AP Physics C: Electricity & Magnetism LC Circuits

Hello, everyone, and welcome back to www.educator.com.0000

I'm Dan Fullerton, and in this lesson, we are going to talk about LC circuits.0003

Our objectives include applying Faraday’s law to a simple LC series circuit0008

to obtain a differential equation for charge as a function of time.0013

To solve the differential equation.0016

To calculate the current in capacitor voltage as a function of time.0019

Finally, sketching graphs of the current through the voltage across the capacitor for the simple LC series circuits.0022

Let us take a look at our LC circuit.0029

We have an inductor and capacitor and typically when we analyze these,0032

what we are going to assume is when the circuit is first turned on, our capacitor is charged and our inductor is not.0036

The interplay of the capacitor and the inductor is going to create an oscillating system very similar in fashion to simple harmonic motion,0043

basically, looking at making the electrical version of a pendulum.0051

Let us see if we can analyze the circuit a little bit.0056

As we look at charge in here, let us first define in our circuit the direction for current flow, let us call that our current.0061

We have our potential across our capacitor DC ± recognizing that the electric field in here0067

is going to go from the positive side to the negative side.0074

Over here on our inductor, we have the potential across it which is L DI DT, that is our positive and our negative side.0079

Again, it is very important to note that the electric field in here is 0, the electric field inside the conductor.0089

With that, we can use Faraday's law to start to analyze our circuit.0097

The integral / the closed loop of E dot DL = - the derivative of the magnetic flux with respect to time, which as we look here is just going to be - L DI DT.0103

Since, we know that the voltage across our capacitor VC is Q/C, we can rewrite this as –Q/ C as we go this way around our circuit,0125

no electric fields and no contribution to the left hand side, over from our inductor is going to be equal to our - L DI DT.0137

Or rearranging this a little bit, we could say that Q/ C - L DI DT = 0.0149

But Q and our charge is changing and DI DT is also a moving change in charge.0161

These are functions of the same variable.0169

Let us see if we can put this all in terms of the same variable.0172

We are going to do that by recognizing first, that our current I is - DQ DT.0176

Negative because we are discharging that.0184

DI DT therefore, must be the opposite of the second derivative of Q with respect to T – D² Q/ DT².0188

I will rewrite our equation, we have Q/ C - L times the second derivative of Q with respect to time = 0.0201

Or in a more standard version of writing this, we can write this as D² Q/ DT² + Q/ LC = 0.0220

We have got a function, a differential equation where the second derivative of something + that something gives you 0.0236

Only way I know of that you can have an answer to a function like that is if you are looking at something like a cos or sin.0244

Remember, the derivative of sin is the cos, derivative of cos is the opposite of the sin.0251

The second derivative of something + that something can give you 0, right away sin is a real function.0256

Let us also go and let us define ω as 1/ √ LC.0264

This is actually ω² Q.0272

When we do that, we are going to solve this equation, what we are going to find is0275

that the form that fits this equation is some constant A times the cos of ω T, where ω is 1/ √ of LC + B sin ω T.0280

Now we can use our boundary conditions to figure out exactly what these A and B are.0300

One of the boundary conditions that we know is that the charge at T = 0 on our capacitor is Q0, our initial charge.0305

If T = 0, that sin function, sin 0 is 0, this whole term goes away.0318

T equal 0 cos of 0 is 1, that means A must be equal to Q0.0324

This implies then that A = Q0.0330

To look at our second boundary condition, when you recognize that our current I which is DQ DT is going to be,0335

as I go through this and take our derivative, A cos ω T, derivative of cos is opposite of the sin.0346

I'm going to get - ω A sin ω T from that first term + derivative of sin is cos, we will have ω B cos ω T.0352

With the boundary condition that the current at time T = 0 has to equal 0, everything is in that capacitor.0369

At T = 0, this whole first term goes away, that means cos 0,1 ω B must equal 0.0379

Ω is 1/ √ LC that can equal 0.0388

Therefore, B must be 0.0391

Putting all of this together, we can write our solution.0396

Therefore, Q of T must be A cos ω T, where A is Q0 so that is going to be Q0 cos ω T.0402

Since B is 0, this term goes way.0415

Here is our solution for the charge as a function of time.0418

Notice it varies as a function of time.0425

It is going to follow a simple harmonic motion type pattern.0427

We know Q of T = Q0 cos ω T, therefore the potential across our capacitor is just Q/ C, so that is going to be Q₀/ C cos ω T.0435

If we want to take a look at current as a function of time, I of T is just - DQ DT, which is going to be ω Q₀ sin ω T.0457

If we wanted to write this with our ω equal to 1/ √ of LC, oftentimes you will see this piece written as I of T = Q₀/ √ of LC sin ω T.0476

Let us graph these so you can really see the sign, we will make sure.0504

When I do this, if I look at the current as a function of time or the potential as a function of time for the circuit,0507

we get the sign in this pattern where we see the current just ahead of the voltage or depending on how you are looking at it.0514

Time you could say voltage ahead of current, just depends on your perspective.0522

But we see we have this repeating pattern of current and voltage as we have the electrical signal oscillating inside our simple LC circuit.0526

Assuming that it is a completely efficient circuit, it would go on forever and ever.0536

Now in reality, you have a dampening effect as energy is always lost in some of these circuits.0541

Eventually those appear out to 0.0547

But for a short term ideal situation, you have an ongoing harmonic oscillator, an electrical oscillator.0549

As we look at the angular frequency, ω = 1/ √ of LC, if we want our frequency in Hertz,0557

we can take a look at frequency is ω / 2 π which is going to be 1/ 2 π √ LC.0565

Hopefully that gives you a bit of insight into LC circuits as electrical oscillators.0579

Thank you so much for watching www.educator.com and make it a great day.0584