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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# LC natural response derivation 1

We begin the derivation of the natural response of the LC circuit, by modeling it with a 2nd-order differential equation. Created by Willy McAllister.

## Want to join the conversation?

- Is it correct to say that in this LC circuit, the capacitor is also the voltage source? Would it be different if we have a separate voltage source instead?(2 votes)
- Hello Emmen,

Both the inductor and the capacitor will appear to be the voltage source. However, they never do so at the same time. Think of this as an energy storage problem. The energy constantly sloshes back and forth between the two devices.

In radio terms we call this a tank circuit as in water sloshing back and forth in a tank. The rate of energy exchange is known as the resonant frequency. The LC circuit is often found in radio frequency system as part of the station tuning mechanism.

Things get really interesting when you add separate voltage sources to this LC circuit. In general the LC circuit loves voltage sources that operate at the natural frequency. It will try to short out anything else.

Regards,

APD(4 votes)

- Hello ,

I have a problem understanding why the voltage on the capacitor and the inductor is the same , at least it shouldn't be the same just as you close the switch .

Any help ?(2 votes)- Hello Hazem,

Before time zero the capacitor was charged to some voltage and the inductor is "empty" which is to say, no current is flowing in the inductor. At time zero the switch is closed. Instantaneously the voltage on both devices is equal. The voltage then oscillates as Willy describes in the video.

Recall that the voltage on an inductor is defined as:

V = Ldi/dt

This allows an instantaneous voltage jump when the switch is closed as the change in current is instantaneous (square wave).

Or stated another way, the inductor likes to keep current constant. Before the switch was closed the current was zero. Immediately after the switch is closed the inductor current is still zero. The only way this can happen is if the inductor's voltage instantaneously rises to the capacitor voltage.

Please leave a comment below if you have any questions.

Regards,

APD(2 votes)

- At2:40Why is the current direction "reversed"? What is the reason for the minus sign in the equation?(2 votes)
- The negative sign is a small accident of how i is defined in the main LC circuit diagram on the left.

At1:58I draw a capacitor over on the far right. The i-v equation for a capacitor is defined assuming the current, i_c, is coming down into the positive voltage terminal of the cap. With this current direction the capacitor equation is i_c = C dv/dt.

Now look at the direction of i_withnosubscript in the LC circuit on the left side of the screen. Notice it points in the opposite direction of i_c. It is flowing up out of the positive terminal of the cap. That means -i = i_c.

In the main drawing on the left there is no variable called i_c, there is just i. To write the capacitor i-v relationship correctly using i as the current variable you substitute in -i in place of i_c. This is how the negative sign gets introduced.(3 votes)

- At5:40, 2nd derivative for the numerator is d^2i but in the denominator it is dt^2? Why is that?(1 vote)
- Good question, I never thought about that before. I don't know why the notation for second derivative is like that, but that's the convention. All the 2's are just part of the notation. They are not really math operators. You don't square anything when taking a second derivative.(2 votes)

- In the previous video (intuition #2), the current was flowing into the same direction, and yet the equation for the capacitor did not need the minus sign. What is the reason for that?(1 vote)
- The missing minus sign is an error I made in LC natural response intuition 1. When constructing the differential equation in this derivation video the minus sign is required to get the math correct. In the intuition video I was a little too quick. Sorry for the confusion.(1 vote)

## Video transcript

- [Voiceover] In this
video, we're gonna begin the derivation of the LC Natural Response, the response of an
inductor capacitor circuit. This is a difficult derivation, but it really pays off in the end. There's a really fun surprise at the end, and that is, this is
where sine waves are born. We're gonna end up with sine
waves at the end of this. And that's a really nice
result because these are everywhere in electronics,
and in the natural world. We're gonna start out by saying, we're gonna put some charge
on this capacitor here. And we're gonna have the
current in this circuit, i, we're gonna have that start at zero. So i zero is zero, but this is the current through the circuit, and
that means that current exists over here as well. Let's put a switch in this circuit. And we're gonna close that
switch at t equals zero. So we have one variable as
i, and the other variable we're gonna use is v, and
v is this voltage here, after the switch closes. So we wanna find i and v. And for this we're gonna
focus on finding i. Once we find i it's
straightforward to find e. So this is gonna be our
independent variable, is the current. So let's begin the analysis. At the moment, t equals zero,
the charge that exists here is gonna start flowing in the circuit, and both voltage and current
are gonna start to change. So let's write some expressions, let's start out by writing
some things that we know are true about our two components. And we'll start with the capacitor. So we'll start by writing an
expression for our capacitor, and there's a little sign
flip that happens here, so we gotta be a little bit careful. So if I have v on a capacitor,
and i in a capacitor, I would say that i
capacitor equals c, dv, dt. Now if we look here, vc
is the same as v here, positive sign at the top,
positive sign at the top, negative sign at the bottom,
so vc and v are the same. But i, we have to be careful,
i is in the opposite direction of the current that I picked
for independent variable, so that's upside down. So i, there's a negative
sign we get to flip here, equals negative c, dv, dt. So that's the IV equation
for our particular circuit. We have that little extra negative sign. And now I wanna write the
integral form of the capacitor IV equation, which is v equals one over c times the integral of idt. And remember our minus sign. This is the important
minus sign to bring along. Okay so let's do the
voltage across the inductor, the inductor voltage is same
variable v, is L, di, dt. So there's no extra minus sign here. This is assigned according
to the sign convention for passive components. So now we have an expression
for v on the capacitor, and we have an expression
for v on the inductor, and we know that that's the same voltage. So let's set these two things together, so we can say that L, di, dt, equals minus one over c, times
the integral of idt. All we did was set the
two identical voltages equal to each other. And now I'm gonna manipulate this a bit, it says that L, di, dt, plus one over c, the integral of idt equals zero. We just gathered terms on one side, and we said the other side became zero. Now since I'm not used to having integrals inside equations, I'd
rather have derivatives, since I have some experience
with differential equations, let's, if we take a derivative
of this whole equation, so here's what we're gonna do. We're gonna take a derivative
with respect to time, of this whole thing here, and that gives us, we
get the second derivative of the first term, so we get L, second derivative of i
with respect to time, plus one over c, and then, the derivative of the integral of idt, turns into just i, and
the derivative of zero, on this side, is zero. So this is now the differential
equation for the LC circuit. And it's called, it has a name, it's called a second order homogeneous ordinary differential equation. It's a differential
equation because it has derivatives in it, it's
homogeneous because it only has derivatives of i with
respect to t and nothing else. The indicator is that this
side is equal to zero, there's no forcing term
over here on this side. So when you can write
the equation this way, we say it's homogeneous. And it's called a second order equation, because it has this second
derivative right here. So now we've set up our
second order homogeneous ODE, and in the next video,
we'll go about solving this. We're gonna go through it step by step.