WEBVTT physics/ap-physics-c-electricity-magnetism/fullerton
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Hello, everyone, and welcome back to www.educator.com.
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I'm Dan Fullerton, and in this lesson, we are going to talk about LC circuits.
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Our objectives include applying Faraday’s law to a simple LC series circuit
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to obtain a differential equation for charge as a function of time.
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To solve the differential equation.
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To calculate the current in capacitor voltage as a function of time.
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Finally, sketching graphs of the current through the voltage across the capacitor for the simple LC series circuits.
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Let us take a look at our LC circuit.
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We have an inductor and capacitor and typically when we analyze these,
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what we are going to assume is when the circuit is first turned on, our capacitor is charged and our inductor is not.
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The interplay of the capacitor and the inductor is going to create an oscillating system very similar in fashion to simple harmonic motion,
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basically, looking at making the electrical version of a pendulum.
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Let us see if we can analyze the circuit a little bit.
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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.
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We have our potential across our capacitor DC ± recognizing that the electric field in here
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is going to go from the positive side to the negative side.
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Over here on our inductor, we have the potential across it which is L DI DT, that is our positive and our negative side.
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Again, it is very important to note that the electric field in here is 0, the electric field inside the conductor.
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With that, we can use Faraday's law to start to analyze our circuit.
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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.
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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,
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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.
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Or rearranging this a little bit, we could say that Q/ C - L DI DT = 0.
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But Q and our charge is changing and DI DT is also a moving change in charge.
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These are functions of the same variable.
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Let us see if we can put this all in terms of the same variable.
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We are going to do that by recognizing first, that our current I is - DQ DT.
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Negative because we are discharging that.
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DI DT therefore, must be the opposite of the second derivative of Q with respect to T – D² Q/ DT².
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I will rewrite our equation, we have Q/ C - L times the second derivative of Q with respect to time = 0.
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Or in a more standard version of writing this, we can write this as D² Q/ DT² + Q/ LC = 0.
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We have got a function, a differential equation where the second derivative of something + that something gives you 0.
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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.
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Remember, the derivative of sin is the cos, derivative of cos is the opposite of the sin.
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The second derivative of something + that something can give you 0, right away sin is a real function.
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Let us also go and let us define ω as 1/ √ LC.
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This is actually ω² Q.
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When we do that, we are going to solve this equation, what we are going to find is
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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.
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Now we can use our boundary conditions to figure out exactly what these A and B are.
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One of the boundary conditions that we know is that the charge at T = 0 on our capacitor is Q0, our initial charge.
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If T = 0, that sin function, sin 0 is 0, this whole term goes away.
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T equal 0 cos of 0 is 1, that means A must be equal to Q0.
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This implies then that A = Q0.
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To look at our second boundary condition, when you recognize that our current I which is DQ DT is going to be,
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as I go through this and take our derivative, A cos ω T, derivative of cos is opposite of the sin.
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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.
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With the boundary condition that the current at time T = 0 has to equal 0, everything is in that capacitor.
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At T = 0, this whole first term goes away, that means cos 0,1 ω B must equal 0.
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Ω is 1/ √ LC that can equal 0.
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Therefore, B must be 0.
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Putting all of this together, we can write our solution.
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Therefore, Q of T must be A cos ω T, where A is Q0 so that is going to be Q0 cos ω T.
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Since B is 0, this term goes way.
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Here is our solution for the charge as a function of time.
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Notice it varies as a function of time.
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It is going to follow a simple harmonic motion type pattern.
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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.
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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.
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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.
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Let us graph these so you can really see the sign, we will make sure.
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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,
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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.
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Time you could say voltage ahead of current, just depends on your perspective.
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But we see we have this repeating pattern of current and voltage as we have the electrical signal oscillating inside our simple LC circuit.
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Assuming that it is a completely efficient circuit, it would go on forever and ever.
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Now in reality, you have a dampening effect as energy is always lost in some of these circuits.
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Eventually those appear out to 0.
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But for a short term ideal situation, you have an ongoing harmonic oscillator, an electrical oscillator.
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As we look at the angular frequency, ω = 1/ √ of LC, if we want our frequency in Hertz,
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we can take a look at frequency is ω / 2 π which is going to be 1/ 2 π √ LC.
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Hopefully that gives you a bit of insight into LC circuits as electrical oscillators.
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Thank you so much for watching www.educator.com and make it a great day.