Main content

## Electrical engineering

### Course: Electrical engineering > Unit 2

Lesson 4: Natural and forced response- Capacitor i-v equations
- A capacitor integrates current
- Capacitor i-v equation in action
- Inductor equations
- Inductor kickback (1 of 2)
- Inductor kickback (2 of 2)
- Inductor i-v equation in action
- RC natural response - intuition
- RC natural response - derivation
- RC natural response - example
- RC natural response
- RC step response - intuition
- RC step response setup (1 of 3)
- RC step response solve (2 of 3)
- RC step response example (3 of 3)
- RC step response
- RL natural response
- Sketching exponentials
- Sketching exponentials - examples
- LC natural response intuition 1
- LC natural response intuition 2
- LC natural response derivation 1
- LC natural response derivation 2
- LC natural response derivation 3
- LC natural response derivation 4
- LC natural response example
- LC natural response
- LC natural response - derivation
- RLC natural response - intuition
- RLC natural response - derivation
- RLC natural response - variations

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# RLC natural response - intuition

An intuitive description of the natural response of a resistor-inductor-capacitor (RLC) circuit. Written by Willy McAllister.

## Introduction

In this article we take an intuitive look at the natural response of a resistor-inductor-capacitor circuit $(\text{RLC)}$ . This is the last circuit we'll analyze with the full differential equation treatment, which we will do in two follow-up articles.

The $\text{RLC}$ circuit is representative of real life circuits we can actually build, since every real circuit has some finite resistance. This circuit has a rich and complex behavior that finds application in many areas of electrical engineering.

### What we're building to

To understand the $\text{RLC}$ natural response in an intuitive sense, we think about how charge moves around the circuit over time. If we place a starting charge on the capacitor and then close the switch, that charge will slosh back and forth from one plate of the capacitor to the other, passing through the inductor and resistor in both directions. Each cycle of oscillation will be a little less than the previous, because energy is lost when the moving charge heats up the resistor.

The $\text{RLC}$ electrical circuit has a mechanical analog: the swinging pendulum. This is a good way to envision what is happening in the circuit.

## Predict the natural response

For this discussion, assume the resistor value is relatively small, like a few ohms. This prediction is similar what we did for the LC natural response. This time we add a small resistor, which is more representative of real life circuits.

Let's say the capacitor has an initial voltage, ${V}_{0}$ , which means it is storing some charge, $q$ . Assume the charge was put there by some external circuit, not shown. Because the switch is open, there is no initial current in the inductor, and no current in the capacitor or resistor, either. So the charge is just sitting there on the capacitor doing nothing.

What is going to happen when the switch closes and we let the circuit do "whatever it wants"? That behavior is what we call the $q$ .

*natural response*. We are going to reason through this by tracking what happens to the charge,The amount of $q$ is set by the product of the initial voltage on the capacitor and the value of the capacitor, $q=\text{C}{\textstyle \phantom{\rule{0.167em}{0ex}}}{v}_{\text{C}}$ . Starting out, all the charge is sitting still on the capacitor. The total amount of charge, $q$ , is constant, it does not change during the natural response. (We can track where it is by observing the voltage on the capacitor.)

### "Put a charge on the capacitor"

When we say "put a charge on the capacitor," it means to put some amount of $+q$ on the top plate and the exact same amount of $-q$ on the bottom plate, creating a charge separation. In the long run, at the end of the natural response, all of that separated charge will have flowed around and found an opposite sign charge to hang out with, becoming neutral. The charge doesn't vanish, but the charge

*separation*goes away.As we craft our prediction, we track the $+q$ , and know that the same amount of $-q$ is moving in the opposite direction. Try to "see" the motion of the charge in your mind as we go through this discussion.

### Close the switch

Now we close the switch and let the $\text{RLC}$ circuit do its "natural" thing.

The inductor starts with $0$ current and $0$ volts. The resistor also has $0$ current, so by Ohm's Law, there is $0$ volts across the resistor.

The closed switch all of a sudden provides a closed path for the $+$ charge on the top plate to search out the $-$ charge on the bottom plate, (and vice versa, not shown).

All of a sudden the inductor and resistor together "see" the capacitor voltage, ${v}_{\text{C}}={\text{V}}_{0}$ . This voltage will create a current in the inductor and resistor. Where does that current come from? It comes from the charge on the capacitor, of course. The charge is pulled by the electric force of attraction towards the opposite charge on the other plate.

The resistor now has a current flowing through it, and Ohm's Law tells us there will be a voltage drop across $\text{R}$ . We assumed $\text{R}$ was small, so the voltage drop will be small, too. Nevertheless, the resistor does get a little warm as it dissipates a little bit of power.

The inductor has a current, so it starts storing energy in its surrounding magnetic field. That stored energy is going to come back out of the magnetic field in a moment. (The voltage across the inductor is a little bit lower than ${v}_{\text{C}}$ due to the small voltage drop across the resistor.)

Over at the capacitor, current flows out from the top plate, goes through the resistor, through the inductor, and around to the bottom capacitor plate. If $q$ is going down, then $q=\text{C}{\textstyle \phantom{\rule{0.167em}{0ex}}}v$ tells us ${v}_{\text{C}}$ has to be going down, too.

Eventually, we come to a state where the amount of 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 at or near $0$ . The energy stored in the inductor's magnetic field tends to keep the current flowing. (Current does not abruptly drop to $0$ when the inductor voltage reaches $0$ . Inductors "don't let" sharp changes in current happen.)

Even after the voltage falls to $0$ , 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. The voltage will be negative, and a little bit less than the original ${v}_{\text{C}}(0)$ where the capacitor started. Remember the resistor? It is draining energy from the circuit, so the peak negative voltage isn't quite as high as the starting point. Charge stops moving for a brief moment when the voltage peaks out, so the current falls to $0$ .

The previous image is nearly the same as where we started. The current is back to zero, and the voltage is at a (slightly lower) peak value. 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:

At the end of one cycle we are back where we started, but with some energy removed from the system. The charge will continue to slosh back and forth between the top and bottom capacitor plates, losing a little energy each time, until the system eventually comes to rest.

## Mechanical analog

The $\text{LC}$ circuit is analogous to a mechanical oscillator, the frictionless swinging pendulum. The $\text{RLC}$ circuit has a similar mechanical analog, The addition of the resistor to the $\text{RLC}$ is equivalent to adding air resistance to make the pendulum dissipate energy and slow to a halt.

As a pendulum swings back and forth, friction due to air resistance dissipates energy, and each swing gets shorter and shorter until the pendulum finally stops moving. If air resistance is low, the pendulum swings for a long time before it stops. If it is very high, the pendulum makes just one slow drop to the bottom center and stops. At one precise value, the pendulum will fall to the bottom center as fast as it can, without overshooting and coming back.

Our $\text{RLC}$ circuit will display the same kinds of behavior as its current and voltage swing back and forth. (Another good mechanical analog is a weight hanging from a spring. If you pull the weight down and let it go, its up-and-down motion is similar to the pendulum's back-and-forth.)

### Closing thought

Do you remember we assumed the resistor was relatively small? A small resistance allows the system to swing back and forth for a while. What do you think will happen if the resistor is larger? (Hint: how long would a pendulum swing if there was more friction in the bearing?)

In the next two articles we will discover precisely how the $\text{RLC}$ works when we do a formal derivation of the natural response. We'll be able to predict the oscillation frequency, and will see how quickly the signal fades away.

## Summary

We followed the charge moving around in an $\text{RLC}$ circuit over time. We started with a charge on the capacitor and closed the switch. The charge flowed back and forth from one plate of the capacitor to the other, passing through the inductor and resistor in both directions.

As the current passes through the inductor, it stores energy in the magnetic field surrounding the inductor. That energy returns to the circuit by pushing charge along.

Each cycle of oscillation is a little lower than the previous, because of the energy lost as the moving charge heats up the resistor.

The swinging pendulum is a mechanical analog of the $\text{RLC}$ electrical circuit. It helps you envision what is happening in the circuit.

## Want to join the conversation?

- I'm tempted to presume that if the circuit was made up of real-world components, some energy would be lost as heat in the inductor as well, as the energy alternates between being stored in the magnetic field and being used to direct charge; is this correct? And if so, how efficient is the change between those states with real-world components (i.e. what percentage of energy is lost during a given transition)?(4 votes)
- Hello Doctor,

You are correct there will be losses in a real inductor. There are three general reasons for this:

* The wire has resistance. There will be an I^2R loss causing the wire to heat up.

* It takes energy to flip the magnetic domains. This leads to heating of the core.

* There will be current flowing where it should not be. Please research eddy currents. Again this leads to heating of the core.

As a general statement a device such as a transformer (two coupled inductors) is in the neighborhood of 95% efficient.

Regards,

APD(14 votes)

- Regarding the mechanical analogue,

If we have: x~q; v~i; a~(di)/(dt); k~(1/C); m~L; ~R;

where force due to air resistance(damping)~v

(the ~ sign has been used to indicate analogousness)

then the math works out the same for both cases.

But if the damping agent is something like friction, its magnitude would be constant(unlike air resistance, which would be proportional to v). So I don't understand how both are same(in terms of the math involved).(2 votes)- Thanks for pointing this out. Friction is a complex phenomenon, so this analogy is something i should rewrite it in terms of air resistance, as you suggest.(5 votes)

- When the switch closes, current starts to increase and the capacitor voltage starts to decrease. An increasing current means, by Ohm's law, an increasing voltage across the resistor. Sign convention for passive components says, the right side of the resistor will be at higher voltage than the left side, since current is entering it from the right side.

So, when the switch closes, the capacitor voltage starts to decrease, and the resistor voltage starts to increase.

Hence, there will come a time when both the voltages will meet at the same point (i.e. they will be equal and opposite). At that point, KVL says, the voltage across the inductor is zero (vL = vC - vR). This implies current in the inductor, at that point, has slope zero, using the inductor i-v equation. This means the current reaches its peak value (i.e. starts to bend over) way before the capacitor voltage reaches zero.

When we had just LC circuit, the reason for current to "bend over" was the capacitor voltage getting reversed. But in RLC, the reason is the total (or effective) voltage of the resistor and the capacitor getting reversed.

So the current starts to slow down way before vC = 0.

If we make the analogy now, it means that the pendulum starts to slow down (or decelerate) way before it reaches the bottom. How does that make sense?(2 votes)- Compare a frictionless pendulum (LC) with one that has some friction (RLC).

Both of them start with the weight pulled over to one side (Vo).

Let the frictionless pendulum go and watch it swing once over to the other side. Remember what the velocity was at every point.

Now do the same for the pendulum with friction, and compare the velocity profiles. At every point in the swing, the pendulum with friction will be moving slower than the frictionless pendulum. You called this deceleration, but I think it is just moving slower than the frictionless pendulum.

The same thing happens in the RLC circuit where current is analogous to velocity. With resistance (friction) the current is always lower than the frictionless LC circuit.

Take a look at the graphs of i and v near the end of the derivation article (https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-natural-and-forced-response/a/ee-rlc-natural-response-derivation). Compare the timing of the peaks and zero-crossings of i and v. You will see they don't necessarily line up with the RLC circuit. Does the timing of the peaks match your intuition about when they should occur?

p.s. I did a substantial rewrite of these RLC articles, posted here:

http://spinningnumbers.org/a/rlc-natural-response-intuition.html

http://spinningnumbers.org/a/rlc-natural-response-derivation.html

http://spinningnumbers.org/a/rlc-natural-response-variations.html(1 vote)