0:00:00.020,0:00:04.094
>> We'll do that same example now only with

0:00:04.094,0:00:08.370
the actual values for the circuit[br]elements and the source here.

0:00:08.370,0:00:10.560
So again, the first thing we do is convert

0:00:10.560,0:00:13.800
the time-domain source to[br]its phasor representation

0:00:13.800,0:00:19.020
and then calculate the impedances[br]associated with the capacitor,

0:00:19.020,0:00:20.625
resistance and inductor here.

0:00:20.625,0:00:24.800
So this the phasor[br]representation V we'll call it

0:00:24.800,0:00:31.110
cap V or phasor V is[br]equal to 10E to the J30.

0:00:31.360,0:00:33.920
Now, let's be explicit here.

0:00:33.920,0:00:37.710
Omega is 1,000 radians per second.

0:00:38.200,0:00:41.345
Radians per second.

0:00:41.345,0:00:46.790
Z_C is equal to one over J Omega C[br]which is equal to one over

0:00:46.790,0:00:49.190
J times Omega which is

0:00:49.190,0:00:53.915
1,000 times C which is 10 times[br]10 to the minus six to six micro,

0:00:53.915,0:01:02.550
10 microfarads and that then[br]equals negative J 100 Ohms.

0:01:02.550,0:01:06.940
That's a negative J 100 Ohms.

0:01:06.940,0:01:10.670
Z_R is just a value of[br]the resistor which is

0:01:10.670,0:01:16.610
50 Ohms and Z_L is equal to J[br]Omega L which is equal to J times

0:01:16.610,0:01:19.670
Omega which is 1,000 times L which is

0:01:19.670,0:01:27.075
75 millihenries and that[br]gives us a positive J 75.

0:01:27.075,0:01:30.515
So this is a J 75 there.

0:01:30.515,0:01:36.455
This is R which is equal to the impedance[br]that is equal to 50 ohms and this,

0:01:36.455,0:01:42.395
the impedance of the capacitor[br]is a negative J 100.

0:01:42.395,0:01:45.350
We'll go ahead and define the current

0:01:45.350,0:01:47.870
flowing through here or the phasor[br]representation of the current going

0:01:47.870,0:01:51.319
through there call it phasor I and let's

0:01:51.319,0:01:56.465
write a Kirchhoff's voltage law equation[br]going around that loop.

0:01:56.465,0:01:59.695
Once again, adding up the voltage drops.

0:01:59.695,0:02:02.150
So, a voltage increase will be a negative,

0:02:02.150,0:02:10.250
that would be a negative 10 E to the J[br]30 plus going across this capacitor,

0:02:10.250,0:02:12.290
the voltage drop across that capacitor will

0:02:12.290,0:02:14.780
be the current I times the impedance.

0:02:14.780,0:02:17.660
So, that will be I times[br]the impedance which is

0:02:17.660,0:02:24.050
a negative J 100 plus the voltage drop[br]across the resistor which is going

0:02:24.050,0:02:30.575
to be I times 50 plus the voltage drop[br]across that inductor is going to be

0:02:30.575,0:02:33.320
I times the impedance of[br]that inductor which is

0:02:33.320,0:02:38.835
J 75 and the sum of those[br]terms then equals zero.

0:02:38.835,0:02:42.075
We want to solve for I[br]the current in this circuit.

0:02:42.075,0:02:45.450
So we'll factor out the I[br]from these common terms

0:02:45.450,0:02:50.300
here and bring this term here on[br]over other side and we get then

0:02:50.300,0:02:57.135
that I times a negative J 100 plus 50

0:02:57.135,0:03:06.870
plus J 75 is equal to[br]a positive 10 E to the J 30.

0:03:06.870,0:03:12.845
We have a negative J 100 plus J 75[br]that gives us a negative J 25 plus 50.

0:03:12.845,0:03:17.570
Dividing both sides by that term[br]then gives us that I is equal

0:03:17.570,0:03:23.830
to 10 E to the J 30 divided by,

0:03:23.830,0:03:30.065
writing the real part first 50 minus J 25.

0:03:30.065,0:03:32.270
You get out your calculator.

0:03:32.270,0:03:37.850
First of all, let's convert[br]just this denominator here to polar form.

0:03:37.850,0:03:40.715
It's a little bit easier to see[br]what's going on here if we do so.

0:03:40.715,0:03:48.230
In polar form, that's equal to 55.9 E

0:03:48.230,0:04:00.650
to negative J 26.57.

0:04:00.650,0:04:06.170
Now, we can see there the division,

0:04:06.170,0:04:13.550
it's 10 E to the J 30 divided[br]by 55.9 E to the minus J 26.57.

0:04:13.550,0:04:18.399
So 10 divided by 55.9[br]gives us the magnitude.

0:04:18.399,0:04:20.630
The magnitude of the numerator[br]divided by the magnitude of

0:04:20.630,0:04:25.815
the denominator gives us 0.179.

0:04:25.815,0:04:28.145
Now, the phase E,

0:04:28.145,0:04:31.245
it will be E to the J 30 in the numerator,

0:04:31.245,0:04:35.515
E to the minus J 26.57 in the denominator.

0:04:35.515,0:04:41.150
We've already demonstrated that when[br]dividing numbers involving exponents,

0:04:41.150,0:04:45.830
it's the exponent of the numerator[br]minus the exponent in the denominator.

0:04:45.830,0:04:51.320
So it's J 30 minus and minus[br]it becomes plus a 26.57 that

0:04:51.320,0:05:01.815
gives us a positive J 56.57 degrees.

0:05:01.815,0:05:08.455
This entire thing here is[br]the phasor representation of the current.

0:05:08.455,0:05:11.150
We can now take this back[br]into the time domain

0:05:11.150,0:05:13.400
by recognizing that the[br]amplitude of the current

0:05:13.400,0:05:17.870
is going to be the magnitude of[br]this phasor or let's be explicit now.

0:05:17.870,0:05:24.310
I of T then is equal to amplitude of 0.179.

0:05:24.670,0:05:27.380
It will be a cosine term.

0:05:27.380,0:05:32.075
We came in as a cosine will come back[br]to the time domain as a cosine term.

0:05:32.075,0:05:35.480
Cosine, the frequency[br]doesn't change that's still

0:05:35.480,0:05:42.335
1,000 T. But now the phase of the[br]current is equal to 56.57 degrees.

0:05:42.335,0:05:49.560
So plus 56.57 degrees.

0:05:49.560,0:05:56.315
That then is the time domain[br]expression for the current.

0:05:56.315,0:05:59.765
The voltage, the source driving[br]this circuit is oscillating at

0:05:59.765,0:06:03.980
1,000 radians per second and the current[br]is oscillating 1,000 times per second.

0:06:03.980,0:06:05.630
It's going back and forth changing at

0:06:05.630,0:06:08.905
the same frequency that[br]the source is going into it.

0:06:08.905,0:06:11.585
The source has an amplitude of 10 volts.

0:06:11.585,0:06:17.900
The current flowing has[br]an amplitude of 0.179 Amps.

0:06:17.900,0:06:20.630
So, the amplitude of the current[br]is smaller than the amplitude of

0:06:20.630,0:06:24.440
the voltage and finally[br]let's look at the phase.

0:06:24.440,0:06:30.170
The phase angle of the voltage source[br]is 30 degrees, positive 30 degrees.

0:06:30.170,0:06:35.920
The angle of the current is 56.57 degrees.

0:06:35.920,0:06:40.910
So, we'd say then that the current[br]has been shifted ahead

0:06:40.910,0:06:47.640
of the source by the 26.57 degrees.

0:06:49.550,0:06:51.990
Here are plots of those,

0:06:51.990,0:06:53.895
the voltage in the red,

0:06:53.895,0:07:04.050
V of T is equal to 10 cosine[br]of 1,000 T plus 30 degrees.

0:07:04.050,0:07:09.725
I of T we have now found[br]to be equal to 0.179,

0:07:09.725,0:07:20.265
cosine of 1,000 T plus 56.57 degrees.

0:07:20.265,0:07:22.640
So here's the voltage.[br]It's got an amplitude of

0:07:22.640,0:07:28.670
10 volts and it has been shifted[br]to the left by 30 degrees.

0:07:28.670,0:07:31.940
The current has an amplitude of

0:07:31.940,0:07:39.080
0.179 and it has been shifted[br]to the left 56.57 degrees.

0:07:39.080,0:07:42.740
So the difference, I can[br]draw this very well.

0:07:42.740,0:07:44.770
I don't know if I can or not.

0:07:44.770,0:07:47.870
That distance right there is

0:07:47.870,0:07:52.430
a phase shift of the current[br]ahead of the voltage.

0:07:52.430,0:07:55.429
We say that the current[br]is leading the voltage,

0:07:55.429,0:07:59.140
the current peaks out[br]before the voltage peaks.

0:07:59.140,0:08:02.690
The current crosses the zero before[br]the voltage crosses the zero.

0:08:02.690,0:08:06.710
The current is leading the voltage by that

0:08:06.710,0:08:11.285
much which corresponds to[br]again that 26.57 degrees.

0:08:11.285,0:08:17.600
Just one note about what appears[br]to be an inconsistency of units.

0:08:17.600,0:08:19.900
Not only does it appear to be[br]an inconsistency of units,

0:08:19.900,0:08:21.560
it isn't inconsistency of units.

0:08:21.560,0:08:25.850
What are the units of omega,[br]the radial frequency?

0:08:25.850,0:08:31.220
That's 1,000 radians per second.

0:08:31.220,0:08:36.200
Yet we're specifying the[br]phase shift of 30 degrees.

0:08:36.200,0:08:38.485
We're specifying it in degrees.

0:08:38.485,0:08:41.669
To be consistent and frankly,

0:08:41.669,0:08:44.450
we're going to leave it in this form[br]I think simply because it's

0:08:44.450,0:08:47.270
a lot easier for us to visualize.

0:08:47.270,0:08:49.835
We have a better intuitive feel for

0:08:49.835,0:08:53.490
phase shifts in terms of degrees than[br]we do radians but if wherever we're

0:08:53.490,0:08:57.110
doing calculations we'll need[br]to convert E to the this into

0:08:57.110,0:09:01.310
degrees per second or[br]change this to radians.

0:09:01.310,0:09:04.115
Each course the change of radius[br]is going to just be simply

0:09:04.115,0:09:10.520
the 30 degrees times Pi divided by 180.

0:09:10.520,0:09:12.560
So that phase shift there,

0:09:12.560,0:09:14.870
the 30 degree phase shift represents

0:09:14.870,0:09:18.680
Pi sixth phase there

0:09:18.680,0:09:21.650
and somebody could do the same for[br]the phase shift and the current.

0:09:21.650,0:09:23.735
But the point is let's[br]just drive this home.

0:09:23.735,0:09:26.660
This circuit here, the circuit that we just

0:09:26.660,0:09:31.100
analyzed did not change the frequency.

0:09:31.100,0:09:33.200
The frequency of the current flowing in

0:09:33.200,0:09:35.720
here is the same as[br]the frequency of the voltage.

0:09:35.720,0:09:39.085
What did change was the[br]amplitude and the phase.

0:09:39.085,0:09:42.710
Using phasor analysis we were[br]able to determine the phase

0:09:42.710,0:09:47.330
on the amplitude and phase of the current.