WEBVTT

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

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

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The question of the day is,

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how can you calculate the
voltage and current in

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a circuit that has
inductors and capacitors?

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In these cases, we're talking about

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a time domain circuit because
we changed something,

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we move this switch from point
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
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
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
closed at time t equal to zero,

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connecting this resistor and
capacitor to the voltage source.

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We discovered that the
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
37 percent of its original value.

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

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rose to the value Vs. At
the time constant tau,

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it had reached 63 percent
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
but the voltage cannot,

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the voltage changes slowly.

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

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the capacitor acts like a short
circuit because it starts

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to charge quickly and the current

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continues to dump charges onto
the capacitor until it reach

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steady state when it's all charged
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
and you'd see that it would

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charge and discharge and charge
and discharge and so on.

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

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

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the initial conditions before
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
the current and voltage there.

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

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and in fact the current
because the switch is

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open is zero and the voltage
across the capacitor is zero also.

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Then we need to consider
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
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
but the voltage cannot.

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So i of zero is not going to
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
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
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
acts like an open circuit.

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The current is now zero
and the voltage goes to

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

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The next thing that we do is
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
the form e to the minus t over tau,

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just remember that e to
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
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
voltages at DC and infinity,

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and e to the minus t over tau
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
zero minus VS times this exponent,

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which gives us the form
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
zero, it was time t0.

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Everything would be the
same except that we would

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substitute t minus zero
for both the cases here.

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So in general, the first thing that we
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
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
a short circuit when it's charging,

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and an open circuit when
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
this equation right here,

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or if the switch closed
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
here which shows that the

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current is capacitance times
the derivative of voltage,

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and that will give you
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
closes at t equal to zero,

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

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If you recall, the current
here changes gradually,

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

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reaching 63 percent of
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
change quickly across an inductor,

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it started out at zero and it jumped
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
its value at the time constant L over

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

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

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The voltage changes instantly
but the current changes slowly.

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When there is a sudden change
across an inductor,

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the inductor acts like an open circuit
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
a square wave across it.

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

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

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We start with the initial condition
before the switch has closed,

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we find out that the
current and the voltage

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before the switch has closed
are both zero once again.

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Then we look to see what
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
the current and voltage.

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Remember that the current
does not change suddenly,

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

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at t equals zero minus
or zero in this case.

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

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The inductor is open when the
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
is going to be, and in this case,

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the inductor is going to
be a short circuit because

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it is now able to fully take all of
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
which for an inductor is L over R,

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and again remember that
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
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
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
analyze the voltage and current,

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

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You then find the final
conditions and the time

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constant and plug them
into these equations.

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If we want to find the voltage
instead of the current,

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remember that the voltage is

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the inductance times
the derivative of the current.

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Now let's consider a general's
Thevenin solution procedure.

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In this case, we break this circuit
where we are looking at

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the capacitance or inductance
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
in all of the equations.

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So that's an evaluation of
the question of the day.

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How can you calculate the
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
solution procedures that I gave

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you and practice those
with some of the examples.