Why do capacitors not discharge everything immediately in LC circuits?

[mymodded]

TL;DR

why do capacitors not discharge entirely immediately in LC circuits?

I just started taking LC circuits, and I was wondering, when capacitors charge without a resistance, they charge immediately, so when they discharge in an LC circuit, why don't they "send" all of their charge immediately? Is it because that's how capacitors work in general, or is it because of the induced current in inductors or what?
sorry if the question is a bit dumb.

 

[phinds]

In order for a cap to discharge immediately, it would have to dump a lot of current into the circuit. What does an inductor do when you try to pump a lot of current into it immediately?

 

[mymodded]

In order for a cap to discharge immediately, it would have to dump a lot of current into the circuit. What does an inductor do when you try to pump a lot of current into it immediately?

it induces an opposing current

 

[kuruman]

it induces an opposing current

And then what happens in an ideal LC circuit with no Ohmic losses? When the charge on the capacitor eventually becomes zero does it remain zero?

 

[Averagesupernova]

it induces an opposing current

It opposes a changing current. Inductors are happy continuing to pass a non-changing current. Try to increase it or decrease it and it does what it has to in order to counter this.

 

[phinds]

it induces an opposing current

And doesn't that answer your question?

 

[DaveE]

Have you studied calculus yet?

 

[mymodded]

And doesn't that answer your question?

I wanted to make sure that this is the reason and not because that's how capacitors work in general

 

[mymodded]

Have you studied calculus yet?

Yes

 

[DaveE]

OK, then,

Capacitors operate based on this equation: $i(t)=C \frac{dv(t)}{dt}$ or $v(t)=\frac{1}{C}( \int_0^t i(t) \, dt + i(0))$

Inductors operate based on this equation: $v(t)=L \frac{di(t)}{dt}$ or $i(t)=\frac{1}{L}( \int_0^t v(t) \, dt + v(0))$

So, for initial conditions of no current and some voltage on the capacitor, the capacitor voltage is also the inductor voltage. According to the equation $\frac{di(t)}{dt}=\frac{v(t)}{L}$, the current will start to increase (it can't change instantly, but it's slope $\frac{di(t)}{dt}$ can). Then the increasing current will start to discharge the capacitor.

There are some good descriptions online. Like this one:
https://www.khanacademy.org/science...-response/v/ee-lc-natural-response-intuition1

 

[mymodded]

OK, then,

Capacitors operate based on this equation: $i(t)=C \frac{dv(t)}{dt}$ or $v(t)=\frac{1}{C}( \int_0^t i(t) \, dt + i(0))$

Inductors operate based on this equation: $v(t)=L \frac{di(t)}{dt}$ or $i(t)=\frac{1}{L}( \int_0^t v(t) \, dt + v(0))$

So, for initial conditions of no current and some voltage on the capacitor, the capacitor voltage is also the inductor voltage. According to the equation $\frac{di(t)}{dt}=\frac{v(t)}{L}$, the current will start to increase (it can't change instantly, but it's slope $\frac{di(t)}{dt}$ can). Then the increasing current will start to discharge the capacitor.

There are some good descriptions online. Like this one:
https://www.khanacademy.org/science...-response/v/ee-lc-natural-response-intuition1

thanks a lot

 

[Herman Trivilino]

I just started taking LC circuits, and I was wondering, when capacitors charge without a resistance, they charge immediately, so when they discharge in an LC circuit, why don't they "send" all of their charge immediately?

Because of the inductor's magnetic field. In a LC circuit the capacitor wouldn't charge instantly, either.

Resistance is just one type of impedance. Inductors also have impedance even if they have negligible resistance.

 

[sophiecentaur]

Because of the inductor's magnetic field. In a LC circuit the capacitor wouldn't charge instantly, either.

It may be worth introducing the Energy situation. The inductor's magnetic field has Energy and the E field across the capacitor has Energy. The rate of transfer of Energy depends on the induced emf in the inductor. Total energy remains the same unless a resistive element is included in the model. That will cause the oscillations to damp down.

However, a real physical circuit (of any size) will radiate a certain amount of EM Power during the oscillations. So you can't rely on any of these problems being answerable is you ignore the resistive / radiative loss. You can get apparent paradoxes.

 

[DaveE]

So you can't rely on any of these problems being answerable is you ignore the resistive / radiative loss.

But you can get a really useful approximation for underdamped cases. Of course, that all depends on the question you're asking and the damping.

I'm a big fan of first finding the easy approximate answer and then asking the question "do I care about more accuracy?" (i.e. estimate the size of the errors). Very often the answers is no. When the answer is yes, you already know a lot about the form of the result or the approach you'll take next.

I'll keep repeating my favorite quote from one of my EE professors until y'all are sick of hearing it:
"Engineering is the art of approximation" - R. D. Middlebrook

Granted, if it's a math class, maybe you shouldn't listen to me.

 

[Herman Trivilino]

Granted, if it's a math class, maybe you shouldn't listen to me.

Okay, but only while in class.

 

[sophiecentaur]

I'm a big fan of first finding the easy approximate answer and then asking the question "do I care about more accuracy?"

I totally agree.EM is multi-layered. I had an advantage of a 'strict' basic physics education. We would accept a model and get familiar with the basic equations. No messing about with photons, electrons and more subtle levels in lower school. If we asked about stuff like that we were told that we'd address it later - and we did. Demanding to 'understand' stuff when you haven't got the basics is to demand instant gratification and it doesn't exist in Science. I have a theory that many of the questions we read in these threads are introduced as distractions from the rigour of the maths involved. We used to try that on in class and we didn't get away with it. PF threads often deviate because 'people who know' things get tempted to introduce them far too early. This often just confuses an OP and they leave the thread because of too much information.

Engineers grow up using the 'near enough is good enough' approach; it's the only way forward in a highly complex topic. After that level has been sorted out, they take the next step.