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 4

In this final step of the derivation, we find two initial conditions and use them to come up with a sinusoidal solution for the LC natural response. Created by Willy McAllister.

## Want to join the conversation?

- Is this another case of solving a problem in the set of real numbers through complex numbers or the complex solution is related to another dimension of the phenomenon?

I'm thinking, for instance, about electromagnetic waves with a couple of orthogonal fields, electric and magnetic. In this case the real solution and the imaginary solution could describe this situation. In our LC model we have an equilibrium between a magnetic field, (inductor), and an electric field (capacitor).

Nature is amazing, but mathematics is astonishing!(2 votes)- I lean towards your first explanation. Imaginary numbers appear in the middle of a journey to a real solution. Once we have the abstract i-v equations of L and C with their derivative terms, the physical phenomena that gave rise to those equations is left behind. The appearance of the imaginary is self-contained within the math.(5 votes)

- so, we've solved out the natural response of LC circuit's current at t=0, am I correct?

but if we found out what current equals to the initial condition (where t=0) then don't we only know the current at t=0?? I'm confused...(2 votes)- You are talking about the initial conditions I believe. These initial conditions can change, but the general solution would still stay the same and work out. Go ahead and try it!(2 votes)

- Sorry but4:20would'nt that be '-A^2*wcoswt'?(1 vote)
- A_2 is a constant. It comes out in front of the derivative of sin(wt), (without being squared).(1 vote)

## Video transcript

- [Voiceover] So now we're
gonna use the initial conditions to figure out our values, our
two constant values A1 and A2. That is in our proposed
solution for current for the LC circuit. So one thing we need to do, because this is a second order equation. We need to have two initial
conditions for the variable that we're studying here. So we're studying I. Right now we have one
initial condition for I, and because we have a
second order equation. That means we need two ic's for I. So we have one initial
condition right here, and what we'd like to
know is what is di dt at time equal zero? So the other piece of
information we have is this v. V knot at time equal zero. So let's use that, and we'll just plug that straight into
the inductor equation. so the inductor equation at t equals zero. The voltage across the inductor is v knot, and that equals L times di dt. Alright and that means the
di dt equals v knot over L. So now I have two initial conditions in terms of I. There's one, and there's one there. And we can use these now
to go after A1 and A2. First off let's plug in
I for time equals zero, and then see if we can work
our something over here. So that means that time equals zero. The current is zero and that
equals A1 cosign omega knot times, t is zero, times zero plus A2 times sine of omega knot times zero, and what does this evaluate to? Okay this is sine of zero
and sine of zero is zero, and cosine of zero is one. So that comes up with zero equals A1. Okay, and A1 equals zero
means that this entire term of our solution just dropped out. Alright let me write what we end up with. I equals A2 sine omega knot t. This whole term here just
dropped out of the solution. So here's our proposed solution down here. Now we need to go after A2, let's do that. As you might suspect, we're gonna use our second initial condition to do that. So to use our initial
condition we need di dt. So let's take di dt of this. So we're gonna take d dt
of this whole equation, and on the left side we'll get di dt, and on the other side we'll get d dt of A2 sine omega knot t. Okay so far so good. Let's roll it down again. So let's take that derivative. We get di dt equals, A2
comes out of the derivative, and the derivative of sine
omega knot t with respect to t is omega knot times cosine omega knot t, and we apply our initial condition. Let's go to t equals zero, and we know the di dt was v knot over L equals A2 omega knot cosine
of omega knot times zero, and cosine of zero goes to one, and so we can solve for A2. A2 equals V knot over L, and omega knot goes down here. So now we solve for our
second adjustable parameter and we can write I. I was A2 sine omega knot t. So let's fill it in for A2. I equals A2 is v knot over L omega knot times sine omega knot t, and I wanna go back now. I wanna write this a little bit different. I wanna go back and plug in
our value for omega knot. So if we remember, we said
omega knot equals one over LC and the square root of that whole thing. So now L omega knot equals square root of one over LC times L, and that equals square
root of L squared over LC, and that equals square root of L over C. Lastly I'll write one over L omega knot equals square root of C
over L, just the reciprocal, and now we can write I equals square root of C over L times V knot sine of omega knot t, and that is the solution
for the natural response of an LC circuit. It's in the form of a sine wave and the frequency is
determined by omega knot. Which is the two component values, and the amplitude is
determined by the energy we started with. Which is represented here by V knot, and the ratio of the two components again. So this is why I said at the beginning that this is where sine waves are born.