor that appears to be 'shorted", there will always be resistance in coils and wires in the circuit of an inductor which can be modeled as a series resistance, but the time constant is going to be no more than seconds or minutes for most inductor coils.

 

[waross]

But it seemed to me there should be an analogous scenario for an inductor where the magnetic field persists.

A special case is the Nuclear magnetic Resonance machines used in research.
These machines use an extremely high strength magnetic field.
The coil windings are super conducting and are cooled with liquid Helium.
The resistance of the windings is zero, and so the time constant is infinite.
Once a current (about 20 Amps (as I recall) is established, the current continues to flow in the closed loop with no external energy supplied.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!