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# LC natural response

Intuitive description of the natural response of an inductor-capacitor circuit. Written by Willy McAllister.
We develop an intuition for the natural response of the inductor-capacitor, $\text{LC}$, circuit.
After we get a good mental image of what's going on, the next article is a formal derivation of the $\underset{―}{\text{LC}}$ natural response.

### What we're building to

Circuits with two energy storage elements (capacitors or inductors) are called second-order systems. In second-order systems, the voltages and currents rock back-and-forth, or oscillate. This article is an intuitive description of how this happens.
Second-order systems are the source of sine waves in electronic circuits.

## First-order systems

Up to now we've looked at first-order circuits, $\underset{―}{\text{RC}}$ and $\underset{―}{\text{RL}}$, that have one energy-storage element, $\text{C}$ or $\text{L}$. The natural response of first-order circuits has an exponential shape that "slumps" to its final value. The energy in its storage element is dissipated by the resistor.

## Second-order systems

Now we look at a circuit with two energy-storage elements and no resistor. Circuits with two storage elements are second-order systems, because they produce equations with second derivatives.
Second-order systems are the first systems that rock back and forth in time, or oscillate. The classic example of a mechanical second-order system is a clock with a pendulum. In electronics, the classic system is the $\text{LC}$ circuit.
We want to find the natural response of this circuit. This is what a circuit does when there is no external driving force. Natural response is always an important part of the total response of a circuit.

## Predict the natural response

Let's say the capacitor has an initial voltage, which means it is storing some charge, $q$. We assume there is no initial current in the inductor (and therefore, no current in the capacitor, either). What is going to happen when the switch closes and we let this circuit do "whatever it wants"? We are going to reason through this by tracking what happens to the charge, $q$.
The amount of $q$ is set by the product of the initial voltage on the capacitor and the value of the capacitor, $q=\text{C}\phantom{\rule{0.167em}{0ex}}v$. $q$ does not change during the natural response. Starting out, all the charge is sitting still on the capacitor.
Now we release the circuit by closing the switch to let it do its "natural" thing.
The inductor starts with $0$ current. All of a sudden it "sees" the initial voltage, $v={\text{V}}_{0}$. This voltage will generate a rising current in the inductor, and it starts storing energy in its surrounding magnetic field.
Where does that current (flowing charge) come from? It comes from the capacitor, of course.
Over at the capacitor, current flows out from the top plate, goes through the inductor and goes around to the bottom capacitor plate. If $q$ is going down, then $q=\text{C}\phantom{\rule{0.167em}{0ex}}v$ tells us $v$ has to be going down, too.
Eventually, we reach a state where the charge on the top plate is the same as the bottom plate. The voltage across the capacitor therefore falls to $0$.
Over at the inductor there is a current flowing, even though the voltage is $0$, because the energy stored in the inductor's magnetic field keeps the current flowing. (Current does not abruptly drop to $0$ when the voltage reaches $0$.)
The inductor current continues to move charge from the top plate of the capacitor to the bottom. Now there is more positive charge on the bottom plate than the top, so the voltage actually reverses sign and becomes negative.
As charge builds up on the bottom plate, it repels against the arrival of new charge from the inductor current (electrostatic repulsion). The inductor current bends over and starts to drop back towards $0$.
After a little while, the voltage will reach a peak negative value when all the charge has flowed to the bottom plate. The voltage will be the negative of whatever the capacitor started at. Charge stops moving for a brief moment, so the current crosses $0$.
The previous image is almost identical to where we started. The current is back to zero, and the voltage is at a peak value. The peak happens to be the negative of where we started. We can go back to the start of the story and tell it again just the same, except with charge moving from the bottom plate of the capacitor back to the top. Here's the end result after one full cycle:
The rate of oscillation (the frequency), is determined by the value of $\text{L}$ and $\text{C}$. We will discover how that works when we do the formal derivation of the $\text{LC}$ natural response in the next article.

## Mechanical analogy: pendulum

A swinging pendulum is a mechanical analog for an $\text{LC}$ circuit.
Voltage $v\left(t\right)$ is the analog of the position. We measure position of the pendulum as it moves away from the center point. The distance is $0$, $\left(v=0\right)$, when the pendulum is hanging straight down, and goes to $v=+{\text{V}}_{0}$ or $-{\text{V}}_{0}$ at either extreme position.
Current $i\left(t\right)$ is the analog of velocity. The pendulum moves its fastest at the mid-point $\left(i={\text{I}}_{max}\right)$. It is motionless, $\left(i=0\right)$, for an instant at either end of its swing.
The initial voltage of $+{\text{V}}_{0}$ corresponds to how much we pull the pendulum to the right before letting go.
Letting go of the pendulum corresponds to closing the switch. What happens next is the natural response. If the pivot point is frictionless and there is no air resistance, the pendulum swings forever.
The $\text{LC}$ circuit (and the pendulum) trade voltage and current back and forth in a sine wave pattern. Both voltage and current are sine waves, and we can see a timing difference of $1/4$ of a cycle between them.

## Summary

We explored an intuitive description of the natural response of an $\text{LC}$ circuit (a second-order system). Both the voltage and current have a sine wave pattern with time.

## Want to join the conversation?

• In the predict the natural response section it says "(Current does not abruptly drop to 000 when the voltage reaches 000.)"
How is this true since Ohm's law states I=V/R?

Also, do the electrons on one side of the capacitor simply run through the inductor to the other capacitor plate? and then back again? So the inductor just acts as a spring that pulls the elctrons back and forth from one plate to another?

I hope this makes sense, if there is a video on inductors please point me in the right direction I'm having a hard time understanding.

Thanks
• Hi KCThomas,

Notice that we are transferring energy between the inductor and the capacitor. There is no resistor in this circuit! Consequently, Ohm's law does not apply. In fact, as described, the energy will slosh back and forth between the components forever...

Personally I had a hard time understanding inductors. After all, an inductor is just a coil of wire. Let me tell you a story. One day I was measuring the current flow though an inductor. I had a 12 VDC battery, an ammeter, and large inductor. To save time I held the wires in my hand. The test went well and I got a good current measurement of 0.5 A.

--Did I mention I was hold both end of the inductor one with my right and one with my left hand? ---

Anyway, all was well until I went to disconnect the circuit. That was the day I learned that an inductor is not a simple piece of wire. I assure you it was a painful lesson!

An inductor stores energy in a magnetic field. The amount of energy is related to the construction of the inductor and the amount of current flowing in the device. This energy is released when the current flow is reduced. Back to my story. There was energy stored in the inductor. And I had foolishly placed my hands across the device. When I disconnected the circuit I released the stored energy. It manifested itself as a high voltage shock that brought a tear to my eye.

Inductor and capacitors both store energy but they do is in different ways. One magnetically and the other electrostatically. When connected in the circuit as described in this article the energy constantly transfers back and forth between the two devices.

A mechanical analogy that may help is the mass and spring. Both store energy, one as kinetic and the other as tension. Please take moment to consider how this analogy relates back the capacitor and inductor.

BTW, if we add a damper to our mechanical system we will have the equivalent of a resistor.

Regards,

APD
• Is the pendulum system only an analogy or both system are equivalent? I mean, in the pendulum system we have a relationship between kinetic and potential energy in a conservative field, ( we simplify by doing air resistance, torsion of rope and so on equal to zero).

Is the magnetic field some kind of potential energy of the electrical system?

My ask have his source in this problem : I can't to reach an intuition about electromagnetic fields... Gravitational fields, instead, are so easy to "imagine"!

Thank you very much for this course!
(1 vote)
• The pendulum system is a mechanical analog of the LC circuit. Voltage on the electric side corresponds to position away from 0 on the mechanical side. Current corresponds to the velocity of the pendulum. The two systems can't be "equivalent" because they are made of totally different things. However, we can use the same mathematics to solve both systems to derive sinusoidal motion. That's what makes this such a powerful and useful analogy, (more interesting than "just an analogy").

If you dig deeper, you can find other relationships in this analog. The magnetic field in the inductor causes the current to keep flowing even after the voltage goes to zero. The inductance (L) is analogous to the inertia of the mass at the end of the pendulum.

This is a great analogy because you can visualize something you've seen before (a swinging pendulum) to help you understand something you are just learning about (LC).