What happens if we suddenly open the switch in a simple LC oscillating circuit

Thread Starter

anony12345ous

Joined Aug 1, 2019
4
Let's say we have an LC oscillating circuit with a switch in the circuit (that doesn't have any resistance), I wanted to understand what will happen if we suddenly open the switch. Since inductor current changes rapidly in the circuit, does it cause sparks in the switch itself or the capacitor can protect it by storing that energy as charge (high amount of charge flows and gets accumulated in the capacitor as the switch is opened) !!
 

oz93666

Joined Sep 7, 2010
669
So ... this is the situation with the switch closed , the energy is bouncing back and forth between the cap and inductor . in the first (time) diagram the energy is all in the cap ... current is zero , if the circuit is open then the cap cannot discharge , no current flows all the energy stays stored in the cap .
in the second diagram the energy is in the inductor if the switch is suddenly open this will tend to stop current in the inductor , resulting in a collapse in field in inductor , and a large voltage build up , a spark will want to cross the switch , energy is dissipated in the spark ....I think
 
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Thread Starter

anony12345ous

Joined Aug 1, 2019
4
So ... this is the situation with the switch closed , the energy is bouncing back and forth between the cap and inductor . in the first (time) diagram the energy is all in the cap ... current is zero , if the circuit is open then the cap cannot discharge , no current flows all the energy stays stored in the cap .
in the second diagram the energy is in the inductor if the switch is suddenly open this will tend to stop current in the inductor , resulting in a collapse in field in inductor , and a large voltage build up , a spark will want to cross the switch , energy is dissipated in the spark ....I think

Yeah, makes sense to me, thanks !! But then again, instead of charges going through the air causing sparks, why don't they just sit on the capacitor plate charging/discharging (depending on the direction of current when the switch is opened) the capacitor plates. Basically inductor pushing the charges into one of the capacitor plates and those charges repelling the charges on the other plate, just storing all (or maybe some) of the inductors energy into the capacitor. And if the sparks were there, would they discharge the capacitor too ??
 

BobaMosfet

Joined Jul 1, 2009
809
Yeah, makes sense to me, thanks !! But then again, instead of charges going through the air causing sparks, why don't they just sit on the capacitor plate charging/discharging (depending on the direction of current when the switch is opened) the capacitor plates. Basically inductor pushing the charges into one of the capacitor plates and those charges repelling the charges on the other plate, just storing all (or maybe some) of the inductors energy into the capacitor. And if the sparks were there, would they discharge the capacitor too ??
You're not grasping the physical nature of an electron. think of a straw full of marbles. the marbles can only move if they have somewhere to go. When you open a switch, the resistance becomes so high past a certain point that the electrons cannot jump the air-gap (not enough voltage exists to maintain the transfer), and since they cannot move in either direction, there is now current flow and now voltage flow.

You need also to understand that voltage and current go hand in hand, in a reciprocal relationship (Ohm's Law). Altering one alters the other.
 

crutschow

Joined Mar 14, 2008
23,374
instead of charges going through the air causing sparks, why don't they just sit on the capacitor plate charging/discharging (depending on the direction of current when the switch is opened)
There would be no sparks if you open the switch at the instance when the capacitor is charged to the maximum voltage and all the energy is stored in the cap.
Otherwise if any current is flowing through the inductor, its energy must be dissipated when the inductor current is forced to stop and much of that will occur in the spark across the switch.
The rest of the energy at that point will stay on the capacitor.
And if the sparks were there, would they discharge the capacitor too ?
There will be some charge/discharge (depending upon the current direction when the switch is opened) of the capacitor by the spark current until the inductive energy is dissipated and the current stops.

Below is the LTspice simulation of an LC oscillating circuit to demonstrate the opening of a switch during the cycle.
(R1-C2 are just to damp the parasitic oscillations due to stray capacitance after the switch opens.)
D1 and D2 simulate the voltage of the spark gap (in reality the spark voltage would be much higher).
As you can see, the inductor current (blue trace) goes rapidly to zero when the switch S1 is opened at 310μs.
The inductor voltage (red trace) jumps to the spark voltage (about 110V here) until the inductor current goes to zero.
The capacitor voltage (yellow trace) stays essentially at its voltage when the switch opens.
(The capacitor voltage does rises about a volt during the spark gap time (measurement at bottom) but that would be much less with the much shorter time of the actual spark.)

upload_2019-8-1_9-22-2.png
upload_2019-8-1_9-27-0.png
 

WBahn

Joined Mar 31, 2012
24,704
Yeah, makes sense to me, thanks !! But then again, instead of charges going through the air causing sparks, why don't they just sit on the capacitor plate charging/discharging (depending on the direction of current when the switch is opened) the capacitor plates. Basically inductor pushing the charges into one of the capacitor plates and those charges repelling the charges on the other plate, just storing all (or maybe some) of the inductors energy into the capacitor. And if the sparks were there, would they discharge the capacitor too ??
What you are trying to describe is charging up just one plate of a capacitor. If you run some numbers, you'll see that you can't get very much charge built up that way before the electric field preventing any further charge coming from the inductor is too great to overcome and the current instead will arc across the switch.
 

Thread Starter

anony12345ous

Joined Aug 1, 2019
4
There would be no sparks if you open the switch at the instance when the capacitor is charged to the maximum voltage and all the energy is stored in the cap.
Otherwise if any current is flowing through the inductor, its energy must be dissipated when the inductor current is forced to stop and much of that will occur in the spark across the switch.
The rest of the energy at that point will stay on the capacitor.
There will be some charge/discharge (depending upon the current direction when the switch is opened) of the capacitor by the spark current until the inductive energy is dissipated and the current stops.

Below is the LTspice simulation of an LC oscillating circuit to demonstrate the opening of a switch during the cycle.
(R1-C2 are just to damp the parasitic oscillations due to stray capacitance after the switch opens.)
D1 and D2 simulate the voltage of the spark gap (in reality the spark voltage would be much higher).
As you can see, the inductor current (blue trace) goes rapidly to zero when the switch S1 is opened at 310μs.
The inductor voltage (red trace) jumps to the spark voltage (about 110V here) until the inductor current goes to zero.
The capacitor voltage (yellow trace) stays essentially at its voltage when the switch opens.
(The capacitor voltage does rises about a volt during the spark gap time (measurement at bottom) but that would be much less with the much shorter time of the actual spark.)

View attachment 182990
View attachment 182991

Yeah this makes sense, thanks a lot :)
 

Thread Starter

anony12345ous

Joined Aug 1, 2019
4
What you are trying to describe is charging up just one plate of a capacitor. If you run some numbers, you'll see that you can't get very much charge built up that way before the electric field preventing any further charge coming from the inductor is too great to overcome and the current instead will arc across the switch.

Okay yes understood, thanks a lot :)
 
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