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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In this lesson we are going to talk about RC circuits, specifically about the transient analysis.
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What happens as a function of time in these circuits?
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Our objectives include calculating and interpreting the time constant τ of an RC circuit.
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Sketch or identify graphs of stored charge voltage for the capacitor or resistor.
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Write expressions that describe the time dependence of the stored charge voltage or current for elements in RC circuit.
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Analyze the behavior of circuits containing several capacitors and resistors.
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As we get into this lesson, please understand this one is going to be pretty math heavy.
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The transient analysis of RC circuits is one of the more challenging portions of the E and M course.
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This may get in depth a little bit, probably a great time to pause, to go back, to take really good notes.
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You may have to come back to this one a couple of times.
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It is not easy stuff and the first time you see it, it looks like there is a lot a math involved.
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What you will find as we go through is you will see a bunch of the same patterns repeating and repeating.
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But the first time you see it, it can look a little bit scary.
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Do not be daunted, you can get through it.
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Let us take a look at what happens when we charge an RC circuit.
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Here I have a basic RC circuit, we have our source of potential difference resistor.
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We have defined our current direction, our capacitor with the voltage across our capacitor, and the time T = 0 we are going to close the switch.
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It also had graphs down here of current in the circuit, charge on our capacitor,
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and the voltage across the capacitor that we are going to be filling out as we go through here.
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We have done these before, but let us take a minute before we get heavy into the math.
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Just to think about what these are going to look like.
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We know the current in the circuit initially is going to be high because the capacitor acts like a wire.
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So we are going to have all the current I = VT/ R.
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We are going to start here at this high level of current and by the time we get roughly to 5 τ,
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the current is going to dwindle down to less than 1% of its initial value because our capacitor starts to act like an open.
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Our graph is going to have an exponential decay, something like that.
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The charge on our capacitor is going to start at 0, it is uncharged.
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After a long time that being again 5 τ, it is going to be CVT.
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We will have an exponential rise toward that asymptote.
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The voltage across our capacitor starts at 0 acting like a wire and over time approaches VT.
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Our graph should look something like that.
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Our goal here is going to be to mathematically describe those by actually figuring out what happens rather than just quick estimations.
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To do that, let us start by looking at Kirchhoff’s voltage law.
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We are going to do that, I'm going to start down here and go clockwise around our circle.
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As I look at our potential drops, I see - VT first + IR + the voltage across our capacitor, brings us back to our starting point.
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All of that must equal 0.
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We also know here that C = Q/ V, therefore, V = Q/ C.
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I can rewrite my equation now as, let me arrange this a little bit to VT is equal to IR + I’m going to replace VC with Q/ C.
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This implies then though, Q and I are related because I = DQ DT.
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We can write this as VT is equal to R × DQ DT, replacing I with DQ DT + we still have our Q/ C.
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We have a differential equation, we have Q and the derivative of Q in the same equation.
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How do we deal with something like that?
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The first thing I’m going to do is called separation of variables.
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I'm going to try and get all the variables of the same type on the same side.
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I need to get Q and DQ together.
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This is going to take a little bit of algebraic manipulation.
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The reason I know how to do this is I have done it quite a few times.
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You really just have to practice it and dive in, and do it again and again and again.
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Let us rearrange this, I'm going to get DQ DT all by itself by dividing both sides by R.
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I will have VT/ R is equal to, we have our DQ DT + Q/ RC.
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Rearranging this again, we are trying to get RQ together .
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We are going to write this as DQ DT is equal to VT/ R -Q/ RC, just a quick algebraic manipulation,
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which implies then that DQ DT =, I can multiply C down here to combine these on the right hand side.
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They give us a common denominator of RC so I would get C VT -Q/ RC.
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Getting those variables separated, I have DQ/ C VT -Q must equal DT/ RC.
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Just a little bit more algebraic manipulation.
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I have my Q on one side, I got my variable T on the other, that means I can go when I can try and integrate both sides.
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Not try, we are going to.
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Integrate both sides, we will take the left hand side first.
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The integral and our variable of integration is Q, which is going to go from some value 0 initial charge on our capacitor is 0,
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to some final charge Q of DQ/ CVT – Q, must equal the integral of the right hand side DT / RC.
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Our variable of integration on the right hand side T goes from some initial time T = 0 to some final time which we are going to call T.
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As we integrate this, the left hand side we have got a problem of the form DU/ U.
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The integral of DU/ U is nat log of U.
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We have got DU / -U so we are going to get - the log of, we have CVT - Q evaluated from 0 to Q.
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The right hand side of, 1/ RC is a constant, we are just going to get T/ RC evaluated from 0 to T.
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A little bit more algebra, this implies then that, we will substitute in our values, our limits here.
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- the log of CVT -Q - the opposite of the log which is going to be + the log of CVT -0, which is just CVT = T / RC.
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If we got a difference in the logs, log of that + log of that,
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we can say that we have the log of CVT - Q/ CVT equal to, we got that negative sign over here to the right, - T / RC.
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We can simplify this left hand side a little bit too.
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That means that the log of, that could be 1 -Q/CVT = - T/ RC.
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I’m starting to run out of room, let us carry this over onto our next screen here.
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We have the log of 1 -Q/ CVT = - T/ RC.
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This log is troubling, how do we get rid of a log?
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We take E and raise it to that power.
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E ⁺log of 1 -Q/ CVT is what we are going to do.
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Raise both sides to that, so E ⁺log of 1 -Q/ CVT must be equal to E ⁻T/ RC,
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doing the same thing on both sides to maintain that equality.
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E ⁺log of something is just that something.
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Our left hand side gets a little simpler.
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We have 1 - Q/ CVT =, right hand side E ⁻T/ RC.
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Let us see if we can get to that one on that side and do a little bit of rearrangement to say that Q/ CVT = 1- E ⁻T / RC,
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which implies then that getting just Q by itself, Q is a function of time is equal to CVT × (1 - E ⁻T/ RC).
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We were able to solve for the charge on the capacitor as a function of time.
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When you do these types of problems, you are going to see very similar forms come up again and again.
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Some constant multiplied by either 1 – E ⁻T/ RC or just that constant × E ⁻T / RC.
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You are going to see this come up again and again and again,
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to the point where you can almost predict the answer before you go and actually solve it.
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If that is our charge, let us see if we can find the voltage across C.
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VC = Q/ C, we just found Q was CVT × 1 - E ⁻T/ RC ÷ C.
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Our C make a ratio of 1, we get that the voltage across our capacitor or potential difference is just VT × 1 - E ⁻T/ RC.
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How about current flow?
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The trick to getting current flow then is realizing that I is the current is equal to the derivative of the charge.
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Current I is DQ DT which is the derivative with respect to T of our Q which was CVT × 1 - E ⁻T/ RC
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which is going to be, we can pull our constant out, - CVT × E ⁻T/ RC × -1 / RC E ⁺U DU or DU is –T/ RC.
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The E ⁺U DU, the derivative is -1/ RC, excuse me.
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That is going to be, CVT / RC × E ^-¬T/ RC, which implies then,
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our capacitance is C is going to make a ratio of 1, that our current then = VT/ R E^ -¬T/ RC.
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By the way, VT/ R that was our initial current.
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I is actually equal to its initial value × E ¬⁺T/ RC.
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We were able to find the values for the current, the voltage, and the charge on the capacitor,
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all as functions of time, much more exactly.
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You see that the exponential relationship over and over again.
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Or again, RC is your time constant τ.
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Sometimes you will actually even see this written as E ⁻T / RC is written as E ^-¬T/ τ.
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Let us take a look at the opposite side of the storage, discharging an RC circuit.
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Just like we did in our steady state analysis, we have pulled our source of potential difference out of the circuit.
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We have our charged up capacitor and at the time T = 0, we are going to close the switch.
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What happens?
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We know that our current starts out as V/ R because initially that capacitor
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is going to act like a source of potential difference, it discharges.
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We are going to start over here with our current with VC / R and that is going to decay down to 0.
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Our charge starts at Q0 and also decays down.
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Our voltage DC starts at its initial value, let us call that V0 and is also going to decay down.
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We have done that before, now let us fill in the detail.
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What happens in between?
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We use KVL again, Kirchhoff’s voltage law.
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This time I'm going to start here and go this way, counterclockwise.
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As I write that, that will be -VC + IR = 0, which implies then because we know capacitance = Q/ V, therefore V = Q/C.
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That Q/ C is equal to IR, just moving it to the other side as well.
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In this case, we know that I is the rate of change of charge with respect to time
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but now we are discharging, we are going the opposite way.
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Do not forget we got to have a negative there, I = – DQ DT.
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We have Q/ C equal, we still have our R, we will put -R in there because our next piece is that DQ DT.
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The - coming from our I = –DQ DT.
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This implies then, let us do our separation of variables again.
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DQ ÷ Q is going to be equal to - DT/ RC, that piece was a little simpler.
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But we can integrate both sides now, we integrate the left hand side.
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The integral of DQ/ Q starting from some Q = 0 all the way to Q =, - the integral from T = 0 to T of DT/ RC.
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We call that Q₀ to some final value Q.
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Q = Q₀, whatever the Q initial is to its final value, that is a little bit better.
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Which implies then that the log of Q, the integral of DU/ U is the natural log,
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so the log of Q going from some its initial value Q0 to some final value Q.
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What we are going to have over here is - T/ RC.
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Taking this left hand side and putting in our limits, we have the log of Q/ Q0 is going to be equal to - T/ RC,
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which implies that if we raise both of these to the E, that the left hand side becomes Q/ Q₀.
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The right hand side becomes E ¬⁻T/ RC, and solving for just Q, Q = its initial value Q₀ E ^-¬T/ RC.
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There we see that exact function, that exponential decay.
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Once again, same sort of form, we are looking at some constant × E ⁻T/ RC or constant ×(1- E ⁻T/ RC).
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Just like we did before, let us take a look at voltage.
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Voltage is Q/C which is going to be in this case, Q₀ E ^-¬T/ RC ÷ C.
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But we know that our initial value of voltage was Q₀/ C.
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Really this is just saying that VC = its initial value V₀ × E ^¬T/ RC.
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Here we see our function that shows the same thing, the same shape that exponential decay.
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Let us take a look and see if we can do current just to maintain some symmetry here.
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As I look at current, we have I = – DQ DT, which is - the derivative with respect to time of
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we had previously Q₀ E ^-¬T/ RC, which is -Q₀ E ^-¬T/ RC × -1/ RC, which will be Q₀/ RC E ⁻T/ RC.
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We can go a little bit further with that because we know that I initial, I0 was V₀/ R which is Q₀ / RC.
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This is just saying then that current is a function of time is its initial value I₀ E^ -¬T/ RC.
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Once again, there is our exponential decay.
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A pretty heavy stuff in there, and not easy to do the first time you go through it.
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It take some time but keep working on that until you can do it repeatedly.
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It is an expectation for the AP Physics C E and M exam but you can do that.
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Not easy, give it some time, give it some practice.
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Let us talk for just a minute more about that time constant again.
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The time constant in RC circuit τ = RC is the time when the quantity has reached 1 – E⁻¹ or 63% of its final value, just like we said.
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By τ time constant it is more than 99% of the way to its final value.
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If you want to be sure that your circuit is in pretty close to a steady state condition, wait at least 5 time constants then you will be there.
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Let us finish this up by, let us do a bunch of old AP problems that involve this process because really it is all about practice.
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We will start with the 2013 exam free response number 2.
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You can go download it, take a minute, print it out.
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Read it over, give it a shot, then come back here when you are done and hit play again, then we will keep on.
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This exam question starts off with a circuit and says indicate the position to which the switch should be moved to charge the capacitor.
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I'm going to draw where it should be, the answer is going to be B,
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The charge that the capacitor so that you keep the source of potential difference in the loop there.
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But I'm going to draw that as something like this.
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We have got our source of potential difference, our resistor R.
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We come from here over to our capacitor and there we go.
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There is our basic circuit.
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It asks on the diagram, draw a voltmeter that is properly connected to the circuit, the line to measure the voltage across the capacitor.
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Here we need to remember that voltmeters are connected in parallel.
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I'm just going to put a nice happy little voltmeter right there in parallel with my capacitor and I should be all set for part A.
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Here it says, it gives us some information of graph.
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Your partner has does some stuff with the stop watch, it asks you to determine the time constant of the circuit from the slope of the linear graph.
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How are we going to do that?
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Let us take a look, as we start to examine what data we have, we have got time and voltage.
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I'm going to look at the relationship for this as we are discharging the capacitor,
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realizing that V is equal to its initial value × that exponential decay portion E ⁻T/ RC.
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I want to see how I can get this into a linear graph based on time and the potential data that we have.
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If I take the log of both sides, I can say that the log of potential is equal to, the log of that is going to be log of V₀ + - T/ RC.
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Or with a little bit more manipulation, the log of V will equal -1 / RC × T + log of V₀.
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Why did I put it that way?
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I’m trying to make it match the form for a line, the equation of a line that says Y = MX + B.
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Really, I'm saying that we are going to have log V on the Y.
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Our X is going to be T, our Y intercept is going to be log of V₀ which means that the slope of that line that we end up with is going to be -1/ RC.
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We made it match that form.
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It says use the rows in the table of the data to record any quantities that you indicated that are not given,
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label each row used and include units.
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We have got time which is going to go on our X, we do not have is the log of V.
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I would add a row in there for the log of the potential and put that data in there right under the voltage.
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The units of that are going to be log volts.
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Take the value that is in there for B, take the nat log of it and make one more row and you should be all set for part B2.
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Next we go on to part C and we are actually going to be graphing this.
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Let us give ourselves a bunch of room here.
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For part C, let us see if we can do a quick approximation of the graph.
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Of course, you guys will do a much better job plotting points carefully.
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We have got time in seconds over here and we have got the log V and log volts over there.
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On our X, we go 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, something like that.
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Heading up, we need to make it to about 6V, it looks like something like that 1234567.
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When we go and actually plot our points, I end up with something that looks roughly,
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I'm going to draw my rough line first and then fill in the points so I'm not plotting them.
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But you get the general idea of what the shape of that is going to look like.
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And I have got a couple of points scattered above and below the line.
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I’m drawing the best fit line something kind of like this.
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That would be the data, draw a straight line that best fits your data points.
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For part D, from your line in part C, obtain the value of the time constant τ from the circuit.
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And we are going to use what we already talked about.
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We are going to use the fact for part D.
00:26:08.200 --> 00:26:16.800
Get the slope which is Y2 –Y1/ X2 – X1.
00:26:16.800 --> 00:26:29.500
As we pick a couple points on our line, that data points, points on our line is, in my case I ended up with something right around -0.079.
00:26:29.500 --> 00:26:43.300
That slope is also has a meaning, it is -1/ RC, which implies that is -1/ τ = -0.079.
00:26:43.300 --> 00:26:56.300
Therefore, τ = 1/ 0.079 or about 12.7 s in my calculation.
00:26:56.300 --> 00:27:04.700
Let us come check out part E, E1 in the experiment that the capacitor had a capacitance of 1.5 µf,
00:27:04.700 --> 00:27:08.300
calculate an experimental value for the resistance R.
00:27:08.300 --> 00:27:15.400
That is easy, if τ is 12.7 s and that is RC, it means RC = 12.7 s.
00:27:15.400 --> 00:27:37.000
Therefore, R = 12.7 s/ C, which it gave us as 1.5 µf, 1.5 × 10⁻⁶ F which is about 8.47 mega ohms with my values from the slope.
00:27:37.000 --> 00:27:45.300
Part E2, let us go to black here.
00:27:45.300 --> 00:27:49.500
On the axis that we made already, use a dashed line to sketch a possible graph
00:27:49.500 --> 00:27:54.900
if the capacitance was greater than 1.5 µf but the resistance R was the same.
00:27:54.900 --> 00:27:58.100
Justify your answer.
00:27:58.100 --> 00:28:08.500
Let us see, the capacitance was greater then RC would go up if capacitance was greater since C goes up.
00:28:08.500 --> 00:28:13.400
Our τ, our time constant is also going to go up.
00:28:13.400 --> 00:28:20.900
The slope is the inverse of τ.
00:28:20.900 --> 00:28:26.200
Our slope would have to decrease.
00:28:26.200 --> 00:28:38.900
I would draw a line that has a shallower slope, something like that perhaps.
00:28:38.900 --> 00:28:44.400
That is part E2 right there and that is how I would answer that one.
00:28:44.400 --> 00:28:51.700
I think that would leave you in pretty good stead on a problem like that.
00:28:51.700 --> 00:29:00.300
Let us move on and take a look at the 2012 exam free response number 2.
00:29:00.300 --> 00:29:07.600
In this one, we have an experimental setup where a student is measuring resistivity of slightly conductive paper.
00:29:07.600 --> 00:29:10.200
It gives us the thickness and we have got some data there.
00:29:10.200 --> 00:29:17.700
The first thing we are asked to do is use the grid to plot a linear graph of the data points from which the resistivity can be determined.
00:29:17.700 --> 00:29:23.200
Include labels and scales for both axis and draw the straight line that best represents the data.
00:29:23.200 --> 00:29:28.100
The way I would start with this, you have got data for resistance and length.
00:29:28.100 --> 00:29:30.500
Let us see what we need to graph there.
00:29:30.500 --> 00:29:38.000
I'm thinking R = ρ L / A, there is a relationship that includes the variables that we have,
00:29:38.000 --> 00:29:45.000
which by the way is ρ L/ the area, is just the width × the thickness.
00:29:45.000 --> 00:29:53.200
Then, resistance = ρ/ WT × L.
00:29:53.200 --> 00:29:59.200
If we want this to fit the equation of a line that is similar to Y = MX.
00:29:59.200 --> 00:30:08.000
If we plot L on our X axis or on our Y axis, the slope should be ρ/ WT.
00:30:08.000 --> 00:30:26.800
Let us take a look here and see if we can sketch in that graph.
00:30:26.800 --> 00:30:30.700
Over here, we are going to have our resistance in ohms.
00:30:30.700 --> 00:30:34.600
Here we will have our length in meters.
00:30:34.600 --> 00:30:45.500
We will have you guys pick appropriate, I have 200 J, 400 J, appropriate intervals there, labeled well 600 K.
00:30:45.500 --> 00:30:55.400
For length in meters, let us say that is 0.1, that is 0.2, and fill in some values in between.
00:30:55.400 --> 00:31:12.000
When I go and do this and actually plot the data, I get something that, with a couple of data points above, below.
00:31:12.000 --> 00:31:15.700
You are drawing your best fit line there.
00:31:15.700 --> 00:31:19.300
That looks remotely like what you should be finding there.
00:31:19.300 --> 00:31:24.800
In part B, says using the graph, calculate the resistivity of the paper.
00:31:24.800 --> 00:31:29.500
We have done the hard work there because we just said the slope is ρ/ ω T.
00:31:29.500 --> 00:31:51.400
If slope is ρ/ WT, that implies then that the resistivity ρ = WT × the slope which is 0.02 × our thickness 10⁻⁴.
00:31:51.400 --> 00:31:58.400
Our slope Y2 – Y 1/ X2 – X1, picking some points from your line not data points,
00:31:58.400 --> 00:32:04.100
wherever they happen to be, whatever your slope is, figure that out.
00:32:04.100 --> 00:32:17.600
I ended up with a value that for the whole thing of the resistivity of about 8.75 ohm meters.
00:32:17.600 --> 00:32:24.900
It should be somewhere in that ballpark.
00:32:24.900 --> 00:32:27.800
We are changing things up a little bit in this problem.
00:32:27.800 --> 00:32:33.500
A student uses those resistors R4 and R5 to build a circuit using wire, a 1.5V battery,
00:32:33.500 --> 00:32:37.500
an uncharged capacitor, and an open switch as shown.
00:32:37.500 --> 00:32:41.100
Calculate the time constant of the circuit.
00:32:41.100 --> 00:32:47.400
I really do not like the way that is drawn, circuit schematics tend make me happy, warm and fuzzy inside.
00:32:47.400 --> 00:32:50.200
I'm going to draw that in a way that is a little more comfortable for me.
00:32:50.200 --> 00:32:53.100
We will do that for part C.
00:32:53.100 --> 00:32:59.100
With a source of potential difference 1.5V, there is the positive and negative side.
00:32:59.100 --> 00:33:11.600
We then have a couple resistors in parallel, R4 and R5.
00:33:11.600 --> 00:33:20.900
We have our capacitor C and we have a switch.
00:33:20.900 --> 00:33:24.000
Something like that.
00:33:24.000 --> 00:33:29.500
Looking at that as an equivalent, we have two resistors in parallel.
00:33:29.500 --> 00:33:33.600
If we wanted to do that to find the equivalent resistance, R4 and R5,
00:33:33.600 --> 00:33:45.300
the equivalent resistance for R4 and R5 is just going to be R4 × R5/ R4 + R5.
00:33:45.300 --> 00:34:00.800
I should say that that is going to be our equivalent for all of those, we have 370,000 ohms or 440,000 ohms R5/ the sum of those two.
00:34:00.800 --> 00:34:07.800
What is it going to be, 781000 ohms.
00:34:07.800 --> 00:34:13.800
Which gives you about 200 kl ohms or 200,000 ohms.
00:34:13.800 --> 00:34:26.900
If we want to find the time constant of the circuit τ, our time constant is RC, is going to be our 200,000 ohms,
00:34:26.900 --> 00:34:42.800
the equivalent resistance of our circuit × our capacitance 10 µf or 10 ×⁻⁶ F, which is about 2 s.
00:34:42.800 --> 00:34:46.900
Part D, at time T = 0, the student closes the switch.
00:34:46.900 --> 00:34:52.100
On the axis, sketch the magnitude of the voltage across the capacitor
00:34:52.100 --> 00:34:57.600
and the voltages VR4 and VR5 across each resistors functions of time.
00:34:57.600 --> 00:35:00.700
The nice thing there is these are going to have the same voltages across them.
00:35:00.700 --> 00:35:05.200
That is going to be just one plot because they are connected together on the ends.
00:35:05.200 --> 00:35:11.900
Label each curve according to the circuit element and on the axis explicitly label any intercepts, things like that.
00:35:11.900 --> 00:35:16.100
Let us make ourselves another nice, happy, little graph.
00:35:16.100 --> 00:35:38.000
Part D, something like that, where we have our potential and time.
00:35:38.000 --> 00:35:41.100
If we start uncharged and then we are going to close that circuit,
00:35:41.100 --> 00:35:47.300
we know that the voltage across the capacitor is going to have an exponential increase to some final value,
00:35:47.300 --> 00:35:51.000
that is going to be the value of the voltage of our circuit.
00:35:51.000 --> 00:36:04.500
We can draw an asymptote here with about 1.5V where our voltage across our capacitor is going to do something like that.
00:36:04.500 --> 00:36:12.100
Where right around here, let us call that around approximately 5 τ or about 10s.
00:36:12.100 --> 00:36:16.200
That is easy.
00:36:16.200 --> 00:36:20.900
VR4 and VR5, are all of a sudden going to drop down.
00:36:20.900 --> 00:36:24.600
They no longer have that voltage across them so they start at 1.5V
00:36:24.600 --> 00:36:28.200
and they are going to have the corresponding exponential decay.
00:36:28.200 --> 00:36:34.400
I think I can draw it a little prettier, that was not the best job ever.
00:36:34.400 --> 00:36:48.900
Something perhaps kind of like that, where this is the voltage across R4 which is equal to the voltage across R5.
00:36:48.900 --> 00:36:51.300
That should cover you for that one.
00:36:51.300 --> 00:37:02.700
The general graph above the voltage across our resistors and across your capacitor as a function of time.
00:37:02.700 --> 00:37:05.600
Let us take a look now at the 2007 exam.
00:37:05.600 --> 00:37:07.900
Take a minute, printout, look it over.
00:37:07.900 --> 00:37:11.600
Hit the pause if you need to, come back.
00:37:11.600 --> 00:37:16.800
E and M number 1 from 2007.
00:37:16.800 --> 00:37:22.000
Here we have another RC circuit, it says a student sets up that circuit in the lab.
00:37:22.000 --> 00:37:29.800
The values of the resistance and capacitance are as shown but the constant voltage delivered by the ideal battery is not known.
00:37:29.800 --> 00:37:34.000
At time T = 0, the capacitor is uncharged and the student closes the switch.
00:37:34.000 --> 00:37:38.100
It gives us a graph of the current as a function of time measured using a Computer System
00:37:38.100 --> 00:37:41.700
and you get that nice, happy, little graph that they show you.
00:37:41.700 --> 00:37:47.700
For part A, using the data, calculate the battery voltage E.
00:37:47.700 --> 00:38:02.500
The current at time T = 0, we can see right from the graph is about 2.25 milliamps or 0.00225 amps.
00:38:02.500 --> 00:38:09.800
To find the battery voltage, that is just going to be equal to IR in our circuit by Ohm’s law
00:38:09.800 --> 00:38:26.500
which will be our current 0.00225 amps × our resistance, we have got 550 ohms there, for a voltage of about 1.24.
00:38:26.500 --> 00:38:33.300
Part B, calculate the voltage across the capacitor at time T = 4s.
00:38:33.300 --> 00:38:36.400
Let us take a second and draw our circuit.
00:38:36.400 --> 00:38:38.800
Here we go, there is E.
00:38:38.800 --> 00:38:57.000
We have got our switch, we have got our resistor that is 550 ohms, we have our capacitor which is 4000 µf.
00:38:57.000 --> 00:39:04.300
What I'm going to do is, I'm going to go through and do a KVL around the loop, starting right there going clockwise.
00:39:04.300 --> 00:39:23.600
I would write that - E +, we will define to the right is the direction of our current - E + IR + the voltage across our capacitor = 0.
00:39:23.600 --> 00:39:29.500
Therefore, DC = E – IR.
00:39:29.500 --> 00:39:42.000
But we know that the current at 4s I at T = 4s is equal to 2.35 milliamps.
00:39:42.000 --> 00:39:44.900
Roughly right from the graph.
00:39:44.900 --> 00:40:02.600
That then means that VC is going to be equal to, we have got our 1.24V – our I 0.00035 amps × our resistance 550 ohms.
00:40:02.600 --> 00:40:11.200
So I would get a VC of around 1.05V when I put that into my calculator.
00:40:11.200 --> 00:40:13.300
That should cover part B.
00:40:13.300 --> 00:40:23.400
Moving on to part C, calculate the charge on the capacitor at T = 4s.
00:40:23.400 --> 00:40:47.900
Q on the capacitor is charge × voltage, which is going to be our 4000 µf, 4000 × 10⁻⁶ F × our potential 1.05V or about 0.0042 C.
00:40:47.900 --> 00:40:50.500
It looks like we are making some headway here.
00:40:50.500 --> 00:40:55.200
Let us take a look at part D, give ourselves more room here in the next page.
00:40:55.200 --> 00:41:01.400
It asks us to sketch a graph of the charge on the capacitor as a function of time.
00:41:01.400 --> 00:41:10.000
Let us draw our axis in here.
00:41:10.000 --> 00:41:19.400
Our Y axis, our T axis.
00:41:19.400 --> 00:41:23.300
There we have T, there is our Q.
00:41:23.300 --> 00:41:26.900
What is that going to look like, the charge on the capacitor as a function of time?
00:41:26.900 --> 00:41:29.800
We can do this over and over by now.
00:41:29.800 --> 00:41:34.000
We are going to have that exponential rise as it charges up.
00:41:34.000 --> 00:41:39.700
There is part D, let us take a look at E.
00:41:39.700 --> 00:41:44.400
Calculate the power being dissipated as heat on the resistor at 4s.
00:41:44.400 --> 00:42:12.500
Power = I² R which is going to be our current 4s at 0.00035 amps² × our resistance 550 ohms or about 6.74 × 10 ⁻5W.
00:42:12.500 --> 00:42:23.400
Onto F, the capacitor is not discharged but it is dielectric of constant K κ = 1 is replaced by dielectric of constant K = 3.
00:42:23.400 --> 00:42:25.600
It triple the dielectric constant.
00:42:25.600 --> 00:42:30.700
The procedure is repeated, is the amount of charge on one plate of the capacitor at 4s greater than,
00:42:30.700 --> 00:42:35.400
less than, or the same as before, and we have to justify our answer.
00:42:35.400 --> 00:42:42.900
I would start by looking at the formula for capacitance K ϵ₀ A/ D.
00:42:42.900 --> 00:42:47.400
But in this case, we know that K triples.
00:42:47.400 --> 00:42:50.200
All those other things are not going to change.
00:42:50.200 --> 00:42:52.300
It is the same capacitor geometry.
00:42:52.300 --> 00:43:06.900
Therefore, we can say that our capacitance is going to triple which implies then that we are going to have a greater amount of charge.
00:43:06.900 --> 00:43:21.700
We can look at that as, with 3 κ, Q3 is going to be E × 3, the initial capacitance, × 1 – E ⁻T/ RC.
00:43:21.700 --> 00:43:24.100
RC is our initial capacitance still.
00:43:24.100 --> 00:43:39.400
That would be 3 EC × 1 - E⁻⁴/ 3 × our resistance 550 × our capacitance 4000 × 10⁻⁶.
00:43:39.400 --> 00:43:44.100
Just to give you a feel for the numbers, complies then that that is going to be,
00:43:44.100 --> 00:44:00.800
Q3 will be, 3EC 5454 µf × about 1.24V or 6.76 × 10⁻³ C which is 0.00676 C.
00:44:00.800 --> 00:44:08.300
Which by the way, as we suggested already when we said it was greater,
00:44:08.300 --> 00:44:23.900
it is greater than what we initially had the initial 0.0042 C.
00:44:23.900 --> 00:44:33.300
Plug in right through these, let us move on and take a look at the 2006 exam free response number 2.
00:44:33.300 --> 00:44:36.400
Take a minute, find the problem, download it, print it out.
00:44:36.400 --> 00:44:38.500
Give it a try, hit pause, comeback.
00:44:38.500 --> 00:44:41.100
If you have had a few minutes to look it over and we will go through this one.
00:44:41.100 --> 00:44:47.300
This one is a bit of a doozy.
00:44:47.300 --> 00:44:52.600
The circuit that they show you has a capacitor of capacitance C, a power supply of EMF given E,
00:44:52.600 --> 00:44:55.800
two resistors R1 and R2, and a couple of switches.
00:44:55.800 --> 00:45:00.400
Initially the capacitor is uncharged when both switches are open.
00:45:00.400 --> 00:45:04.900
Switch S1 then gets close at time T = 0.
00:45:04.900 --> 00:45:10.900
Part A says, write a differential equation that can be sought to obtain the charge on the capacitor as a function of time T.
00:45:10.900 --> 00:45:14.000
We started out the lesson with something like this but let us do it.
00:45:14.000 --> 00:45:16.300
A, let us take a look.
00:45:16.300 --> 00:45:23.100
We have our source of EMF, our potential difference.
00:45:23.100 --> 00:45:27.300
I’m just going to redraw this a little bit so it is a little easier to see.
00:45:27.300 --> 00:45:40.000
R1, we have to capacitor and it should look like that initially based on where the switches are.
00:45:40.000 --> 00:45:43.200
That both are switches are open and S1 gets close to T = 0.
00:45:43.200 --> 00:45:46.000
Right at T = 0, this is basically our functioning circuit.
00:45:46.000 --> 00:45:48.100
Makes it look a little simpler.
00:45:48.100 --> 00:45:52.800
We also know of course, τ was going to be RC in the circuit.
00:45:52.800 --> 00:46:02.900
Let us do KVL around the loop again, - E + IR1 + VC = 0.
00:46:02.900 --> 00:46:14.400
Which implies then, since VC = Q/ C, that we have - E + IR1 + Q/ C = 0.
00:46:14.400 --> 00:46:19.600
We also know that I is DQ DT, that is charging up.
00:46:19.600 --> 00:46:32.300
Then, we get - E + R1 DQ DT + Q/ C = 0.
00:46:32.300 --> 00:46:33.700
There is our differential equation.
00:46:33.700 --> 00:46:38.400
We have Q and the derivative of Q in the same equation.
00:46:38.400 --> 00:46:40.800
That should cover us for part of A.
00:46:40.800 --> 00:46:43.900
Typically, that is as far as they ask you to go on most questions.
00:46:43.900 --> 00:46:46.300
In this one, they ask you actually go and solve it.
00:46:46.300 --> 00:46:51.500
In part B, solve the differential equation to determine the charge on the capacitor as a function of time.
00:46:51.500 --> 00:46:53.900
Exact same thing we have been doing.
00:46:53.900 --> 00:47:01.200
Chances are you could probably even, based on this tell me what the answer is without even going through the derivation.
00:47:01.200 --> 00:47:04.800
However, to get full credit you probably got to go through and do all the work.
00:47:04.800 --> 00:47:07.900
Let us do that.
00:47:07.900 --> 00:47:10.500
Let us pull the R1, divide everything by r1.
00:47:10.500 --> 00:47:23.300
We have - E / R1 + DQ DT + Q/ R1 C = 0.
00:47:23.300 --> 00:47:37.600
A little bit of rearrangement which implies then that DQ DT = E/ R1 -Q/ R1 C, which implies that DQ DT =,
00:47:37.600 --> 00:47:41.300
we will get a common denominator here, multiply this one by C.
00:47:41.300 --> 00:47:43.900
We have our 1C in the denominator.
00:47:43.900 --> 00:47:53.300
That will give us EC -Q/ R1C.
00:47:53.300 --> 00:48:06.500
Then separating our variables, this implies that DQ/ EC -Q = DT/ R1C.
00:48:06.500 --> 00:48:11.600
Which implies then that DQ and is going to switch the order here with a negative sign,
00:48:11.600 --> 00:48:20.200
multiplying through Q - EC = - DT/ R1C.
00:48:20.200 --> 00:48:23.000
Now we can integrate both sides.
00:48:23.000 --> 00:48:28.500
We will integrate the left hand side from some Q = 0 to final value Q.
00:48:28.500 --> 00:48:34.200
The right hand side will be from some T = 0 to final value T
00:48:34.200 --> 00:48:41.300
which is going to give us the left hand side integral of the DU/ U will be the nat log of U.
00:48:41.300 --> 00:48:55.900
We will get something that looks like the log of Q - EC evaluated from 0 to Q = - T / R1C.
00:48:55.900 --> 00:49:13.000
Which implies then that the log of Q - EC - the log of - EC = - T/ R1C.
00:49:13.000 --> 00:49:18.200
The left hand side, the log of the difference is equal to the log of the quotient.
00:49:18.200 --> 00:49:21.400
The difference of logs is the log of the quotient.
00:49:21.400 --> 00:49:40.400
That will give us the log of Q - EC/ -EC = Q – T/ R1C, raising both of these to the E.
00:49:40.400 --> 00:49:43.000
E ⁺log of that just gives us that piece.
00:49:43.000 --> 00:49:53.600
We will have Q - EC/ -EC = E ^¬T/ R1C.
00:49:53.600 --> 00:50:08.500
Multiply it through by that EC so we get Q - EC is equal to – EC E ^¬T/ R1C.
00:50:08.500 --> 00:50:24.600
Or getting Q by itself, Q = EC × 1 - E ^-¬T/ R1C.
00:50:24.600 --> 00:50:28.800
You earn your points on that one.
00:50:28.800 --> 00:50:35.500
Or we could also take a look at that Q = EC 1 - E -T / R1C.
00:50:35.500 --> 00:50:39.700
I think that works.
00:50:39.700 --> 00:50:47.500
Let us give ourselves more room before we move on to part C.
00:50:47.500 --> 00:50:54.800
For part C, it says determine the time at which the capacitor has a voltage of 4V across it.
00:50:54.800 --> 00:51:09.400
To do that, we know Q, let us see if we can solve for V and then we can back out the time from the voltage.
00:51:09.400 --> 00:51:30.200
If Q = CV which is equal to EC × 1- E ^-¬T/ R1C, that implies then that V is just equal to E × 1 – E ^-¬T/ R1C.
00:51:30.200 --> 00:51:42.400
Or V/ E we will pull out that -1, = –E ^-¬T/ R1C.
00:51:42.400 --> 00:51:46.600
Which implies then that, let us switch our negative signs around.
00:51:46.600 --> 00:52:07.100
1 - V / E = E ^-¬T/ R1C, which implies then that the log of 1 - V/ E = – T/ RC.
00:52:07.100 --> 00:52:18.600
Or getting T all by itself, T = – RC × the log of 1 - V/ E.
00:52:18.600 --> 00:52:21.700
We can substitute in our values to find that time.
00:52:21.700 --> 00:52:36.800
T is equal to -4700 × 0.06 × our log of 1 - 4/ 12.
00:52:36.800 --> 00:52:50.600
Put that all on my calculator and I come up with a time of about 114.3s.
00:52:50.600 --> 00:52:57.600
After switch S1 has been closed for a long time, switch S2 now gets close in new time T = 0.
00:52:57.600 --> 00:53:04.900
For part D, we have got a new T = 0 configuration which looks kind of like this.
00:53:04.900 --> 00:53:13.800
We have got our source of potential difference, over here we have got our resistor R1.
00:53:13.800 --> 00:53:20.800
We come down here, we have our capacitor C.
00:53:20.800 --> 00:53:27.800
We come down here, we have got R2.
00:53:27.800 --> 00:53:32.500
There we go, those two are in parallel.
00:53:32.500 --> 00:53:39.800
Sketch graphs of the current I1 and R1 vs. Time and of the current I2 and R2 vs. Time.
00:53:39.800 --> 00:53:42.900
Clearly label which is I1 and I2.
00:53:42.900 --> 00:53:44.100
I guess that is not so bad.
00:53:44.100 --> 00:54:03.400
Let us draw our axis here.
00:54:03.400 --> 00:54:07.800
As I look at this, it looks like I1, the current through R1 is going to start at 0,
00:54:07.800 --> 00:54:14.100
it is going to go to some center point to some final value toward an asymptote.
00:54:14.100 --> 00:54:20.600
It looks like I2 is going to start at a maximum current when we first do that and approach the same point,
00:54:20.600 --> 00:54:23.700
so from slightly different directions here.
00:54:23.700 --> 00:54:30.700
As I take a look at that because those resistors are equal, they are going to split the voltage across them and end up with the same current.
00:54:30.700 --> 00:54:34.600
I would probably draw this something kind of like this.
00:54:34.600 --> 00:54:40.500
I would expect that I1 to go like this.
00:54:40.500 --> 00:54:45.400
I will label my axis here, there is our current, here is our time.
00:54:45.400 --> 00:54:55.200
I want to do something like that and I2 to come in from about the same point and do something like that,
00:54:55.200 --> 00:55:01.900
where they are getting closer and closer to each other but not quite meeting.
00:55:01.900 --> 00:55:09.500
Those should be symmetric but we have got to think you have got the right idea of what that graph should look like.
00:55:09.500 --> 00:55:11.400
You can perhaps a little closer to that.
00:55:11.400 --> 00:55:15.300
And that should cover part D.
00:55:15.300 --> 00:55:24.200
Alright, that finishes up that problem and let us see if we can do one more.
00:55:24.200 --> 00:55:29.900
Let us look at the 2003 AP physics C E and exam free response number 2.
00:55:29.900 --> 00:55:33.800
As always, we will have you take a minute, look it up, print it out if you can.
00:55:33.800 --> 00:55:41.900
Give it a shot, come back here, and we will see how this one goes.
00:55:41.900 --> 00:55:48.100
In a lab, you can connect a resistor and a capacitor with unknown values in series with a battery of EMF 12V.
00:55:48.100 --> 00:55:51.400
You include a switch in a circuit, and when the switch is closed the circuit is completed.
00:55:51.400 --> 00:55:58.400
You measure the current through the resistor as a function of time, as they would show you in a plot below.
00:55:58.400 --> 00:56:03.600
Using common symbols, draw the circuit that you have constructed that does this.
00:56:03.600 --> 00:56:06.400
Show the circuit before the switch is closed and include whatever other devices
00:56:06.400 --> 00:56:10.300
you need to measure the current through the resistor to obtain that plot.
00:56:10.300 --> 00:56:13.400
Label the components in the diagram.
00:56:13.400 --> 00:56:21.500
That is not so bad, we need to set up an RC circuit where we can measure the current through the capacitor and the switch S1.
00:56:21.500 --> 00:56:24.100
We are showing it before the switch has been closed.
00:56:24.100 --> 00:56:30.100
We have got our source of potential difference E and we will label that.
00:56:30.100 --> 00:56:38.400
We will label that, we would make it nice and clear.
00:56:38.400 --> 00:56:52.200
We have some switch S1, we have a resistor R, we have a capacitor C.
00:56:52.200 --> 00:57:01.100
Somewhere in this series configuration, I will put E in here.
00:57:01.100 --> 00:57:06.400
Just to be safe, I would probably go through and actually write capacitor, resistor, switch, ammeter.
00:57:06.400 --> 00:57:08.500
I will let you guys do that.
00:57:08.500 --> 00:57:14.000
Having obtained the curve shown, determine the value of the resistor at two placed in the circuit.
00:57:14.000 --> 00:57:22.800
The way I would do that one is I would first look at the point where you have got T = 0 because you can use a ohms law R = VI.
00:57:22.800 --> 00:57:27.200
At T=0, the voltage across the capacitor is 0.
00:57:27.200 --> 00:57:36.300
Let me write that at T = 0, VC = 0, which implies that the voltage across R is equal to EMF,
00:57:36.300 --> 00:57:54.500
which implies that the resistance is just going to be the EMF ÷ the current which is 12V/ 0.01 amps or about 1200 ohms.
00:57:54.500 --> 00:57:59.500
C, what capacitance you need to insert in the circuit to give that result?
00:57:59.500 --> 00:58:10.600
I would take a look here and say, our time constant RC must be 4s from the formula
00:58:10.600 --> 00:58:13.800
that they gave us for the current as a function of time.
00:58:13.800 --> 00:58:34.800
Therefore, the capacitance must be 4s/ R which is 4s/ 1200 ohms or about 3.3 × 10⁻³ F.
00:58:34.800 --> 00:58:38.600
Let us check part D, give ourselves some more room here
00:58:38.600 --> 00:58:44.000
because it looks like we got a significantly different portion of the question.
00:58:44.000 --> 00:58:49.100
For part D, you are now asked to reconnect the circuit with a new switch so as to charge
00:58:49.100 --> 00:58:56.900
and discharge the capacitor and be able to get a graph like that when you are switching between positions A and B.
00:58:56.900 --> 00:59:01.300
Draw that schematic and label anything you might need.
00:59:01.300 --> 00:59:06.700
It looks like in part A, it is charging in the switches at A and at part B it is discharging.
00:59:06.700 --> 00:59:09.500
There are bunch of ways you can draw this.
00:59:09.500 --> 00:59:11.900
Electrically they should all be equivalent though.
00:59:11.900 --> 00:59:17.900
I would start off with something like our battery.
00:59:17.900 --> 00:59:21.800
Be good and label this, our battery with an EMF.
00:59:21.800 --> 00:59:26.500
Let us put this switch over here and we will call that position A.
00:59:26.500 --> 00:59:34.600
We will put the switch maybe right here, we will call that S1.
00:59:34.600 --> 00:59:40.600
We will have another position for it down here called B.
00:59:40.600 --> 00:59:45.200
As we go through here, we need our resistor in the circuit.
00:59:45.200 --> 00:59:59.300
We need our capacitor, they want to measure the current flow so we will put an ammeter in here in line with that in series.
00:59:59.300 --> 01:00:01.400
We will connect there so different position B.
01:00:01.400 --> 01:00:04.200
All we have is this loop to discharge.
01:00:04.200 --> 01:00:05.400
I will just continue that up here.
01:00:05.400 --> 01:00:09.600
In position A, this does not make any difference, it is an open circuit.
01:00:09.600 --> 01:00:11.900
In position B, the whole left hand side is missing.
01:00:11.900 --> 01:00:17.100
We have got our battery and we have got our ammeter here.
01:00:17.100 --> 01:00:22.100
We have got our capacitor.
01:00:22.100 --> 01:00:36.200
If we want to measure the voltage across all of that, let us put our voltmeter in parallel with our capacitor.
01:00:36.200 --> 01:00:45.300
We have got our resistor up there and there is our switch.
01:00:45.300 --> 01:00:48.400
I think that should cover all the requirements.
01:00:48.400 --> 01:00:54.100
Hopefully, you got pretty good feeling for transient analysis of RC circuits.
01:00:54.100 --> 01:00:59.000
It takes some time, you might want to go back and do some of these analyses a couple of times
01:00:59.000 --> 01:01:02.100
until it starts to make sense to you, until you start doing the patterns.
01:01:02.100 --> 01:01:04.900
Thank you very much for watching www.educator.com.
01:01:04.900 --> 01:01:07.000
We will see again real soon, make it a great day everybody.