L 5 3 and 5 4 (Natural Response of RC and RL Circuits)

This lecture describes how to calculate the time domain voltages and currents in an RC or RL circuit. See chapters 5-3 and 5-4 in Introduction to Circuits by FT Ulaby, MM.Maharabix, C.M.Furse, 3rd edition. For ECE1250 at the University of Utah

>> 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.