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Now that you've been introduced to the inductor, let's take a look at a simple

circuit with a resistor and inductor in series, just like the RC circuit that we

just analyzed. So here's the circuit.

We're going to take, put the resistor in series with an inductor.

there's a battery V0, over here, and then there's a switch.

And we're going to assume that the switch is closed at t equals 0.

Now, there's going to be a current flowing, I of t, after the switch is

closed, and I can calculate the inductance voltage from L dI/dt.

Now, we just found that the current as a function of time is given by this

expression. So now I want to consider the same two

cases as we did before. Let's say that we open and close the

switch repeatedly, and let's do that very rapidly, in this case.

And what we're doing is we want to measure the voltage across the inductor.

Now, if I do it rapidly, that voltage is about equal to V0.

And the way I can figure that out is I take this equation for I of t, and I have

to take the derivative with respect to time, and when I do that, then the

constant term goes away. I bring a minus R over L out in front,

the minus signs cancel. So I just get V0 zero over L times this

exponential function, and if I multiply that by L then the L's cancel.

so I just get V0 e to the minus R over Lt.

That's this function here. So, if I divide through my V, V0, then

that function starts off at value one, at t equals 0 and then it decays away after

a few time constants, to near zero. Now, in this case, the time constant is L

over R. That has units of time, and that's the so

called RL time constant for this circuit. So, if I take and I normalize the time by

that time constant, I can plot along here without dimensions.

Now, if I open and close the switch rapidly, the voltage never has time to

det, to decay from the initial voltage of V0.

If I do it very slowly, then the voltage has time to die down.

And so, rapidly changing signals find their way to the output of the circuit.

A slowly changing signal is going to give me near, nearly a zero Voltage at the

output. Now, remember again, that after one time

constant, the exponential reaches only about 37% of the initial value, that's at

t equals L over R. Okay, now let's take and like we did

before, I'm going to reverse the position of the inductor and resistor in this

circuit. Now this is the same solution up here, it

doesn't matter the order of these if you go back and look at the, the way the

solution was arrived at. Kirchhoff's law doesn't, didn't care

which term came first. I could interchange them.

But now, I'm measuring the output here as VR, and that's just I of t times R.

So here's I of t. I just multiply by R, and this is

going to cancel. And I'm going to find this kind of a

curve. Now, it starts off, this, this, at t

equals 0, this whole exponential term is unity, e to the 0 is 1, and so this

becomes 1, 1-1 is 0. So this starts off near 0.

Now as t becomes larger, the exponential term gets smaller.

E to a large negative number approaches 0.

So, this whole second term disappears as time goes on.

And so, I just get V0 over R. So, if I take VR divided by V0.

VR of course is I times R, so if I multiply this times R the R is cancelled

and then if I divide by zero and then this thing I am left with this factor

here. And so, what you see is that, that factor

starts off at zero and then as time goes on several time constants later, the

exponential term goes away, and this approaches one.

So now, if I assume that I'm going to rapidly close and open the switch, the

current, and therefore the voltage across the resistor is never going to have a

chance to get very big. So, a very rapidly changing signal, a

high frequency signal, will not appear at the output.