LC natural response example

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 the square root of capacitance over inductance times the starting voltage ( v_0 ) times sine ( \omega_0 t ).

And ( \omega_0 ) is the natural frequency, and we said that ( \omega_0 ) was equal to the square root of ( \frac{1}{L \times C} ), and ( v_0 ) is the starting voltage on the capacitor. Our original assumption was that ( i ) was ( 0 ) to begin with, and that's the expression for the current.

So, what I want to do now is a specific example, and we'll do that down here. First, I'm going to install a switch into our circuit so that we can add some energy, and it won't go anywhere.

Then, for ( C ), ( C ) is going to be equal to a quarter of a farad, and ( L ) is going to be equal to one henry. All right, and on the capacitor, I'm going to put enough charge to bring this up to ( 10 ) volts before we close the switch.

Then, at ( t = 0 ), we'll close the switch, and what we want to find is, what is ( i(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_0 ). ( \omega_0 = \frac{1}{\sqrt{L \times C}} ), so that equals ( \sqrt{\frac{1}{1 \ \text{henry} \times \frac{1}{4} \ \text{farad}}} ), which equals ( \sqrt{4} ) or ( 2 ).

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 = \sqrt{C} ) which is ( \frac{1}{4} \ \text{farad} ) divided by ( 1 \ \text{henry} ) times ( v_0 ). ( v_0 ) was ( 10 ) volts times sine ( \omega_0 t ). Sine ( \omega_0 ) is ( 2t ).

And finally, ( i = \sqrt{\frac{1}{4}} ) or ( \frac{1}{2} ), so that's ( \frac{1}{2} ) times ( 10 ) is ( 5 ).

Thus, ( i = 5 \sin(2t) ). So for this specific circuit, that's the answer for this current here.

Now, I want to show you what that actually looks like. So this is a plot of ( i(t) = 5 \sin(2t) ) and this is what a sine wave looks like with time.

So, the axes are time in seconds, and this axis goes up to ( 5 ) amperes and then down to ( -5 ) amperes and continues on that way, and it basically goes on forever.

What I want to 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 want to find ( v ). What's ( v )? That's what we're looking for.

So, we know that ( i = 5 \sin(2t) ), and to find ( v ), one of the easy ways to find ( v ) is to use the inductor equation.

We know that ( v = L \frac{di}{dt} ); that's just the basic inductor ( i , v ) equation, right? Let's see what happens here.

( v = L ) is ( 1 ) times ( \frac{di}{dt} ), so that's ( \frac{d}{dt}(5 \sin(2t)) ). Now, let's take that derivative.

Okay, I'll go up here. ( v = 5 ) comes out of the derivative. The derivative of ( \sin(2t) ) is ( 2 \cos(2t) ).

All right, so our voltage solution is ( 10 \cos(2t) ). 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: this is ( v(t) ) and that we decided was equal to ( 10 \cos(2t) ). It starts at a value of ( 10 ) at ( t = 0 ) and then forms a cosine wave. It 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(t) ) in blue and ( v(t) ) in orange. One of them is a sine wave; the current is a sine wave, and the 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!