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### 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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# LC natural response example

We solve the voltage and current of an example LC circuit with given values for L and C, and an initial charge on the capacitor. Created by Willy McAllister.

## Want to join the conversation?

- do you make the var names up or r they allways used(4 votes)
- The variable names I used in this video are fairly common choices for this problem. There are no strict rules for naming, but you should always try to pick simple variable names that give your reader the best chance of following along with your thoughts.(7 votes)

- Are there other waveforms in RC and L/R circuits other than sine waves , square waves , and sawtooth waveforms? Also, what are the uses of the waveforms . And would you elaborate further on 1 RC time (63%) and 1 L/R time (63%)? thankyou. OBEARCC(3 votes)
- To add to what APD said, you should look into fourier series, as square waves are often constructed as the summation of different waves and you will soon find that any wave is simply the addition and subtractions of fundamental waves!(3 votes)

- At4:04, why did you skip talking about the frequency of the natural response?(3 votes)
- Good point. I mentioned the amplitude, but not the frequency. Wish I had.(2 votes)

- Can we obtain the same sine waves using power sources instead of LC components arranging some electronic switching at desired frequency?(2 votes)
- Yes. A common way to generate a sine wave is to use a digital circuit to create a 0101010... pattern (a square wave) followed by filtering circuits to remove all but the main fundamental frequency. One of the main applications of L and C components is to build those filters. We study the LC circuit because it is the simplest possible way to generate a sine, and other ways of doing it can be viewed as an alternative implementation of a basic LC.(3 votes)

- I understand most of the videos but would like to do more exercises. Where should I look at? Do you have recommendation for textbooks? Thank you!(2 votes)
- One of the textbooks I used for reference is Foundations of Analog and Digital Electronic Circuits, Agarwal and Lang. Chapter 13 is all about Sinusoidal Analysis and impedance. You might be able to find this book online.(2 votes)

- Is there like a review lesson that would explain some of the abbreviations?(1 vote)
- The three basic components R L C are introduced in this video: https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/circuit-elements/v/ideal-circuit-elements

and in this article: https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/circuit-elements/a/ee-ideal-circuit-elements

The unit of inductance is the henry, (H). The unit of capacitance is the farad, (F).(1 vote)

- I made a graph of the current equation so we can visualize how the values of C and L affect the current over time.

https://www.desmos.com/calculator/cnoooncemh(1 vote)- Cool! It's interesting to see how the magnitude of the sine wave also changes when you change L and C.(1 vote)

- What is the square root of (C/L ) equal to ?(1 vote)
- See the video at3:00.

sqrt(C/L) = sqrt (0.25 F / 1 H) = 1/2

If you are asking about the units of sqrt(C/L) you can figure them out from the equation for i.

i = sqrt(C/L) V sin wt

The sin wt term has no units. V has units of volts, and i has units of amperes.

That means sqrt(C/L) has to be able to convert volts to amperes.

Ohm's Law tells us i = V/R.

That means sqrt(C/L) has units of 1/Ohms.

That might look kind of strange, but it's true.(1 vote)

## Video transcript

- [Voiceover] So in previous videos, we worked out an expression for the current i in an
LC circuit like this, and what we found was that i is square root of
capacitance over inductance, times the starting voltage, V knot, times sine omega knot t. And omega knot is the natural frequency, and we say that omega knot was equal to square root
of one over L times C. And V knot is the starting
voltage on the capacitor. And our original assumption was that i was zero to begin with, and that's the expression for the current. So what I wanna do now
is a specific example, and we'll do that down here. First I'm gonna install a switch into our circuit so that
we can add some energy and it won't go anywhere. Then for C, C is gonna be
equal to a quarter of a farad, and L is gonna be equal to one henry. All right? And on the capacitor, I'm
gonna put enough charge to bring this up to 10 volts
before we close the switch. And then at time equals
zero, we'll close the switch. And what we wanna find is, what is i of t? We have L, we have C, we
have a starting voltage, and so now we have everything we need to work out the current. Let's do that, okay? So first off we'll do omega knot. Omega knot equals one over
the square root of LC, so that equals square root of one over one
henry times one quarter farad, which equals square root of four, or equals two, and that's in units of radians per second. That's the natural frequency. We know the natural frequency right here, and now we can work out the rest of it. So we can just fill in i
equals square root of C, which is one quarter farad, divided by one henry, times V knot. V knot was was 10 volts times sine omega knot t. Sine, omega knot is two t. And finally, i equals, this is the square root of
one quarter or one half, so that's one half, times 10 is five, i equals five sine 2 t. So for this specific circuit, that's the answer for this current here. Now I wanna show you what
that actually looks like. So this is a plot of i of
t equals five sine two t. And this is what a sine
wave looks like with time, so the axes are time in seconds, and this axis goes up to five amperes and then down to minus five amperes and continues on that way, and
it basically goes on forever. Now what I wanna do next is work out the voltage in the circuit. We didn't talk about the voltage yet. So I'll sketch the circuit again here. Here's L and C. And this is the voltage right here, voltage across both guys. We've already worked out i. And now we wanna find v. What's v? That's what we're looking for. So, we know that i equals five sine two t, and to find v, one of
the easy ways to find v is to use the inductor equation. We know that v equals L di/dt. That's just the basic
inductor i v equation, right? And let's see what happens here. v equals L is one, times di/dt. So that's d/dt of 5 sine two t. Now let's take that derivative, okay? I'll go up here. v equals five comes out of the derivative. Sine two t. The derivative of sine of two t is two times cosine of two t. All right? So our voltage solution
is 10 times cosine two t. So something interesting
just happened here. Let me show you. We started with the current
being a sine function, and we eventually took the
derivative of that sine function and now we have a cosine function. So we went from sine
to cosine for voltage. That means the voltage doesn't look quite exactly like the current, okay? Let me show you a plot of the voltage. Here's the voltage here. This is v of t. And that we decided was
equal to 10 cosine two t. Starts at a value of 10,
the time equals zero, and then forms a cosine
wave that goes up and down between plus and minus 10 volts. Now let me show you what it looks like when we plot both i and
v on the same graph, and we can see the timing
relationship between them. So this is a plot of i of t in blue, and v of t in orange. One them is a sine wave,
the current is a sine wave, and voltage is a cosine wave. So this is what we were going for. This is the natural
response of an LC circuit, and we had two specific
component values in it. We saw that we came out with
both current and voltage looking like sinusoidal waves. This LC circuit, this is where sine waves
come from in electronics. It's pretty cool.