What happens if you connect floating cap to charged inductor?

[Trotters_Independent2]

Hi all

Please look at the simple circuit I drew in the attachment.



If I were to have a battery charging up an inductor through a closed switch for some arbitrary time and then open that switch.... I'm wondering what would happen (besides possible arching across the switch) and what I'd see on the voltmeter. I know the inductor voltage would transiently rise in an attempt to keep the current through it flowing... but would there be any actual potential difference resulting from this for my ideal voltmeter to detect between the two capacitor plates? Would any energy at all cycle back and forth between the L and C while getting damped by R?

I am suspecting what would happen is just the right C plate voltage will rise with the inductor volts. And since the other C plate is not connected to it (except through a non-conducting dielectric)... it would just remain floating - so it would essentially be like probing a voltage with a voltmeter... but with a single probe and no reference voltage.... therefore the voltmeter would read zero.

Am I wrong? Or in practice.... ie: if I got some real components and built a circuit like this on a breadboard.... might something different happen?

[Kyle_from_somewhere]

What you're asking - if power will rock back and forth between inductor and capacitor, is called a tank circuit.

R doesn't matter here as everything has enough internal resistance to damp the process. we don't have perpetual motion.

Like all circuits, a tank circuit has to be closed for electricity to flow. leave the end floating and nothing happens.

If you build it in real life it won't do anything. You will have better luck if you put voltmeter between floating end of cap and the ground, but even then, the power will dissipate faster than you can observe it.

[Zero999]

Nothing will happen because it's a an open circuit.

[eugene]

It might help to understand that you cannot, in practice, put extra charge on one plate of a capacitor. Effectively, what is possible is to move charge from one plate to the other. Adding charge to one plate without removing an equal charge from the other plate is, for practical purposes, impossible.

[Stray Electron]

Hi all

Please look at the simple circuit I drew in the attachment.

(Attachment Link)

If I were to have a battery charging up an inductor through a closed switch for some arbitrary time and then open that switch....

   That's exactly what happens in the primary side of old breaker operated ignition systems used most engines prior to the early 1970s.  The points close, the battery charges the primary side of the ignition coil, the points (a cam driven mechanical switch) open and the coil voltage quickly rises in an attempt to keep the current flow constant. Capacitance in the circuit would make the voltage ring but resistance inherent in the circuit would quickly dampen the current flow and voltage. The secondary side of the coil magnifies the primary voltage and the secondary is connected to the spark plugs and it serves as the ignition voltage. Automobile Ignition Analyzers (basically a scope capable of very high input voltages) were designed to test the ignition systems and could display both the primary and secondary voltages.

  You can clearly see the dampened ringing pattern in the third picture here http://patrickmsblog.blogspot.com/2010/11/ws3-ocilloscope-patterns.html

[mc172]

Hi Treez

[radiolistener]

When you turn off the switch, your L and C circuit will starts oscillating.

What your Voltmeter can show depends on Voltmeter characteristics. If it allows to catch a short voltage pulse, it can display pulses of oscillations. The voltage also will depends on the wire length on the second terminal (floating, with open end) of the capacitor. 

Since wires and conductors inside coil and capacitor have resistance, the energy of oscillation will drops down. There is also energy loss on electromagnetic wave emission which also leads to amplitude drop down. Higher energy loss, more short time period for amplitude drop down.

So, if your voltmeter is DC or just too slow and unable to detect a short pulse, it will show zero.

[T3sl4co1l]

What happens where -- what does the circuit actually represent?

If this schematic is a lumped element model, it is either incomplete, or contains singularities:

For an ideal switch, turning off into a charged ideal inductor, the inductor will discharge instantaneously -- generating infinite voltage for zero time, with a total area under that curve corresponding to the flux it was charged to.  A Dirac delta.  The voltmeter still reads zero.

If we complete the circuit so it is well-defined and finite, then there is some shunt impedance on the switching node which defines the voltage, and the switch may have finite turn-off speed (in which case, its V-I characteristics during turn-off also matter).


If this schematic instead represents a physical arrangement of components, then it matters very much what those components are, and how they are arranged.  (Indeed, a real switch will most likely arc in this case.)  The voltmeter may or may not read anything, for a number of reasons -- most of which are not described by the diagram as shown, and so are not obvious to the casual observer.

Usually, we like schematics to be somewhere between these two cases: enough components/elements to capture the desired behavior space, while omitting information that isn't relevant to it.

Tim

[fordem]

That's exactly what happens in the primary side of old breaker operated ignition systems used most engines prior to the early 1970s.

A breaker point ignition system is quite a bit different to that...

[Zero999]

People here appear to be overthinking this.

In theory, no current will flow through the capacitor. In reality, if the capacitor has previously been charged, a tiny current will flow, causing the capacitor to discharge through the meter's 10M input impedance. Remember, the node is floating. If the capacitor isn't charged then no current will flow.

[Terry Bites]

If the output is floating, it’s not referenced to ground, no current can be defined in R or C, therefore the node voltage is undefined.
An undefined current flows and an undefined energy will be transferred from L to C.
Zero is well defined mate.

Who knows if the new undefined will become a source of free energy...... I'm taking my pills, its ok. OK!

[Picuino]

If one of the capacitor terminals is not connected, no current can flow through the capacitor. Therefore the capacitor maintains a constant voltage between terminals (for example 0 volts).

   No current flow = No voltage change on the capacitor terminals

When the switch is opened, the coil generates a negative overvoltage on the terminal that is not connected to ground and the same negative voltage will be on the two terminals of the capacitor which, at most, will behave as a small antenna.