WEBVTT

00:00:00.000 --> 00:00:03.060
>> In this video, we're going to
demonstrate the actual process of

00:00:03.060 --> 00:00:05.939
analyzing circuits using
phasors and impedances,

00:00:05.939 --> 00:00:07.905
analyzing circuits in the phasor domain.

00:00:07.905 --> 00:00:09.450
We've already pointed out that in

00:00:09.450 --> 00:00:13.150
the phasor domain or rather
that we can analyze a circuit,

00:00:13.150 --> 00:00:21.060
in the phasor domain by representing
the voltages in terms of phasors,

00:00:21.060 --> 00:00:22.980
the currents in terms of phasors,

00:00:22.980 --> 00:00:27.860
and the impedances or representing
the effects of these capacitors,

00:00:27.860 --> 00:00:30.880
resistors, inductors in
terms of their impedances.

00:00:30.880 --> 00:00:32.604
So first of all,

00:00:32.604 --> 00:00:35.650
let's just represent phasor V,

00:00:35.650 --> 00:00:42.315
the source as having an amplitude
V sub m e to the j Theta.

00:00:42.315 --> 00:00:44.995
So, representing this time domain signal as

00:00:44.995 --> 00:00:48.910
its phasor with an amplitude of V
sub m and a phase shift of Theta.

00:00:48.910 --> 00:00:53.365
Now the impedance of the capacitor
is equal to one over

00:00:53.365 --> 00:00:58.720
j Omega C. The impedance of
a resistor is just equal to R,

00:00:58.720 --> 00:01:01.270
and the impedance of
the inductor is equal to j times

00:01:01.270 --> 00:01:04.810
Omega times L. So we'll
represent that or show that

00:01:04.810 --> 00:01:12.610
here is this impedance is one
over j Omega C. This is just R,

00:01:12.610 --> 00:01:15.840
and this impedance here is j Omega

00:01:15.840 --> 00:01:20.380
L. Now in the phasor domain

00:01:20.380 --> 00:01:25.340
there will be a current flowing
that we will represent as phasor I,

00:01:25.550 --> 00:01:29.930
and with these representations
let's go ahead now and write

00:01:29.930 --> 00:01:34.100
a Kirchhoff's voltage law equation
in terms of these phasers.

00:01:34.100 --> 00:01:36.170
Starting at this point and
going in this direction,

00:01:36.170 --> 00:01:38.000
I'm going to add up the voltage

00:01:38.000 --> 00:01:40.730
drops which means the voltage increase
will be a negative.

00:01:40.730 --> 00:01:45.600
So we'll have negative V sub m e

00:01:45.600 --> 00:01:50.510
to the j Theta plus now going in
the direction of current flow.

00:01:50.510 --> 00:01:54.485
So direct and current flow represents
a voltage drop and it will be a plus.

00:01:54.485 --> 00:01:56.810
The voltage drop across the capacitor is

00:01:56.810 --> 00:01:59.525
just the impedance of
the capacitor times the current

00:01:59.525 --> 00:02:05.470
or one over j Omega C times I,

00:02:05.470 --> 00:02:08.840
plus the voltage drop across the resistor
again going in the direction of

00:02:08.840 --> 00:02:13.210
current flow will be just R times I.

00:02:13.210 --> 00:02:16.010
Plus the voltage drop across
the inductor will be,

00:02:16.010 --> 00:02:21.060
the impedance of the inductor
which is j Omega L times I.

00:02:21.060 --> 00:02:24.375
The sum of those terms equals zero.

00:02:24.375 --> 00:02:27.860
Now, this is not a function of I.

00:02:27.860 --> 00:02:29.900
In fact, this is the constant
associated with the source.

00:02:29.900 --> 00:02:31.910
So let's bring it to the other side

00:02:31.910 --> 00:02:34.820
and factor out the I that is
common in each of those terms,

00:02:34.820 --> 00:02:42.950
we have in that I not equal
to I times one over j Omega C

00:02:42.950 --> 00:02:53.470
plus R plus j omega L is equal
to V sub m e to the j Theta.

00:02:53.470 --> 00:02:56.120
We now solve for I by
dividing both sides of

00:02:56.120 --> 00:02:59.015
the equation by this term in
parentheses, and we have then I,

00:02:59.015 --> 00:03:04.130
the phasor representation of
I is equal to V sub m e to

00:03:04.130 --> 00:03:10.005
the j Theta divided by one over j Omega C,

00:03:10.005 --> 00:03:15.240
plus R plus j Omega L.

00:03:15.240 --> 00:03:18.440
Phasor I the phasor representation is

00:03:18.440 --> 00:03:23.140
calculated by doing that complex division.