>> Hello, this is Dr. Cynthia Furse at the University of Utah,
and today we're going to talk about RC and RL circuits.
The question of the day is,
how can you calculate the voltage and current in
a circuit that has inductors and capacitors?
In these cases, we're talking about
a time domain circuit because we changed something,
we move this switch from point one to point two for example,
at some point t equals zero.
We talked about RC and RL circuits last time,
and we're going to review those again,
and I'm going to give you
a general procedure for being able to analyze these circuits.
So let's review. Last time we talked about
a circuit where the switch closed at time t equal to zero,
connecting this resistor and capacitor to the voltage source.
We discovered that the current went from zero,
and suddenly jumped up,
it changed instantly to Vs over R,
and that the equation became Vs over R,
e to the minus t over tau,
where tau equals RC,
and that is the time constant.
At the time constant tau,
this current has fallen down to 37 percent of its original value.
The voltage on the other hand started out at zero and gradually
rose to the value Vs. At the time constant tau,
it had reached 63 percent of its maximum value,
and here's the equation.
So here's the current and voltage.
What had really happened?
We know that for a capacitor,
the current can change instantly but the voltage cannot,
the voltage changes slowly.
At time t equal to zero when there is a sudden change,
the capacitor acts like a short circuit because it starts
to charge quickly and the current
continues to dump charges onto the capacitor until it reach
steady state when it's all charged up it acts like an open circuit.
This is what it would look like.
This is the voltage across the capacitor.
If we put a square wave on this and you'd see that it would
charge and discharge and charge and discharge and so on.
So now let's consider a general procedure for analyzing this type of circuit.
The first thing that we're going to do is calculate
the initial conditions before anything has happened.
This is when the switch is opened,
we call it t equals zero minus,
and this is before the switch closes,
and we need to find the current and voltage there.
At steady-state, we know that the capacitor is an open circuit,
and in fact the current because the switch is
open is zero and the voltage across the capacitor is zero also.
Then we need to consider the initial condition,
this is when the switch is closed,
that's time t equals zero plus,
it's just barely closed,
and we need to find the current and the voltage.
When the current closes,
this capacitor acts like a short circuit,
the current can change instantly but the voltage cannot.
So i of zero is not going to be the same thing as i0 minus,
it's going to quickly go to Vs over
R. The voltage on the other hand is the same as zero minus,
it's going to stay at zero.
The next thing that we're going to do is find
the conditions at time t equal infinity,
when the capacitor is fully charged up.
In that case, the capacitor acts like an open circuit.
The current is now zero and the voltage goes to
the original voltage source or 10 volts, in this case.
The next thing that we do is evaluate the time constant.
For a circuit like this,
the time constant is RC.
I just made some notes here,
because we're going to be using the form e to the minus t over tau,
just remember that e to the negative one is 0.37,
that's where the 37 percent comes from,
and one minus e to the one is 0.63,
that's where the 63 percent comes from.
The next thing that we do is we use a general voltage equation,
the general form of the voltage across
the capacitor right here is where VC is,
the general form of the voltage across
the capacitor is the voltage at infinity,
plus the difference of voltages at DC and infinity,
and e to the minus t over tau multiplied by the step function.
The step function just looks like this.
Then if we plug in our values,
we're going to get VS plus zero minus VS times this exponent,
which gives us the form that we derived yesterday.
Now let's consider what would happen if
the switch closed at some different time,
it wasn't time t equals zero, it was time t0.
Everything would be the same except that we would
substitute t minus zero for both the cases here.
So in general, the first thing that we do is we find the initial conditions.
We start before the switch closes and then
we find the conditions after the switch closes.
We then find the final conditions and we
remember that the capacitor is a short circuit when it's charging,
and an open circuit when it's fully charged.
We then find the time constant which is RC,
and we plug everything into this equation right here,
or if the switch closed at a different time,
this is the equation we use.
Finally, if we want to find the current,
just use this equation right here which shows that the
current is capacitance times the derivative of voltage,
and that will give you the equations shown here.
Now let's review the RL circuit.
This is again when the switch closes at t equal to zero,
and we want to find out what happens to the voltage and current.
If you recall, the current here changes gradually,
it started out as zero and then it gradually builds up to Vs over R,
reaching 63 percent of its value at the time constant,
the time constant in this case is L over
R. The voltage on the other hand does change quickly across an inductor,
it started out at zero and it jumped up very quickly to the value of Vs.
It has reached and then it sags,
and it reaches 37 percent of its value at the time constant L over
R. So this is what the current and voltage look like for an inductor.
Oops, this is supposed to be an inductor and not a capacitor.
The voltage changes instantly but the current changes slowly.
When there is a sudden change across an inductor,
the inductor acts like an open circuit and when it has reached steady-state,
it acts like a short.
So this is what it would look like.
Here's the voltage that we would see across
an inductor if we put a square wave across it.
Now let's talk about the general solution procedure for RL.
It works very much the same way as a capacitor does.
We start with the initial condition before the switch has closed,
we find out that the current and the voltage
before the switch has closed are both zero once again.
Then we look to see what happens when we first
close the switch at t equals zero plus,
and we need to find the current and voltage.
Remember that the current does not change suddenly,
so the current at time t equals zero is going to be the current
at t equals zero minus or zero in this case.
Then, we look at to see what the voltage is.
The inductor is open when the current first starts to flow,
and so that makes the voltage the
Vs. Now we need
to go to find what's going to happen in
the steady state or the final conditions,
then we look to see what the current is going to be, and in this case,
the inductor is going to be a short circuit because
it is now able to fully take all of the current that's going through it,
so the current at infinity is Vs over R,
and the voltage is zero.
Next we need to consider the time constant which for an inductor is L over R,
and again remember that the same exponent apply.
Then we use this equation for the current.
It looks very similar to the voltage equation for the capacitor.
Plugging all of our things in,
we come up with exactly the same current equation that we had last time.
So here's the general solution procedure.
You start at t equals zero minus and analyze the voltage and current,
and then you convert those over to t equals zero plus.
You then find the final conditions and the time
constant and plug them into these equations.
If we want to find the voltage instead of the current,
remember that the voltage is
the inductance times the derivative of the current.
Now let's consider a general's Thevenin solution procedure.
In this case, we break this circuit where we are looking at
the capacitance or inductance and we look into the circuit,
and then we go back to our equations I just
gave you and replace Vs with V Thevenin,
and replace R with R Thevenin in all of the equations.
So that's an evaluation of the question of the day.
How can you calculate the voltage and current in
a circuit that has inductors or capacitors?
Go back to the two general solution procedures that I gave
you and practice those with some of the examples.