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 RC circuits, circuits that have resistors and capacitors.
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We are going to focus on them in the steady state as opposed to the transient analysis.
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What happens as a function of time, we will come up in our next lesson.
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Our objectives include calculating equivalent capacitance for capacitors in series and parallel configurations.
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We have done that before, we are just going to review it a little bit.
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Describe how stored charges divided between capacitors in parallel.
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Determine the ratio of voltages for capacitors in series.
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Calculate the voltage or stored charge under steady state conditions for capacitor connected to a circuit consisting of a battery and resistor network.
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Finally, sketch graphs of current stored charge and voltage for a capacitor or resistor in one of these RC circuits.
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Very quickly, let us review capacitors in parallel.
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We know capacitor store charge on plates and capacitors in parallel can be replaced with an equivalent capacitor.
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We talked about doing that, the equivalent capacitance for capacitors in parallel is just the sum of the individual capacitances.
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For capacitors in series, the charge on them must be the same and they can be replaced with an equivalent a capacitor, following 1/ the equivalent capacitance is 1/ C1 + 1/ C2, and so on.
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As we get into RC circuit specifically then, RC circuits are comprised of a source of potential difference to make the current flow, a resistor network and 1 or more capacitors.
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We are going to look at RC circuits from the perspective now of what happens when their first turned on and what happens after a long time.
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What is in the middle, we are going to leave in the next lesson.
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At least we are starting out.
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The key to understanding RC circuit performance, uncharged capacitors act like wires.
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Charged capacitors act like opens.
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You got to know these 2 facts, it is going to help you immensely with your analysis of capacitor circuits.
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Let us take a look at charging an RC circuit.
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Here we have a source of potential difference some VT, we have got a resistor R, we have a capacitor C,
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and we need to find the positive and negative side of our capacitor,
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and at time T = 0 we are going to close the switch and see what happens.
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Initially, a capacitor acts like a wire when it is uncharged.
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If we were to take a look at this and we are going to follow a Kirchhoff’s voltage law path around that way,
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I could write that -VT because starting here we see the negative side first.
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+ IR + the voltage across our capacitor is going to equal 0.
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We also know that our capacitor is charge ÷ voltage across our capacitor.
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Therefore, the voltage across our capacitor is Q ÷ C.
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I can rewrite my equation as - VT + IR + Q/ C must equal 0.
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But we also said that time T = 0.
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When we first closed that switch, the charge on our capacitor is going to be equal to 0, it is uncharged.
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That simplifies our analysis a little bit.
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And we have - VT + IR equal 0 which implies that VT = IR.
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Probably not a surprise.
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VC therefore, must equal 0.
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If we were to do our plot of the different things we have down here, initially the current flowing through our circuit is going to be VT/ R.
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We can make 1 point over here for the current through our circuit and call that VT/ R time equal 0.
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We also know the charge on our capacitor at time equal 0 is equal to 0.
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We also know the voltage across our capacitor because Q is 0, we will start at 0.
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There is our initial analysis.
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Let us take a look and assume that the thing is charge up, it has been a long time.
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The long time is something we are going to define here a couple of minutes.
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After a long time, the switch has been closed for a while, the capacitor now acts like an open.
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It does not allow current to flow .
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As T approaches infinity, as it gets very big and we apply Kirchhoff’s voltage law to the same loop,
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go around the same way, I have -VT + IR + VC = 0.
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Since I equal 0, we find that VT = VC.
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No current flowing so we have the same voltage across the source of potential difference that we do across the capacitor.
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Our current went down to 0, it is going to follow a path something like this and exponentially up to this point that we are going to call 5 τ 5 time constant.
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We will talk about exactly what that time is here in a moment.
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The capacitor must be charging up during this time.
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When it is fully charged, we know that it must have a charge of CVT on.
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I’m going to draw an asymptote in here at CVT, when it is fully charged.
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It is going to have an exponential increase toward that value, getting really close to it about 5 τ 5 time constant.
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The voltage across our capacitor started at 0.
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We know after long time, we just determined that, that was going to be equal to VT.
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We are going to follow that same asymptotic relationship getting very close when we get to about 5 τ.
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How about if we were to discharge an RC circuit?
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We have pulled the τ supply or the source of potential difference out of here, our capacitor starts out charged
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and at time T equal 0, we are going to close the switch.
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When we do that, let us go and we will go around our circuit this way for Kirchhoff’s voltage law.
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The first thing I see, if I start down here at this blue point, I see - VC + IR, that brings me back to my initial point = 0.
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We also know that I then is VC/ R with a little bit of rearrangement.
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We also know that charge is going to be CVC.
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Initially, current is VC/ R so I can draw to that in my graph down here, we start at VC/ R for our current flow.
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Our charge is going to be C × VC, which initially CV is 0 for this problem which was CVT.
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We are going to start at a fully charged capacitor, we will call that CV0 for this specific problem.
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The voltage across our capacitor also has to start fully charged at whatever V0 happens to be.
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We let some time elapse after a long time.
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Now a capacitor as it becomes uncharged, it is going to act like a wire.
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At that point, as we go around and write our Kirchhoff's current law equation.
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Starting here, I have -VC + IR equal 0, but since it is acting like a wire, VC equal 0 that implies then that I must be 0 and Q equal 0.
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We are going to get these exponential decay.
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Current is going to DK something like that.
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Our charge is going to decay with that 0 line as an asymptote.
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Our potential is going to do the same basic thing in exponential decay.
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It gets pretty close to about 5 τ so that is what happens as we look at this from a steady state looking at right when the switch is open or closed.
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And right at the other end, after a long time.
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Let us talk a little bit about this time constant τ, squiggly lowercase t or τ.
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The time constant in an RC circuit is equal to the resistance × the capacitance indicates that time
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at which the quantity under observation has achieved 1 – E⁻¹ or roughly 63% of its final value.
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Why is it important?
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After 1 time constant, you are closing in on 2/3 to your final value.
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By 5 time constants, the quantity under observation is within 1% of its final value.
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What you are talking about a long time with asymptotes that can be kind of confusing.
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If you pick 5 τ, you are pretty much all the way to your final value.
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You are within 99% of it.
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We are going to use this time constant quite regularly, τ = RC for an RC circuit.
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Let us take a look at a little bit of an RC analysis here with an example problem.
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What is the current through R2 when the circuit is first connected and we got our capacitor down here C1 of 5 µf.
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What is the current through R2 a long time after the circuit has been connected?
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When it is first connected, that capacitor acts like a wire so we can almost pretend it is not even there.
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And I see a combination circuit where we have got a series portion and a parallel portion.
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I could redraw that if I wanted to, to find the current through our circuit.
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Finding the equivalent resistance for these 2 to make an equivalent series type circuit.
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I’m going to do that, let us call that our 20V.
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There is our 200 ohm R1 and over here we have R2, 3 equivalent.
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That is what our circuit looks like from an electrical perspective, when that capacitor is acting like a wire, when it is first turned on.
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The value of R2,3 is going to be R2 × R3, 400 × 300/ 700, which is at about 171 ohms.
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That allow me to find the total current in the circuit, I = V/ R which is going to be our 20V ÷ our total resistance.
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If this is 171, that is 200, we divide by 371.
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171 + 200 to come up with a current of about 0.0539 amps.
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That means we have got 0.0539 amps walking through there.
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We have that current 0.0539 amps and if this is what I'm going to call my ground here, 0 V.
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0.0539 × 200, that means we are going to drop just over 10 V, about 10.78 V.
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Meaning our potential up here is going to be about 9.22 V.
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We can find the individual current through R2.
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I2, the current through R2 then, is going to be V2/ R2, which will be 9.22 V/ 400 ohm, which is going to come out to be about 0.0231 amps.
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There is our first answer.
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What is the current through R2 when the circuit is first connected?
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0.0231 amps.
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After a long time, or at least 5 τ as an approximation.
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Then, our capacitor acts like an open, it is fully charged.
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Our circuit starts to look really simple then, we have our source of potential difference or 20 V.
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We have R1 over here, 200 ohms.
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Nothing is going to flow through this right branch because we have got an open.
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The only thing we have to worry about here is our 400 ohm R2.
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Now, current = V/ R which is going to be 20 V our total resistance 600 ohms, or about 0.0333 amps.
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There is our current after a long time.
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An example where we are using this steady state analysis of an RC circuit, to solve a slightly more complex problem.
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Let us do another one, here we have a circuit that has a source of potential difference.
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We have got 3 resistors and another capacitor, what is the current through R3 when the circuit is first connected?
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What is the current through R2 a long time after it has been connected?
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Let us look when it is first connected.
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This is uncharged so it acts like a wire.
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Our circuit is going to look something like this.
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We got our source of potential difference, we have got our 100 ohm R1, we have got 200 ohm here, and we have got 200 ohm here.
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I’m going to find the equivalent resistance of these two.
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200 × 200/ 200 + 200 or by inspection I know that that is going to be half the value, since they are identical 2 capacitors in parallel.
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What we are going to have is our source of potential difference,
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our 100 ohm resistor and our equivalent resistance right over here and each is going to be 100 ohm.
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If this is 10 V, pretty easy to see that the voltage is going to be divided evenly between them
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because they got an equivalent equal resistance, that point must be 5 V.
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If that is the case, then that means when over here, this must be 5 V when the circuit is first turned on.
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We can find the current through R3, when it is first turn on.
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It is just going to be V/ R or that 5 V ÷ 200 ohm which is 0.025 amps.
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So there is our current through R3 when the circuit is first connected.
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Let us take a look what happens after it is a long time after it is been connected, to find the current through R2.
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After a long time, this is going to act like an open current through R2, a long time after the circuit has been connected.
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We will have 10 V, we will have R1, and we will have our 200 ohm, 100 ohm R1, we have a 200 ohm R3.
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This is open so it is not even in play at the moment.
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What do we have here?
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Our current is going to be V/ R which is going to be 10 V/ 300 ohm, which is 0.0333 amps, that is our current through R1 and through R3.
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But it asks us, what is the current through R2 a long time after the circuit has been connected?
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A long time after the circuit has been connected through R2, there is nothing because we have got an open branch.
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Its capacitor is open, nothing flows through R2.
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The current through R2 after long time is going to be 0.
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Taking a look at some equivalent capacitances.
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What is the equivalent capacitance of the capacitor network shown here?
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You have got a 5 µf, got 210 µf in parallel, and a 5.
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I can redraw that to try and make something a little bit simpler for me to see as 5 µf,
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we will replace these 2 in parallel with their equivalent.
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The capacitors in parallel add up so that is the equivalent of 20 µf capacitor.
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And down here, we have got our other 5 µf capacitor.
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Now I can use my equivalent capacitance equation for capacitors in series.
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1/ the equivalent capacitance is 1/5 µf + 1/ 20 µf.
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Keeping in mind that micro × to 10⁻⁶ + 1/ 5 µf or 1/ the equivalent capacitance is going to be 450,000, 1/ F,
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which implies then that the equivalent capacitance is going to be 1/ 450,000, which is 2.22 × 10⁻⁶ F or 2.22 µf.
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How about another equivalent capacitance problem?
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When you look at things like this, you usually tend to be a little tricky when you first look at them.
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See if you can simplify and put in a format that looks a little bit more easy on your eyes.
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As I look at this one, 45° angle start to just mess with my brain.
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Let us see if we can straighten this out a little bit.
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I could redraw this, it looks down on the left hand side we have got C to another capacitor C,
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to another capacitor C, and then we come back to some point.
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On the right hand side, we just have 1 capacitor that is the equivalent circuit to that.
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That looks a whole lot easier.
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We have got these 3 C that are in series.
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Let us find that they are equivalent, 1/ C equivalent is going to be 1/ C + 1/ C + 1/ C which is 3/ C.
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Therefore, C equivalent for these 3 is just going to be C/ 3.
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I will redraw this now as, I got a C/ 3 here and that still that my C from over here on the right.
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Capacitors in parallel, we just add them up.
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Our total equivalent capacitance, C + C/ 3 is going to be 4/ 3 C.
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Let us take a look and do a couple of practice AP problems.
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We are going to take a look at the APC E and M free response number 2 problem.
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You can find a link to it here at the top.
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Search it on www.google.com, it is pretty easy to find.
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Take a minute and print out the questions, see if you can do it and then follow along.
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If you get stuck, use that to keep you going or if you get through the whole thing, excellent, you can check your answer.
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Let us take a minute here and look at this problem.
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We have got a circuit that has a couple capacitors in it with a switch.
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Initially switch S is open and the battery has been connected for a long time.
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What is the steady state current through the ammeter?
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As we look at part A here, after a longtime, the current through the capacitor is going to be 0.
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Therefore, you have 0 current through the ammeter.
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Let us write that here.
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After a long time, the current through the capacitor equal 0.
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Therefore, you have 0 current through your ammeter.
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Because at that point, if you were to draw the circuit, it kind of looks like this.
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Ignoring that switch because nothing is going through it while it is open.
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You have got a resistor, a capacitor, and your ammeter.
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You have another resistor, another to capacitor, and the whole thing comes back to your source of potential difference.
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But if it is fully charged up, no current flows.
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Therefore, this one is pretty easy.
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Part B says, calculate the charge on the 10 µf capacitor, this capacitor up here.
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To do that, C = Q/ V which implies then the charge = C × V, which we know is 10 µf, 10 × 10⁻⁶ F × the potential difference across it.
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As we look at that, that is got be 30 V.
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We end up with 300 µc.
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We got out 30 V power supply here, 40 ohm, 20 ohms, 5 µf, 10 µf.
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For part C, calculate the energy stored in the 5 µf capacitor.
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The potential energy there is going to be ½ CV² which is ½ × our capacitance, 5 × 10⁻⁶ F × the square of our potential difference 30 V² to be 2250 mJ, or 2.25 mJ.
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Just highlighting our answers.
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We had 0 current 300 µc and 2.25 mJ.
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Next, it is going to ask us to close the switch and after we come into a new steady state,
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calculate the steady state current through the battery.
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Let us go to a new page, give ourselves some room here.
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I'm going to redraw the circuit now that we have that switch closed, see if we can make that a little bit clearer.
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If I were to draw it, I’m going to draw it in a way that is a little bit more familiar to me.
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We will that our 30 V source of potential difference.
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We have our 5 µf capacitor here.
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We have a 10 µf capacitor here, and we also have over here our resistor network 40 ohm, 20 ohm,
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and we have a connection right there where the switches to be.
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Rearrange a little bit with an equivalent circuit.
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We are trying to find what the current is through the battery.
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As I look at that then, I see we have got 30 V across our 2 resistors here, that should be pretty straightforward.
00:24:07.600 --> 00:24:22.400
I = V/ R which is going to be 30 V/ 60 ohm or 0.5 amps.
00:24:22.400 --> 00:24:30.000
Moving on to part E, calculate the final charge on the 5 µf capacitor.
00:24:30.000 --> 00:24:38.900
We already said charge is capacitance × voltage which will be 5 µf, 5 × 10⁻⁶ F.
00:24:38.900 --> 00:24:50.400
Our voltage here is going to be 20 V across our 5 µf capacitor because we have got a basic voltage divider here.
00:24:50.400 --> 00:24:56.000
We are going to drop 20 V across our 40 ohm resistor and 10 V across our 20.
00:24:56.000 --> 00:25:05.500
That is going to be 100 µc.
00:25:05.500 --> 00:25:14.900
Let us take a look now at part F, calculate the energy dissipated as heat in the 40 ohm resistor in 1 minute, once a circuit has reached steady state.
00:25:14.900 --> 00:25:22.200
It sounds like we need to find the power first and then multiply that power by the time, in order to find the energy expended.
00:25:22.200 --> 00:25:35.000
Power is I² R is going to be our ½ amp² × 40 ohm resistance or 10 W.
00:25:35.000 --> 00:25:42.100
The energy expended is going to be power × time, that is the electric field here.
00:25:42.100 --> 00:25:54.200
Energy which is 10 W × 60 s or 600 J.
00:25:54.200 --> 00:25:57.300
I guess 10 W is not really an answer, it is part of the way to an answer.
00:25:57.300 --> 00:26:02.000
That takes care of the 2010 free response number 2.
00:26:02.000 --> 00:26:14.300
Pretty straightforward, once you realize that an uncharged capacitor acts like a wire and a fully charged capacitor acts like an open.
00:26:14.300 --> 00:26:19.100
Let us take a look at the 2004 exam now, free response number 2.
00:26:19.100 --> 00:26:22.200
Take a minute, find it, look at the problem.
00:26:22.200 --> 00:26:26.100
Download it, give it a shot, and we will come back here and see what we can do with it.
00:26:26.100 --> 00:26:32.800
As I look at this one, we are given a circuit and we are also given a graph of voltage vs. Time.
00:26:32.800 --> 00:26:39.000
The first thing it asks us to do is determine the voltage across resistor R2 right after that switch is closed.
00:26:39.000 --> 00:26:41.800
It usually helps me to draw the circuit again.
00:26:41.800 --> 00:26:44.000
Let us put that up here.
00:26:44.000 --> 00:26:49.300
For part A, we have our voltage source.
00:26:49.300 --> 00:26:55.200
Over here, we have that is R2 according to their notation.
00:26:55.200 --> 00:27:15.100
Here is R1, and as we go a little further in the circuit, we come to our switch S and our capacitor which is C 20 µf.
00:27:15.100 --> 00:27:19.300
I'm going to call that voltage C +- VC.
00:27:19.300 --> 00:27:26.000
At time 0, we are closing that switch.
00:27:26.000 --> 00:27:31.000
At time T = 0, VC equal 0.
00:27:31.000 --> 00:27:33.300
There is no charge on our capacitor.
00:27:33.300 --> 00:27:42.000
Therefore, V across R1 if that is 0, that is got to be 0.
00:27:42.000 --> 00:27:49.200
Therefore, V R2 must have the entire 20 V of them, source across it.
00:27:49.200 --> 00:27:55.900
The voltage across R2 right after the switch is closed must be 20 V.
00:27:55.900 --> 00:28:02.700
Part B, determine the voltage across resistor R2 a long time after the switch is closed.
00:28:02.700 --> 00:28:06.600
For part B, we are looking as T gets really big.
00:28:06.600 --> 00:28:13.000
In that case, VC is going to start approaching the voltage that is across R1.
00:28:13.000 --> 00:28:24.800
The way we could figure that out is that is going to be, we have got 15 kilo ohms here, and we need to figure out what R2 is going to be.
00:28:24.800 --> 00:28:30.700
We can look right at our graph though, and see that the capacitor voltage is 12 V after a long time.
00:28:30.700 --> 00:28:36.800
If this is 12 V here, that means the voltage across R1 here has to be 12 V.
00:28:36.800 --> 00:28:50.500
VC is 12 V and V R1 is 12 V, which means V R2, the remaining of the 20 V must be 8 V.
00:28:50.500 --> 00:28:54.500
That graph just came in mighty handy.
00:28:54.500 --> 00:28:59.800
For part C, calculate the value of resistor R2.
00:28:59.800 --> 00:29:07.800
To do that, I'm going to look again as T approaches infinity, we said that V R1 was 12 V,
00:29:07.800 --> 00:29:26.900
which implies that I R1, you have to have the same voltage drop which is V R1/ R1 is 12 V/ 15,000 ohms.
00:29:26.900 --> 00:29:35.300
Therefore, we get a current of about 0.0008 amps.
00:29:35.300 --> 00:29:44.700
V R2 is 8 V which is I R2 × R2.
00:29:44.700 --> 00:29:56.700
Therefore, R2 must equal V R2/ I R2 or I R1 because there is nothing flowing through your capacitor.
00:29:56.700 --> 00:30:16.300
That is going to be 8 V/ 0.0008 amps from up above, or 10,000 ohm which is 10 kilo ohms.
00:30:16.300 --> 00:30:22.200
We found that resistance 10 kilo ohm.
00:30:22.200 --> 00:30:24.100
What else do we have here?
00:30:24.100 --> 00:30:29.700
Find the energy stored in the capacitor a long time after the switch is closed.
00:30:29.700 --> 00:30:33.100
The energy stored in the capacitor after a longtime.
00:30:33.100 --> 00:30:40.300
U is ½ CV² is just going to be ½ × our capacitance 20 µf.
00:30:40.300 --> 00:31:01.300
20 × 10⁻⁶ F × 12 V² or 0.00144 J, which is 1.44 mJ.
00:31:01.300 --> 00:31:03.500
It looks like we are going to be making a graph.
00:31:03.500 --> 00:31:06.300
Graph current R2 from 0 to 15 s.
00:31:06.300 --> 00:31:09.300
Let us give ourselves some more room here.
00:31:09.300 --> 00:31:24.700
I will make a graph for part E, our axis first.
00:31:24.700 --> 00:31:29.500
It looks like we are graphing current and amps on the Y.
00:31:29.500 --> 00:31:32.000
We got our time in seconds over here.
00:31:32.000 --> 00:31:39.000
We will break it up, 5, 10, 15.
00:31:39.000 --> 00:31:42.900
Graph the current through R2 from 0 to 15 s.
00:31:42.900 --> 00:31:57.500
At T = 0, current through R2 is V R2 ÷ R2, or 20 V/ 10 kilo ohms which is 0.002 amps.
00:31:57.500 --> 00:32:03.600
We are going to have to start up here at 0.002.
00:32:03.600 --> 00:32:24.900
After a long time, as T approaches infinity and I R2 is V R2/ R2, which now is going to be 8 V/ 10 kilo ohms or 0.0008 amps.
00:32:24.900 --> 00:32:40.800
Let us see if I make it thick there at about 0.001, we are really targeting this asymptote down here of 0.0008 amps.
00:32:40.800 --> 00:32:49.000
I would expect our graph to looks something like, time constant looks like that happens pretty quickly.
00:32:49.000 --> 00:33:02.100
We are probably looking at something kind of like that, as we get closer and closer to the longer and longer time.
00:33:02.100 --> 00:33:14.100
Part F, indicate whether the energy stored in the capacitor is greater than, less than, or equal to its value when R2 was in the circuit, explain.
00:33:14.100 --> 00:33:30.100
The potential energy is ½ CV² but the voltage across the capacitor goes up if R2 decreases.
00:33:30.100 --> 00:33:36.000
Therefore, U, our potential energy must be increasing.
00:33:36.000 --> 00:33:44.700
It is got to be greater than.
00:33:44.700 --> 00:33:48.300
Hopefully, that gets you started on RC circuits in a steady state.
00:33:48.300 --> 00:33:53.700
In the next lesson, we will get to RC circuits, talking about their transient analysis.
00:33:53.700 --> 00:34:00.700
What happens between time equal 0 at a long time, trying to put some quantitative numbers around that.
00:34:00.700 --> 00:34:03.000
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