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# LC natural response intuition 2

We can predict the shape of voltage and current in an LC circuit by tracking the motion of charge as it flows back and forth. Created by Willy McAllister.

## Want to join the conversation?

• You said in this video at time that the current will continue to flow in the same direction until it reaches 0. What I don't understand is that how come the current continues to flow even though there is no voltage around the capacitor.
• Hello Usman,

There is energy stored in the inductor. This energy is defined by the flow of current. Its as if the inductor has momentum. Once current is flowing the inductor will do everything is can to keep that flow constant..

You are correct, the capacitor is "empty" is has no energy whatsoever. But at this moment the inductor has all of the energy and it is doing everything it can to keep the current constant. Consequently, the capacitor is driven negative.

Regards,

APD
• Excuse me at , , I thought the main reason there's a current is the charge's higher concentration on one plate of the capacitor , once some charges flow to the other plate and the have equal charges , the flow should cease .
But How exactly does the inductor do to keep a steady current flowing ?? this is so strange to hear do you have further detailed explanation ?
• Once some charge flows to the other plate and the plates have equal charges, the voltage becomes zero (as shown in the sketched response vs. time), but the current does not go instantly to zero. Think about the pendulum analogy for the LC circuit. When the bob (the pendulum's weighted end) swings back to bottom dead center, it's position is back to 0, but the velocity is not 0 (the pendulum does not suddenly stop when it gets to the bottom of the swing, it keeps going.) In the pendulum analogy, the position corresponds to voltage and the velocity corresponds to current.
• The current and voltage versus time graph will keep going on like we plotted it. Does that mean in ideal circuit, current will always keep flowing in this circuit ? What happens in real circuits ?
• The ideal LC circuit rings forever. There is no mechanism for energy to leave the system, so it just keeps going.

In real circuits there is always some small resistance in the circuit. That changes it from an LC to an rLC. Every time charge passes through the little resistor a little bit of energy is lost (the resistor warms up a little bit). Eventually current and voltage die out to zero. The RLC circuit is covered in the next set of articles and videos.

A good analogy to an electrical LC circuit is a bell. If you strike a bell it rings with a pure-sounding sine wave tone. It will ring for a very long time until the vibrations give out due to mechanical resistances in the system.

I suggest you take a look at these revised and improved articles on LC and RLC natural response. https://spinningnumbers.org/t/topic-natural-and-forced-response.html#lc
• Is this circuit kind of like an AC power source?
• The LC circuit is similar to an AC power source in that the both create sine waves. They differ in that the AC power source actually delivers power to whatever it is connected to, whereas the LC circuit can't deliver power in a sustained way. If you connect the LC to something the stored charge immediately drains off and the sine wave stops.

Think of the LC as like a pendulum swinging back and forth. If you leave it alone it keeps going, but if you touch it the pendulum stops.

The AC power source is more like a bicycle being pedaled by a strong athlete. The back wheel goes round and round. If you try to stop it with your hand it will keep going due to the strength of the rider.
• So here we start with a capacitor that is already energized somehow, and an inductor that has zero stored energy, and what follows in the video makes sense.

In my head I've wondered about what would happen if the inductor already had some energy stored in it as well. In particular, (especially since there's no resistor in the circuit) what would happen if we had energy stored in the inductor pushing current clockwise, and the energy stored in the capacitor is trying to push that same current counterclockwise?
At that point are we allowed to use superposition to figure out which one would "win"?
• If the inductor starts with some stored energy, that means there is a current flowing in it. At in the video is where I sketch the starting current. In the video, i(t<0) = 0. If the current is something other than 0, this is where you would sketch in the non-zero value. The rest of the intuitive process goes on from there. You will get the same sine wave shape for current. It will just start at a different point on the sine curve.
• At , I thought the voltage of the capacitor decreases instead of the voltage of the inductor therefore di/dt should not be negative?
• After the switch closes, there's only one voltage (v) in the circuit, across both the capacitor and the inductor. When the capacitor voltage goes down, so does the inductor voltage, because they are connected to each other.

At the point in the video you are thinking about, , the capacitor voltage has dropped a little, to a smaller positive value. Looking over at the inductor, we see that if the voltage is smaller (but still positive), the slope of the current, di/dt, has a smaller (but still positive) value. That's why the current sketched in the plot leans over a little to a less positive slope.
• someone convince me!
not sure why v started from v not since the inductor will prevent a sudden leap voltage.
So I thought that the voltage would gradually become v not.
and I know I'm wrong.. somebody convince me somehow