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 RL circuits, circuits that have resistors and inductors.
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Our objectives include, applying Faraday’s law to a simple RL series circuit to obtain a differential equation for current as a function of time.
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Solving the differential equation for the current as a function of time
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using that separation of variables technique that we have used a couple of times.
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Calculating the initial transient and final steady state currents through any part of that simple series
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in parallel circuit containing an inductor and a couple of resistors.
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Finally, being able to sketch graphs of the current through or the voltage across the resistors
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or the inductor in these simple series and parallel circuits.
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All about circuits that have resistors and inductors in them.
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Let us start off with our analysis of the circuits.
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When the circuit is first turned on, the inductor opposes the current flow and it acts like an open circuit.
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Contrast that to the capacitor, where when you first turn it on, it acted like a wire.
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After a long time, the inductor keeps current going, it acts as a wire acts, as a short.
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Contrast that with the capacitor which after a long time it acted like an open.
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They are opposites.
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After a long time, if you remove the battery, the inductor is going to act as an EMF source to keep the current going.
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You can almost think of an inductor is something that likes to oppose change.
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If there is no current flow in the circuit, it does not want current flow.
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But after you eventually get it going for a while and you stop the current flow, it wants to keep it going.
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As the resistor dissipates power, the current will decay exponentially to 0 as you use up that energy.
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Standard RL circuit diagram or source of potential difference, resistor and our inductor.
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Something important to note here.
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As we go through these analyses, the electric field inside that inductor is 0.
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Let us take a look at the current in the RL circuit.
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The first thing I'm going to do is I'm going to define a couple things.
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Let us talk about our current flow.
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Let us say it is in that direction.
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Let us also note our electric fields in here.
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If this is our source of potential difference, positive and negative side, we know the electric field points that way in there.
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Our electric field is going to go in this direction through our resistor.
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And inside our inductor, keynote, electric field is 0.
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When analyzing these sorts of circuits, lots of people like to use Kirchhoff’s voltage law.
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They like to say, as you go around here - V + IR + the voltage across the inductor equal 0.
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Although you can get the right answer doing that, it is not technically correct because Kirchhoff’s voltage law
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is really just looking at items where you have a non changing magnetic flux.
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Because you have a non changing magnetic flux, we have to go back to some of these first principles.
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We are going to start there and do this the more correct way.
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Let us start by writing our equation, the integral / the closed loop of E ⋅ DL is going to be the opposite of D φ B DT.
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We have been doing that for a while, Faraday’s law.
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When we do this, let us look at our E ⋅ DL side.
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We are going to have - V + our potential drop across R IR, so - V + IR =, and we know that D φ B DT is just going to be - L DI DT.
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We showed that in our last lesson.
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That is how we set up our equation to get going here.
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Let us start to rearrange this a little bit.
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We have to I - V/ R must equal - L/ R DI DT, as we work to separate our variables and our differential equation.
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Which implies then that all we have as we separate these DI/ I - V/ R must equal - R/ L DT.
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The integral from I = 0 to some final I must equal the integral from T equal 0 to some final value T.
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Starting to solve this, DU/ U that form the integral of that is going to be the log of U.
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The left hand side is going to look like the nat log of I - V / R evaluated from I = 0 to some value I =,
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and the right hand side just becomes - R/ LT.
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Expanding out this left hand side, substituting in our limits.
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That implies then that the log of I - V / R - the log of - V / R must be equal to - R/ L × T and the difference of the logs is the log of the quotient.
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Therefore, we could write the left hand side is the log of I - V/ R divided by -V/ R = - R/ LT.
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Once again, this nat log is pestering me.
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Let us raise everything, E raised to those powers on both sides so we can get rid of our natural log.
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And we can write the left hand side as I - V/ R divided by - V/ R must be equal to E ^- R/ L × T.
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Or doing a little bit of algebra here, we will multiply both sides by that - V/ R so that the left hand side becomes I - V/ R = - V/ R E ^- R/ LT.
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Or going to step further, I =, if we add that V/ R to that side, we are going to get V/ R.
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If I factor that out of the right hand side, it would be 1 – E ⁺_R/ L × T.
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I = V / R, that current that you would get if you do not have the inductor × (1 – E ^- R/ LT).
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There is our current flow as a function of time.
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When we talk about RC circuits, we talk about a time constant τ which is equal to R × C.
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And we are talking about RL circuits, our time constant which we still call τ is now L/ R.
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We can write this then in the form I = V/ R 1 - E ^- T/ τ.
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It fits that same basic form of the solution that we saw previously when we are dealing with RC circuits.
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Once again, though we are dealing with something 1 – E ^- T/ the time constant or E × – E ^- T/ the time constant.
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You are going to see that form again and again and again.
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There is the current, let us take a look at the voltage in RL circuits.
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The voltage across our inductor is a function of T, is just L DI DT which is going to be L × the derivative with respect to time of V/ R × 1 – E ^- R/ LT.
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And that is going to be, we can pull our V/R, that is a constant.
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That will be L V/ R × the derivative of 1 - E ^- R/ LT.
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Taking that derivative, we have LV/ R.
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We have - E ^- R/ LT × - R/ L.
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It complies then that the voltage across our inductor as a function of time is just going to be VL / R, R/ L, make a ratio of 1.
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Our negatives cancel out so V × E ^- R/ LT.
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If you want that to put that in terms of τ again, that is a VE ^- T/ τ.
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Those are equivalent statements, as long as τ is L/ R.
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There is the voltage in RL circuits.
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Let us also take a look at the rate of change of current in RL circuits.
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Of course, we could just come back here with our DI DT and look at that because we know what that is.
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But instead, I think it is worthwhile to go to the math once more and get a little bit more practice.
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Let us take a look at the rate of change of current in the RL circuit.
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DI DT is going to be the derivative with respect to time of our V / R × 1 - E ^- R/ L × T,
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which we can pull out V/ R our constant, V/ R × the derivative with respect to T of 1 - E ^- R/ LT.
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Or DI DT = V/ R - E ⁻R/ LT × - R/ L which just gives you V/ L E ^- R/ LT.
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Or in terms of τ, our time constant V/ L E ^- T/ τ.
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Both of those are equivalent again.
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When we only take a look at the graphs of these, we already talked about the general shape of them.
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But looking at current in voltage graphs, let us start with current up here.
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Current is a function of time T, we said we started with 0 current and
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we are going to approach some maximum level that is called Imax and just V/ R.
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If we start at 0, we are going to approach that asymptote and we are going to do that
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in such a manner that it start to get mighty close to it, right about the time we hit 5 τ, 5 time constants or 5 × L/ R.
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At that point we are more than 99% of our way to our final value.
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If we do the same thing, taking a look at voltage here, we are going to start at our maximum voltage,
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Vmax or the voltage of our source of potential difference B for battery.
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Overtime we are going to decay to 0.
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As we do that, our voltage across our inductor, let me bring that down just a little bit better.
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That kind of matches our graph up above.
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We start to get pretty close to that final value around 5 τ.
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There are some more here, why do not we do a couple example problems.
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We will pull out some old AP exams and look at some problems from their that involve these RL circuits.
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Starting with the 2011 exam, let us take a look at E and M problem number 2.
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You can find that at this link or google it.
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Take a minute, print it out.
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Look it over, see if you can solve it yourself.
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Hit the pause button, check it out, then come back here and we can compare solutions.
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We have got a circuit that has a resistor, a capacitor, and inductor in the switch in there.
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It looks like it says an experiment, when the switch is close to position S1 at time T1 and it is left there for a long time.
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If I want to redraw my circuit when we are in that configuration A1, we have got a 9V battery,
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we have a 500 ohm resistor, and we have a 25 mf capacitor.
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With the switch in that position, it looks like that is all we have to worry about and it is been on a long time.
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Calculate the value of the charge on the bottom plate of the capacitor a long time after the switch is closed.
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Key thing to note, usually we are solving for the magnitude of the charge.
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Since, we are looking at the bottom plate of the capacitor, we are looking for the negative charge.
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We have to be careful with your answer, we get our sign right.
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We can use C = Q/ V to solve this, knowing that the charge on one of the plates is CV, which is going to be 0.025 F,
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25 mf × the voltage across our capacitor, the potential which after a long time when it is charged up, there is no current flowing.
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No potential drop across the resistor, all 9V must be spread across that capacitor.
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9V is going to give us then that Q our charge = 0.225 C.
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We want the charge on the bottom so that is going to be -0.225 C.
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Let us take a look at part 2 then, part 2 gives us a graph and it wants us to
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sketch a graph of the magnitude of the charge on the bottom plate of the capacitor as a function of time.
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We are looking at the bottom plate of the capacitor but we are only looking at magnitudes,
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we do not have to worry about signs and drawing our graphs upside down or anything like that.
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Let us sketch out our axis nice and neatly here.
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It looks like this is our time axis in seconds, this is the magnitude of our charge and it gives us a point here T1.
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Before we close that switch, we cannot have any charge on the capacitor, it is initially uncharged.
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The initial part should be really straightforward 0.
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After that, we are going to have rise to our final value of 0.225 C.
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We are going to do that, approaching that asymptote of our maximum charge.
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It should look something like that, where we have an asymptote here of about 0.225 C.
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It gets to this point, and if we want to sketch everything in their, right around T is approximately 5 τ.
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Which in this case, it is going to be 5 RC, which is going to be 5 × R 500 × RC 0.025.
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Or I come up with about 62.5s.
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After that, you get just a touch over a minute you are almost here to your final value.
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I would use that for my graph for part A2.
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Moving on, let us give ourselves some more room on the next page.
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For A3, we are given another graph.
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Sketch a graph through the resistor of the current through the resistor as a function of time.
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Let us make our axis again, our Y and our X.
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Now we have our current flow through the resistor as a function of time.
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We have got to our T1 labeled here.
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As we start off again, our current flow before we turn anything on, that is an open circuit.
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We are not going to have any current flow.
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That piece is pretty straightforward.
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Then, once we turn it on, we are going to initially have our maximum current flow as the capacitor acts like a wire,
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we are going to start at some high value.
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It is just going to be V/ R, 9V/ 500 ohms is going to give us 18 ma.
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We are going to have that exponential decay of which, maybe not the prettiest graph there
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but you get the idea and we start to get really close to our 0 point at around 5 τ.
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A3, let us move on to our next page which says part B.
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An experiment to the capacitors on charge and the switch is close to position S1 at T1
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and it is looped at position S2 at time T2 when we have 105 mc on the capacitor, allowing electromagnetic oscillation in the LC circuit.
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We are going to charge up that capacitor and then close our circuits.
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All we have is a capacitor and inductor.
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The capacitor discharges, sending current in the inductor.
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When it does that, it create a magnetic field in the inductor.
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It does that like that it opposes that.
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As that runs out, it starts putting that magnetic field energy back into current flow.
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You are going to have this oscillating system.
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Let us draw our circuit at this point, let us call it LC circuit because it has an inductor and a capacitor.
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There is our inductor and this is 25 mf here.
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We have got a 5 H inductor over there and their initial charge Q0 when our capacitor is 0.105 C.
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What are we trying to find?
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The energy stored in the capacitor at time T2.
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When that happens, our energy stored in the capacitor we know the charge on it.
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If ½ CV² is the energy of the capacitor, we know C = Q/ V.
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Therefore, V = Q/ C so that can be rewritten as ½ × our C.
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Instead of V², I’m going to have Q² /C².
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Let us put Q/ C².
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Then, UC = ½, we will have Q² /C as we simplify that, which is going to be our 0.105 C² / our capacitance 0.025 F.
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I come up with about 0.221 J stored in that capacitor at that point in time.
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Taking a look at B2, calculate the maximum current that will be present during the oscillations.
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We are going to have the maximum current flow when we have depleted that capacitor.
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When the charge on the plate of that capacitor = 0, max current when QC is equal to 0.
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Therefore, the energy stored in the inductor is going to be 0.221 J.
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No energy stored in the capacitor, it is all in the inductor that point.
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The energy stored in the conductor is ½ LI².
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Then our I² is just going to be 2 × 0.221 J/ our inductance of 5 H.
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We want just the current not the current², so if I take the square root of both sides, I determine that the current is going to be about 0.297 amp.
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There is our maximum current flow in the circuit.
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Let us take a look at part B3, calculate the time rate of change of the current DI DT when the charge on the capacitor plate is 50 mc.
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We are trying to find DI DT when the charge on our capacitor is 0.05 C .
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I’m going to go to Faraday’s law here and state that the integral around the closed loop of E ⋅ DL
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is - the time rate of change of our magnetic flux, which by the way is - L DI DT.
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Then, as we go around our E ⋅ DL, we have Q/C from our capacitor.
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Its contribution to the electric field Q/ C must equal - L DI DT, which implies then that DI DT is going to be equal to -Q/ LC,
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which implies that DI DT is –Q, which is -0.05 C/ 5 H our inductance × our capacitance 0.025 F,
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which implies then that the time rate of change of current at that point is going to be about -0.4 amps/ s.
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I think that covers the 2011 question number 2.
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Let us take a look at our next question moving to 2008.
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The 2008 exam, we are going to look at the E and M portion free response question number 2,
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where again we have a couple configurations of circuits.
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Taking a look at part A1 here, giving us a circuit A and B are terminals to which we can attach different circuit components.
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We are going to calculate the potential difference across R2 immediately after the switch is closed.
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When first off, we have a 50 ohm resistor connecting A and B.
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Let us draw that circuit first.
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It looks like we are going to have our source of potential difference 1500 V.
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We are going to go over here to R1, which is 200 ohms.
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We have got a 300 ohm resistor there, that brings us back to the power supply.
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We have another branch now that has a 100 ohm resistor.
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We are also going to insert for this part of the problem a 50 ohm resistor.
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200 ohms, 300 ohms.
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At this point, we want to know the potential difference across R2.
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I suppose I would label all these here so it is a little easier to see which one is which.
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There is R1, R2, R3, and our special resistor.
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What I would do here, if I wanted to find the potential across R2,
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is recognized that I have 2 resistors here in series that have an equal resistance of 150.
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We just have to add them up and these are in parallel.
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300 in parallel with 150, let us find an equivalent resistance for these 3 resistors.
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I’m going to call that R2,3 equivalent, is going to be, our first resistance 300 × the second 150.
00:25:58.000 --> 00:26:03.600
Adding those up and divided by the sum of those which is going to be 450
00:26:03.600 --> 00:26:09.700
which shall give us an equivalent resistance of 100 ohms for these 3 in that configuration.
00:26:09.700 --> 00:26:15.600
Our total current flow, treating it as a 200 ohm resistor and a 100 ohm resistor in series,
00:26:15.600 --> 00:26:29.600
is going to be V/ R or 1500 V/ our equivalent total resistance 300 ohms, which is going to be 5 amps.
00:26:29.600 --> 00:26:38.600
Now that we have done that, that means that our potential across R2 is going to be the potential that goes through our 100 ohm resistor.
00:26:38.600 --> 00:27:00.600
Equivalent resistor there so V R2 is going to be equal to our 5 amps, the current flow, × our 100 ohms or 500V.
00:27:00.600 --> 00:27:04.500
That is a little better.
00:27:04.500 --> 00:27:10.400
For part 2, we are going to have a 40 mh inductor in place of that 50 ohm resistor.
00:27:10.400 --> 00:27:14.600
Let us take a look at that circuit.
00:27:14.600 --> 00:27:22.400
We have our 1500 V source of potential difference.
00:27:22.400 --> 00:27:29.700
We have our 200 ohm R1 resistor.
00:27:29.700 --> 00:27:40.300
We have our 300 ohm R2.
00:27:40.300 --> 00:28:03.000
Our way over here, we have our 100 ohm R3 and we have our inductor which is a 40 mh inductor.
00:28:03.000 --> 00:28:09.400
We want to know the potential difference across R2 immediately after the switch is closed.
00:28:09.400 --> 00:28:15.900
Immediately after the switch is closed, an inductor is going 2 act like an open.
00:28:15.900 --> 00:28:18.700
It does not want any change in that state.
00:28:18.700 --> 00:28:22.600
The current through the inductor at time T = 0 is 0.
00:28:22.600 --> 00:28:31.300
If this entire circuit here is open, so what we have effectively, we can pretend that this is all gone
00:28:31.300 --> 00:28:34.900
and all we have is this nice simple series circuit.
00:28:34.900 --> 00:28:40.800
If we want to know the voltage drop across R2, we can just do the ratio as a voltage divider
00:28:40.800 --> 00:28:45.000
and I can see pretty easily that that is going to be 900V.
00:28:45.000 --> 00:28:50.000
3/5 of the resistance is R2, therefore, 3/ 5 of the voltage will be dropped across R2.
00:28:50.000 --> 00:28:57.300
Or they can do this mathematically, recognizing that the total equivalent resistance at T equal 0
00:28:57.300 --> 00:29:02.000
is just going to be R2 resistors in series or 500 ohms.
00:29:02.000 --> 00:29:12.700
Therefore, the current is going to be V/ R or 1500 V/ 500 ohms, which is going to be 3 amp.
00:29:12.700 --> 00:29:21.300
3 amp through 300 ohms is going to give us, V = IR Ohm’s law.
00:29:21.300 --> 00:29:37.200
V through R2 = I2 R2 is going to be 3 amps × 300 ohms or 900 V.
00:29:37.200 --> 00:29:40.600
That was not so bad.
00:29:40.600 --> 00:29:49.200
Taking a look at part of 3, now we are going to replace that inductor with an uncharged 0.8 µf capacitor.
00:29:49.200 --> 00:29:53.100
We are getting good at drawing these circuits so let us do it again.
00:29:53.100 --> 00:30:06.600
A3, we have our 1500 V supply, we have a resistor R1 200 ohms.
00:30:06.600 --> 00:30:17.500
We have another resistor R2 300 ohms.
00:30:17.500 --> 00:30:30.900
Moving further to the right, we have our next branch which is R3 which is 100 ohms and we have our capacitor.
00:30:30.900 --> 00:30:43.200
Which we say it was 0.8 µf or 800 nf.
00:30:43.200 --> 00:30:49.100
What we want to find out is what the potential drop is across R2 right after the switch is closed.
00:30:49.100 --> 00:30:55.300
Right after the switch is closed, this is going to act like a wire.
00:30:55.300 --> 00:31:00.500
All we have is another circuit now or we can pretend this does not exist here.
00:31:00.500 --> 00:31:06.100
Let us even erase it out of our picture.
00:31:06.100 --> 00:31:09.200
At T = 0, there is our equivalent circuit.
00:31:09.200 --> 00:31:19.600
The equivalent resistance between the 300 and 100 over here, R equivalent for 2,3 is going to be 300 ×
00:31:19.600 --> 00:31:28.200
100/ their sum 400 or about 75 ohms.
00:31:28.200 --> 00:31:36.100
Our total effective resistance for our circuit is 200 + 75 in series or 275 ohms.
00:31:36.100 --> 00:31:50.900
Our total current flow is going to be V/ R which is 1500 V/ 275 ohms or about 5.45 amps.
00:31:50.900 --> 00:32:03.200
Then V R2 is going to be equal to the current flow × the equivalent resistance, our equivalent 2,3
00:32:03.200 --> 00:32:16.600
which is 5.45 amps × 75 ohms which is about 409 V.
00:32:16.600 --> 00:32:22.500
We answered part A1, 2, and 3.
00:32:22.500 --> 00:32:28.100
For part B, it looks like we are going to be get into the graphing game.
00:32:28.100 --> 00:32:36.900
The switch gets close to time T equal 0, on the axis below sketch the graphs of the current, in R3 vs. Time for each of the 3 cases.
00:32:36.900 --> 00:32:42.500
The case for we have got to resistor there, the inductor, and the capacitor.
00:32:42.500 --> 00:33:03.500
Let me go to the next page to give us lots a room and we will make our graph of current vs. Time.
00:33:03.500 --> 00:33:11.300
There is our time, there is our current, and let us start with the case for we have got that resistor.
00:33:11.300 --> 00:33:19.900
When it is a resistor, we are not going to have any time dependents so our current flow is going to be a constant.
00:33:19.900 --> 00:33:25.500
Let us do that here in blue.
00:33:25.500 --> 00:33:32.500
Here is our current flow through our resistor R3.
00:33:32.500 --> 00:33:39.300
In the case where we happen to have a resistor in that blank spot between A and B.
00:33:39.300 --> 00:33:47.700
When we put the inductor in there, remember the current flow through R3 started at 0 and then it got bigger.
00:33:47.700 --> 00:33:51.000
It follow that path up until it heads for that asymptote.
00:33:51.000 --> 00:33:55.800
For that, when we are going to started 0 it is going to have a higher over all current.
00:33:55.800 --> 00:34:00.300
It is probably going to look something like that.
00:34:00.300 --> 00:34:03.300
There is the case where we have an inductor.
00:34:03.300 --> 00:34:10.100
With the capacitor in there, we start at maximum current where it started looking like a wire.
00:34:10.100 --> 00:34:13.500
After a long time as it becomes charged up, it becomes an open.
00:34:13.500 --> 00:34:22.200
It is going to start in its maximum value of current and it is going to exponentially decay down to 0.
00:34:22.200 --> 00:34:26.100
Something like that, that will be the case where we have our capacitor in there.
00:34:26.100 --> 00:34:37.600
Those would be the sketches I would put in place for current vs. Time through that R3, which is right in line with our mystery element.
00:34:37.600 --> 00:34:44.300
Let us take a look at 1 more free response problem.
00:34:44.300 --> 00:34:47.900
Moving to the 2005 E and M exam free response 2.
00:34:47.900 --> 00:34:51.800
We have got another inductor circuit and another RL circuit.
00:34:51.800 --> 00:34:59.700
Here we have resistors 1 and 2 of resistance R1 and R2 and an inductor connected to a battery and we have got a switch, of course.
00:34:59.700 --> 00:35:06.900
The switch is close at time T equal 0, find the current through resistor 1 immediately after the switch is closed.
00:35:06.900 --> 00:35:13.900
I like to redraw the circuits, it helps me just feel better and get a feeling for what is actually in there to draw it myself.
00:35:13.900 --> 00:35:22.600
We have got our source of potential difference, positive, negative.
00:35:22.600 --> 00:35:40.100
Over here, we have our first resistor R1 ± , we have R2 ±, and we have a second branch coming
00:35:40.100 --> 00:35:51.300
down over here to our inductor ± and there it has some inductance.
00:35:51.300 --> 00:35:54.700
At T equal 0, find the current through resistor 1.
00:35:54.700 --> 00:36:00.000
At time T = 0, the current through our inductor is equal to 0.
00:36:00.000 --> 00:36:02.200
As if this branch does not exist.
00:36:02.200 --> 00:36:05.900
All we have here is a straightforward series circuit.
00:36:05.900 --> 00:36:17.500
The current at that point is going to be V / R is going to be our electromotive force E/ our total resistance R1 + R2.
00:36:17.500 --> 00:36:25.500
There is the current through resistor 1 right after we turn that on.
00:36:25.500 --> 00:36:38.900
For part B, determine the magnitude of the initial rate of change of current DI DT in the inductor.
00:36:38.900 --> 00:36:41.400
The initial rate of change in the inductor.
00:36:41.400 --> 00:36:47.300
What we can do is take a look at Faraday’s law as we make the big loop around here.
00:36:47.300 --> 00:36:49.600
We will skip R2 as we make our loop.
00:36:49.600 --> 00:37:07.500
The integral/ the closed loop of E ⋅ DL = – D φ B DT or as I fill that in, as we go this way, we see -E first + our initial I0 × R1.
00:37:07.500 --> 00:37:26.200
No electric field in the inductor so nothing there and that is going to be equal to R- D φ B DT which by now we know is - L DI DT for our inductor.
00:37:26.200 --> 00:37:51.100
Solving for DI DT, DI DT is going to be equal to, we have E / L we divide that out, E/ L – I is 0 R1/ L,
00:37:51.100 --> 00:38:09.000
Which implies that DI DT is going to be equal to E/ L - we can replace I₀ with what we just found up here E / R1 + R2.
00:38:09.000 --> 00:38:15.700
There is our I₀ × R1/ L.
00:38:15.700 --> 00:38:36.400
With a little bit of simplification, I can factor out an E/ L to say that DI DT is equal to E/ L × the quantity 1 - R1/ R1 + R2.
00:38:36.400 --> 00:38:38.300
We can simplify that further if you want to.
00:38:38.300 --> 00:38:45.600
I think we can make this quantity look like R2/ R1 + R2, but that looks like it is puny simplified to make.
00:38:45.600 --> 00:38:49.200
I think we will call that our answer for part B.
00:38:49.200 --> 00:38:53.500
In many cases here, you are more than welcome to always go through and try clean these up further
00:38:53.500 --> 00:38:58.800
but that should meet the requirements of the problem.
00:38:58.800 --> 00:39:03.600
Let us take a look at part C.
00:39:03.600 --> 00:39:06.400
Let us give ourselves more room on the next page.
00:39:06.400 --> 00:39:13.500
Part C says, determine the current through the battery a long time after the switch has been closed.
00:39:13.500 --> 00:39:24.700
As time approaches infinity a long time, the potential across our inductor is going to be 0 because it is acting like a wire at that point.
00:39:24.700 --> 00:39:35.300
The current through the battery is just going to be your total current V/ R which is going to be E/ R1.
00:39:35.300 --> 00:39:42.900
That is going to completely bypass R2, it is got that short around it.
00:39:42.900 --> 00:39:47.700
Part C, taking a look at part D.
00:39:47.700 --> 00:39:51.800
On the axis below, sketch a graph of current through the battery as a function of time.
00:39:51.800 --> 00:39:54.900
We have done all the hard work there, we only have to do is fill in our graphs.
00:39:54.900 --> 00:40:06.100
Let us do that, let us put our axis up here.
00:40:06.100 --> 00:40:09.200
We have our current through our battery I.
00:40:09.200 --> 00:40:19.500
We have our time T and initially we already found that I 0 was E/ R1 + R2.
00:40:19.500 --> 00:40:26.500
We found that after a long time, you are actually up here at E/ R1.
00:40:26.500 --> 00:40:29.200
That is going to be our asymptote.
00:40:29.200 --> 00:40:39.300
What we have to do is fill in our graph as we approach that asymptote, something like that.
00:40:39.300 --> 00:40:44.900
Finally part E, let us give ourselves some more room again.
00:40:44.900 --> 00:40:51.000
Part E says, determine the voltage across resistor 2 right after you open the switch.
00:40:51.000 --> 00:40:57.500
When you do that, we are going to have a new circuit that looks kind of like this.
00:40:57.500 --> 00:41:12.600
We have got our resistor R2, we have our inductor L, and because the switch is open, the rest of it does not come into play.
00:41:12.600 --> 00:41:24.000
I R2 is going to be equal to the current through our inductor which is just E / R1.
00:41:24.000 --> 00:41:29.600
Exactly what we had right before that because that conductor wants to oppose the change.
00:41:29.600 --> 00:41:32.700
It is going to keep that going as long as it can.
00:41:32.700 --> 00:41:38.600
It is going to keep that current for the time being and it is going to decay as that power is dissipated in the resistor.
00:41:38.600 --> 00:41:54.500
Then V R2 is going to be equal to the current through × the resistance which is going to be our current through R2 × I R2 × R2,
00:41:54.500 --> 00:42:07.900
which we just said I R2/ R1, that is E/ R 1 R2 or E × R2/ R1.
00:42:07.900 --> 00:42:10.500
That should finish them up.
00:42:10.500 --> 00:42:13.100
Hopefully, that gets you a great start on RL circuits.
00:42:13.100 --> 00:42:15.800
Thank you so much for watching www.educator.com.
00:42:15.800 --> 00:42:17.000
Make it a great day everybody.