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>> Hello, this is Dr. Cynthia Furse[br]at the University of Utah,

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and today we're going to talk[br]about RC and RL circuits.

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The question of the day is,

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how can you calculate the[br]voltage and current in

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a circuit that has[br]inductors and capacitors?

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In these cases, we're talking about

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a time domain circuit because[br]we changed something,

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we move this switch from point[br]one to point two for example,

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at some point t equals zero.

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We talked about RC and[br]RL circuits last time,

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and we're going to review those again,

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and I'm going to give you

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a general procedure for being[br]able to analyze these circuits.

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So let's review. Last time we talked about

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a circuit where the switch[br]closed at time t equal to zero,

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connecting this resistor and[br]capacitor to the voltage source.

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We discovered that the[br]current went from zero,

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and suddenly jumped up,

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it changed instantly to Vs over R,

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and that the equation became Vs over R,

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e to the minus t over tau,

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where tau equals RC,

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and that is the time constant.

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At the time constant tau,

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this current has fallen down to[br]37 percent of its original value.

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The voltage on the other hand[br]started out at zero and gradually

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rose to the value Vs. At[br]the time constant tau,

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it had reached 63 percent[br]of its maximum value,

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and here's the equation.

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So here's the current and voltage.

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What had really happened?

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We know that for a capacitor,

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the current can change instantly[br]but the voltage cannot,

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the voltage changes slowly.

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At time t equal to zero when[br]there is a sudden change,

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the capacitor acts like a short[br]circuit because it starts

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to charge quickly and the current

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continues to dump charges onto[br]the capacitor until it reach

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steady state when it's all charged[br]up it acts like an open circuit.

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This is what it would look like.

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This is the voltage across the capacitor.

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If we put a square wave on this[br]and you'd see that it would

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charge and discharge and charge[br]and discharge and so on.

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So now let's consider a general procedure[br]for analyzing this type of circuit.

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The first thing that we're[br]going to do is calculate

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the initial conditions before[br]anything has happened.

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This is when the switch is opened,

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we call it t equals zero minus,

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and this is before the switch closes,

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and we need to find[br]the current and voltage there.

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At steady-state, we know that[br]the capacitor is an open circuit,

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and in fact the current[br]because the switch is

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open is zero and the voltage[br]across the capacitor is zero also.

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Then we need to consider[br]the initial condition,

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this is when the switch is closed,

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that's time t equals zero plus,

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it's just barely closed,

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and we need to find the[br]current and the voltage.

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When the current closes,

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this capacitor acts like a short circuit,

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the current can change instantly[br]but the voltage cannot.

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So i of zero is not going to[br]be the same thing as i0 minus,

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it's going to quickly go to Vs over

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R. The voltage on the other hand[br]is the same as zero minus,

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it's going to stay at zero.

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The next thing that we're[br]going to do is find

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the conditions at time t equal infinity,

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when the capacitor is fully charged up.

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In that case, the capacitor[br]acts like an open circuit.

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The current is now zero[br]and the voltage goes to

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the original voltage source[br]or 10 volts, in this case.

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The next thing that we do is[br]evaluate the time constant.

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For a circuit like this,

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the time constant is RC.

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I just made some notes here,

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because we're going to be using[br]the form e to the minus t over tau,

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just remember that e to[br]the negative one is 0.37,

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that's where the 37 percent comes from,

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and one minus e to the one is 0.63,

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that's where the 63 percent comes from.

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The next thing that we do is we[br]use a general voltage equation,

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the general form of the voltage across

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the capacitor right here is where VC is,

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the general form of the voltage across

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the capacitor is the voltage at infinity,

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plus the difference of[br]voltages at DC and infinity,

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and e to the minus t over tau[br]multiplied by the step function.

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The step function just looks like this.

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Then if we plug in our values,

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we're going to get VS plus[br]zero minus VS times this exponent,

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which gives us the form[br]that we derived yesterday.

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Now let's consider what would happen if

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the switch closed at some different time,

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it wasn't time t equals[br]zero, it was time t0.

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Everything would be the[br]same except that we would

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substitute t minus zero[br]for both the cases here.

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So in general, the first thing that we[br]do is we find the initial conditions.

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We start before the switch closes and then

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we find the conditions[br]after the switch closes.

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We then find the final conditions and we

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remember that the capacitor is[br]a short circuit when it's charging,

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and an open circuit when[br]it's fully charged.

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We then find the time constant which is RC,

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and we plug everything into[br]this equation right here,

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or if the switch closed[br]at a different time,

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this is the equation we use.

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Finally, if we want to find the current,

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just use this equation right[br]here which shows that the

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current is capacitance times[br]the derivative of voltage,

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and that will give you[br]the equations shown here.

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Now let's review the RL circuit.

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This is again when the switch[br]closes at t equal to zero,

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and we want to find out what[br]happens to the voltage and current.

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If you recall, the current[br]here changes gradually,

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it started out as zero and then it[br]gradually builds up to Vs over R,

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reaching 63 percent of[br]its value at the time constant,

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the time constant in this case is L over

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R. The voltage on the other hand does[br]change quickly across an inductor,

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it started out at zero and it jumped[br]up very quickly to the value of Vs.

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It has reached and then it sags,

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and it reaches 37 percent of[br]its value at the time constant L over

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R. So this is what the current and[br]voltage look like for an inductor.

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Oops, this is supposed to be[br]an inductor and not a capacitor.

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The voltage changes instantly[br]but the current changes slowly.

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When there is a sudden change[br]across an inductor,

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the inductor acts like an open circuit[br]and when it has reached steady-state,

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it acts like a short.

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So this is what it would look like.

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Here's the voltage that we would see across

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an inductor if we put[br]a square wave across it.

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Now let's talk about[br]the general solution procedure for RL.

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It works very much the same[br]way as a capacitor does.

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We start with the initial condition[br]before the switch has closed,

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we find out that the[br]current and the voltage

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before the switch has closed[br]are both zero once again.

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Then we look to see what[br]happens when we first

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close the switch at t equals zero plus,

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and we need to find[br]the current and voltage.

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Remember that the current[br]does not change suddenly,

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so the current at time t equals[br]zero is going to be the current

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at t equals zero minus[br]or zero in this case.

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Then, we look at to see[br]what the voltage is.

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The inductor is open when the[br]current first starts to flow,

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and so that makes the voltage the

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Vs. Now we need

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to go to find what's going to happen in

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the steady state or the final conditions,

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then we look to see what the current[br]is going to be, and in this case,

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the inductor is going to[br]be a short circuit because

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it is now able to fully take all of[br]the current that's going through it,

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so the current at infinity is Vs over R,

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and the voltage is zero.

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Next we need to consider the time constant[br]which for an inductor is L over R,

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and again remember that[br]the same exponent apply.

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Then we use this equation for the current.

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It looks very similar to[br]the voltage equation for the capacitor.

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Plugging all of our things in,

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we come up with exactly the same[br]current equation that we had last time.

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So here's the general solution procedure.

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You start at t equals zero minus and[br]analyze the voltage and current,

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and then you convert[br]those over to t equals zero plus.

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You then find the final[br]conditions and the time

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constant and plug them[br]into these equations.

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If we want to find the voltage[br]instead of the current,

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remember that the voltage is

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the inductance times[br]the derivative of the current.

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Now let's consider a general's[br]Thevenin solution procedure.

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In this case, we break this circuit[br]where we are looking at

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the capacitance or inductance[br]and we look into the circuit,

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and then we go back to our equations I just

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gave you and replace Vs with V Thevenin,

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and replace R with R Thevenin[br]in all of the equations.

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So that's an evaluation of[br]the question of the day.

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How can you calculate the[br]voltage and current in

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a circuit that has inductors or capacitors?

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Go back to the two general[br]solution procedures that I gave

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you and practice those[br]with some of the examples.