0:00:00.920,0:00:03.670
>> Our third and final method for finding

0:00:03.670,0:00:07.000
the Thevenin equivalent[br]impedance of the circuit,

0:00:07.000,0:00:12.820
involves applying an external source[br]to the terminals of our circuit,

0:00:12.820,0:00:18.130
after deactivating[br]the any independent sources.

0:00:18.130,0:00:21.930
So this circuit that we're[br]finding the Thevenin circuit for,

0:00:21.930,0:00:27.085
involved or has a single[br]independent voltage source.

0:00:27.085,0:00:29.620
We're going to deactivate[br]that independent source

0:00:29.620,0:00:32.215
again by replacing it with a short-circuit.

0:00:32.215,0:00:40.765
This method then involves applying[br]an external voltage source just V,

0:00:40.765,0:00:43.954
call it V_ex for external,

0:00:43.954,0:00:46.970
which would then cause a current to flow.

0:00:46.970,0:00:49.175
We'll call it I_ex,

0:00:49.175,0:00:50.870
for the external current of the current,

0:00:50.870,0:00:53.050
is due to this external voltage source,

0:00:53.050,0:00:57.014
and Z Thevenin then is going to equal,

0:00:57.014,0:01:02.955
the ratio of V_ex to I_ex.

0:01:02.955,0:01:08.845
This method works under any circumstances,

0:01:08.845,0:01:12.725
whether you have dependent[br]or independent sources.

0:01:12.725,0:01:14.885
This third method will work.

0:01:14.885,0:01:17.000
Sometimes it gets to be[br]algebraically a little bit

0:01:17.000,0:01:20.270
cumbersome involving multiple equations[br]and multiple unknowns,

0:01:20.270,0:01:22.670
but it will always work.

0:01:22.670,0:01:25.774
This method is equivalent to,

0:01:25.774,0:01:29.310
and can maybe something of a visual,

0:01:29.310,0:01:35.120
of a gasoline engine and an exhaust system.

0:01:35.120,0:01:37.835
If you wanted to measure or to model,

0:01:37.835,0:01:45.305
the back pressure due to[br]the resistance of the exhaust system.

0:01:45.305,0:01:47.855
This would be like turning off the engine,

0:01:47.855,0:01:51.230
and then putting your mouth over[br]the exhaust pipe and blowing

0:01:51.230,0:01:54.534
into it and measuring[br]the current that flows,

0:01:54.534,0:01:57.800
then that resistance of[br]the exhaust system would be

0:01:57.800,0:02:01.490
equal to the pressure that[br]you're pushing against it,

0:02:01.490,0:02:04.415
divided by the amount of[br]air that flowed into it.

0:02:04.415,0:02:06.830
So our Thevenin impedance will be,

0:02:06.830,0:02:10.020
we're going to deactivate[br]the source, the independent sources.

0:02:10.020,0:02:11.690
If there were dependent sources in here,

0:02:11.690,0:02:13.830
we would leave them active,

0:02:13.960,0:02:18.945
and having deactivated[br]the independent sources,

0:02:18.945,0:02:21.730
pushing against this or applying a voltage,

0:02:21.730,0:02:25.520
and taking the ratio of[br]the voltage to the current.

0:02:25.520,0:02:27.275
So here's our circuit then.

0:02:27.275,0:02:28.550
We've deactivated the source,

0:02:28.550,0:02:31.415
we're applying the external voltage source,

0:02:31.415,0:02:35.390
and we need to do some algebra and write[br]some equations that will allow us to

0:02:35.390,0:02:39.950
come up with a ratio of the external[br]voltage to the external current.

0:02:39.950,0:02:43.000
With this circuit here,

0:02:43.000,0:02:45.605
and with the source deactivated,

0:02:45.605,0:02:47.500
we see now that this 20 ohm and

0:02:47.500,0:02:51.580
the negative j25 ohm capacitor[br]are in parallel with each other.

0:02:51.580,0:02:54.875
So let's call that Z parallel,

0:02:54.875,0:03:00.040
and by reducing those to[br]a parallel combination,

0:03:00.040,0:03:01.825
we'll have these two,

0:03:01.825,0:03:05.105
the parallel equivalent[br]in series with that,

0:03:05.105,0:03:09.100
and we'll then be able to write[br]the expression for the external current,

0:03:09.100,0:03:12.140
will just be equal to this Thevenin[br]or this external voltage,

0:03:12.140,0:03:16.450
divided by the series[br]combination of those two,

0:03:16.450,0:03:19.390
of this plus the parallel equivalent.

0:03:19.390,0:03:23.150
So Z parallel is equal to,

0:03:23.150,0:03:26.934
negative j25 times 20,

0:03:26.934,0:03:31.580
divided by 20 minus j25,

0:03:31.580,0:03:40.680
and that again works out[br]to be 12.2 minus j9.76.

0:03:40.680,0:03:42.970
So we have then,

0:03:42.970,0:03:46.535
just redrawing to make obvious[br]what we're doing here.

0:03:46.535,0:03:50.200
You've got this inductor j50,

0:03:50.900,0:03:54.270
and we have this equivalent impedance,

0:03:54.270,0:04:00.430
of 12.2 minus j9.76.

0:04:02.660,0:04:05.334
This is our external voltage,

0:04:05.334,0:04:08.520
and there'll be[br]an external current flowing.

0:04:09.610,0:04:13.915
We can now write an expression for I_ex.

0:04:13.915,0:04:18.445
The external current is just[br]equal to the external voltage,

0:04:18.445,0:04:20.950
divided by the sum of those,

0:04:20.950,0:04:34.380
which is j50 plus 12.2 minus j9.76,

0:04:34.380,0:04:39.940
and the ratio V_ex over I_ex then,

0:04:39.940,0:04:42.730
can be gotten by multiplying[br]both sides of the equation by

0:04:42.730,0:04:46.750
this denominator and dividing[br]both sides by I_ex, and we get then,

0:04:46.750,0:04:51.430
that V_ex over I_ex is equal

0:04:51.430,0:04:59.685
to 12.2 plus j40.2,

0:04:59.685,0:05:05.690
and that of course is[br]our Thevenin equivalent voltage.

0:05:05.690,0:05:07.250
So we've seen now,

0:05:07.250,0:05:12.180
three different methods of[br]calculating the Thevenin impedance,

0:05:12.180,0:05:17.600
and which one works best becomes[br]a matter of art and experience.

0:05:17.600,0:05:19.550
You'll have some opportunities to do enough

0:05:19.550,0:05:21.740
of these and you'll start to get a feel for

0:05:21.740,0:05:27.750
the circumstances that make each of[br]these different methods most applicable.