Why do capacitors not discharge everything immediately in LC circuits?

In summary, capacitors in LC circuits do not discharge immediately due to the interplay between the capacitor's stored electric energy and the inductor's magnetic energy. When a capacitor discharges, it induces a current that flows into the inductor, creating a magnetic field. This magnetic field then stores energy, and when the capacitor fully discharges, the energy in the inductor can cause the current to continue flowing, leading to oscillations. This process results in an oscillatory exchange of energy between the capacitor and inductor, preventing immediate discharge.
  • #1
mymodded
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TL;DR Summary
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.
 
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  • #2
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?
 
  • #3
phinds said:
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
 
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  • #4
mymodded said:
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?
 
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  • #5
mymodded said:
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.
 
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  • #6
mymodded said:
it induces an opposing current
And doesn't that answer your question?
 
  • #7
Have you studied calculus yet?
 
  • #8
phinds said:
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
 
  • #9
DaveE said:
Have you studied calculus yet?
Yes
 
  • #10
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
 
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  • #11
DaveE said:
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
 
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  • #12
mymodded said:
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.
 
  • #13
Mister T said:
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.
 
  • #14
sophiecentaur said:
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.
 
  • #15
DaveE said:
Granted, if it's a math class, maybe you shouldn't listen to me.
Okay, but only while in class. :rolleyes:
 
  • #16
DaveE said:
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.
 
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FAQ: Why do capacitors not discharge everything immediately in LC circuits?

1. What is an LC circuit?

An LC circuit, also known as a resonant circuit or tank circuit, is an electrical circuit that consists of an inductor (L) and a capacitor (C) connected together. The inductor stores energy in the form of a magnetic field, while the capacitor stores energy in the form of an electric field. Together, they can oscillate energy back and forth, creating an alternating current (AC) signal.

2. Why do capacitors not discharge immediately in LC circuits?

Capacitors do not discharge immediately in LC circuits because the discharge process is governed by the circuit's inductance and capacitance. When a capacitor is connected to an inductor, the energy stored in the capacitor is gradually transferred to the inductor, creating oscillations rather than a rapid discharge. The interaction between the electric field of the capacitor and the magnetic field of the inductor leads to a sinusoidal exchange of energy, resulting in a sustained oscillation rather than an instantaneous discharge.

3. What factors influence the discharge time of a capacitor in an LC circuit?

The discharge time of a capacitor in an LC circuit is influenced by several factors, including the values of the inductance (L) and capacitance (C), the resistance in the circuit, and the initial voltage across the capacitor. The resonant frequency of the circuit, given by the formula \( f = \frac{1}{2\pi\sqrt{LC}} \), also plays a crucial role in determining the rate of oscillation and energy transfer between the capacitor and inductor.

4. What happens to the energy in the capacitor during oscillation?

During oscillation in an LC circuit, the energy stored in the capacitor is alternately converted into magnetic energy in the inductor and back into electric energy in the capacitor. As the capacitor discharges, it releases its energy to the inductor, creating a magnetic field. Once the capacitor is fully discharged, the magnetic field collapses, inducing a current that recharges the capacitor in the opposite polarity. This cycle continues, resulting in oscillations until energy is lost due to resistance in the circuit.

5. How does resistance affect the discharge of capacitors in LC circuits?

Resistance in an LC circuit causes energy loss in the form of heat, which affects the oscillations. When resistance is present, it dampens the oscillations, leading to a gradual decrease in amplitude over time. This phenomenon is known as damping. In a purely ideal LC circuit (with no resistance), the oscillations would continue indefinitely. However, in a real circuit with resistance, the energy will eventually dissipate, causing the capacitor to discharge more slowly and leading to a finite number of oscillations before stopping.

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