LR circuit confusion

Thread Starter

Vorador

Joined Oct 5, 2012
87
But isn't the act of removing the voltage source from the inductor an attempt to change it's current and shouldn't an induced voltage then appear across it?
 

Brownout

Joined Jan 10, 2012
2,390
Sure, but remember that voltage across an inductor can change instatainously, so if at the very moment you removed the voltage, you connect an uncharged capacitor in parallel, the voltage will be zero volts, although current continues to flow from the inductor. The capacitor voltage begins to rise as soon as inductor current flows into it. Current through an inductor cannot change instantainously.
 

WBahn

Joined Mar 31, 2012
29,979
But isn't the act of removing the voltage source from the inductor an attempt to change it's current and shouldn't an induced voltage then appear across it?
Not necessarily. The inductor will induce whatever voltage is required in order to keep the same current flowing in that instant (the current can then start changing immediately after that, it just can't changed instantly from one value to another value).

But with the capacitor connected across it the voltage across the inductor can't change instantly, either, because the voltage across a capacitor can't change instantaneously. But there is no such restriction on the current through a capacitor. So at the moment the switch is opened, the current in the capacitor will jump from being zero to being the current in the inductor. Since that satisfies the inductor's requirement that the same current continue to flow, there is no need for it to induce any voltage at all.

But now that there is current flowing in the capacitor, it's voltage will start changing. That voltage will be applied to the inductor and will be of the polarity that results in the inductor current dimishing in magnitude so the current will start decreasing.

Eventually, the current will be reduced to zero and the capacitor will be fully charged. At this point the capacitor voltage starts driving current into the inductor going the other way, which takes charge out of the capacitor resulting it the voltage going down. This will continue until the capacitor is dishcharged and the inductor has the original current in it, but going the other way.

This is one half of the resulting oscillation and now the process repeats until you end up with exactly the same situation you had originally, which is then one full cycle of the oscillation. That's if all you have is an L and a C. If there is also an R, then in each half cycle there is some current flowing in the resistor which is dissipating energy and so the peak inductor current and peak capacitor voltage drop in magnitude with each half cycle until, eventually, all of the energy is dissipated and there is no current and no voltage anywhere.
 

WBahn

Joined Mar 31, 2012
29,979
Sure, but remember that voltage across an inductor can change instatainously, so if at the very moment you removed the voltage, you connect an uncharged capacitor in parallel, the voltage will be zero volts, although current continues to flow from the inductor. The capacitor voltage begins to rise as soon as inductor current flows into it. Current through an inductor cannot change instantainously.
In the example I gave, you don't need to switch in a capacitor. Since the system is in steady state before you open the switch, there is no voltage across the inductor which means no voltage across the capacitor which means that it is uncharged.
 

Brownout

Joined Jan 10, 2012
2,390
In the example I gave, you don't need to switch in a capacitor. Since the system is in steady state before you open the switch, there is no voltage across the inductor which means no voltage across the capacitor which means that it is uncharged.

I didn't assume steady state in my example. I wanted a more general one.
 
Last edited:

WBahn

Joined Mar 31, 2012
29,979
I didn't assume steady state in my example. I wanted a more general one.
Then you've got to deal with the situation that the capacitor that is already in the OP's RLC circuit can have a non-zero voltage across it when you switch in your additional uncharged capacitor.
 

Brownout

Joined Jan 10, 2012
2,390
Then you've got to deal with the situation that the capacitor that is already in the OP's RLC circuit can have a non-zero voltage across it when you switch in your additional uncharged capacitor.

Nope. Since no circuit was posted, I was describing a completely unique and theoritical circuit.
 

WBahn

Joined Mar 31, 2012
29,979
Nope. Since no circuit was posted, I was describing a completely unique and theoritical circuit.
A circuit was posted:

Alright so I think I understand it now, but I have just one more question concerning a source-free parallel RLC circuit problem. The initial data given in the problem is inductor current IL(0)=6A and capacitor voltage Vc(0)=0V.
 

Thread Starter

Vorador

Joined Oct 5, 2012
87
Not necessarily. The inductor will induce whatever voltage is required in order to keep the same current flowing in that instant (the current can then start changing immediately after that, it just can't changed instantly from one value to another value).

But with the capacitor connected across it the voltage across the inductor can't change instantly, either, because the voltage across a capacitor can't change instantaneously. But there is no such restriction on the current through a capacitor. So at the moment the switch is opened, the current in the capacitor will jump from being zero to being the current in the inductor. Since that satisfies the inductor's requirement that the same current continue to flow, there is no need for it to induce any voltage at all.

But now that there is current flowing in the capacitor, it's voltage will start changing. That voltage will be applied to the inductor and will be of the polarity that results in the inductor current dimishing in magnitude so the current will start decreasing.

Eventually, the current will be reduced to zero and the capacitor will be fully charged. At this point the capacitor voltage starts driving current into the inductor going the other way, which takes charge out of the capacitor resulting it the voltage going down. This will continue until the capacitor is dishcharged and the inductor has the original current in it, but going the other way.

This is one half of the resulting oscillation and now the process repeats until you end up with exactly the same situation you had originally, which is then one full cycle of the oscillation. That's if all you have is an L and a C. If there is also an R, then in each half cycle there is some current flowing in the resistor which is dissipating energy and so the peak inductor current and peak capacitor voltage drop in magnitude with each half cycle until, eventually, all of the energy is dissipated and there is no current and no voltage anywhere.
Thank you for such a detailed response! It makes things a lot more clear now! :)

You said that the inductor only induces voltage when it is necessary to keep the same current flowing through it. Does this mean that in a loop consisting of only an ideal inductor, the same current would keep flowing through it forever?
 

WBahn

Joined Mar 31, 2012
29,979
Thank you for such a detailed response! It makes things a lot more clear now! :)

You said that the inductor only induces voltage when it is necessary to keep the same current flowing through it. Does this mean that in a loop consisting of only an ideal inductor, the same current would keep flowing through it forever?
Yes. And such loops exist -- they are called superconducting magnets. I don't know what the record is, but I know that persistent currents have been sustained for years at a time.
 

crutschow

Joined Mar 14, 2008
34,285
If you really want to explore the operation of LC (or other) circuits I suggest you try a Spice type analog simulator. It makes it easy to observe the operation of such circuits as you can readily view the currents and voltages of the circuit while it is operating. LTspice is a good, free simulator that many on these forums use.
 
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