Module 1_9_4 | MIT 8.02SC Physics II: Electricity and Magnetism, Fall 2010


Uploaded by MIT on 29.12.2010

Transcript:

PROFESSOR LEWIN: Problem number four, which is 32-19.
I have an R, I have an L, I have a C, I have E zero
cosine omega t.

300 Ohms, 0.25 Henry, 8 times 10 to the minus 6 Farads, 120
volts, 400 radians per second.
Forget the transients, we'll only look at the
steady state solution.
I equals I zero cosine omega t minus phi.
And the tangent of phi equals omega L minus 1 over omega C
divided by R. I think it's the third time that
you see this now.
It's interesting to write down the reactance of the three
components.
It's 312 Ohms for the capacitor, it's 100 Ohms for
the self inductance, and it is 300 Ohms for the resistor.
And so since Z equals the square root of omega L minus
omega C squared, plus R squared, you can immediately
calculate that this is 367 Ohms. And since I zero equals
E zero divided by Z, you'll find that
this is 0.327 Amperes.
And you find that phi is minus 35 degrees.
So the current is ahead of the driving voltage.
Because if phi is negative, then minus phi is positive, so
the current is ahead of the driving voltage.
That should not surprise you.
Because the capacitor is more potent than the self inductor.
If the self inductor was more potent, it would have the
capability of delaying the current.
But if the capacitor is in charge, it
puts the current ahead.
And that's exactly what you will see, the current is ahead
of the driving voltage.
If we now put volt meters over the R, C, and the L. Then,
you're being told that those volt meters, V R, as I have
drawn them before, V C and V L, there's no sense in drawing
them again.
They are volt meters.
That they now only show amplitudes.
So all phase information is lost. You don't see them shake
back and forth with frequency omega.
There is no phase information.
All they will indicate is the amplitude.

If they could show the instantaneous values, V at any
moment in time, then you would see, when V R is a maximum,
and I repeat myself, because I've said it before, then V L
would be zero, and then V of C would be zero.
The volt meter over the capacitor.
And when V R was zero, that means when I is zero, V L
would read a maximum value, and so would V C also read a
maximum value.
Even though these two would be 180 degrees out of phase.
All that is lost here.
These volt meter are designed in such a way that all they
record are amplitudes.
So the amplitude of the volt meter over the capacitor
equals I zero times the reactance off the capacitor.
The volt meter over to resistor would read I zero
times R. And the amplitude of the volt meter over the self
inductance would read I zero times omega L.
Again, I repeat myself, at the risk of boring you, that these
values do not occur simultaneously.

It is more common when we deal with alternating currents, to
have volt meters that express not the amplitude of the
voltage, but the root mean squared.
Which is simply the amplitude divided by the
square root of two.
There's nothing very sacred about that.
So if you wanted, in this particular problem, to covert
the values to RMS values, all you would have to do is divide
them by the square root of two.
And there's nothing more to it than that.