0:00:01.850,0:00:04.490
All righty,[br]let's work through some examples of

0:00:04.490,0:00:07.020
finding Thevenin equivalent circuits.

0:00:07.020,0:00:08.350
Take a look at this circuit right here.

0:00:08.350,0:00:11.610
We want to determine[br]the Thevenin equivalent circuit

0:00:11.610,0:00:15.190
between the terminals a and b.

0:00:15.190,0:00:17.690
That means we need to find[br]the open circuit voltage and

0:00:17.690,0:00:19.980
we need to determine the Thevenin[br]equivalent resistance.

0:00:21.020,0:00:26.479
Now, you'll notice the open circuit[br]voltage is the voltage across here, but

0:00:26.479,0:00:32.702
because it's open circuit, that means that[br]there's no current flowing through here.

0:00:32.702,0:00:36.732
I equals zero, so if there's no[br]current flowing through that resister,

0:00:36.732,0:00:39.404
there's no voltage drop[br]across that resister.

0:00:39.404,0:00:45.152
And the voltage at the terminals,[br]the open circuit voltage then,

0:00:45.152,0:00:50.190
is, in fact,[br]the voltage across this 30 ohm resistor.

0:00:51.510,0:00:53.360
Again, because there's[br]no voltage drop here,

0:00:53.360,0:00:57.310
the voltage that hit this point is[br]the same as the voltage at this point.

0:00:59.000,0:01:03.495
And again, because there's no current[br]going here, the current source here,

0:01:03.495,0:01:09.220
0.5 amps flowing is driving these[br]two in series at this point.

0:01:10.380,0:01:14.550
And we have then that V open[br]circuit is simply equal

0:01:14.550,0:01:19.350
to 0.5 times, 0.5 is the current,

0:01:19.350,0:01:25.130
times the 30 ohms is equal to 15 volts.

0:01:26.840,0:01:29.330
We now need to determine[br]the Thevenin equivalent resistance,

0:01:29.330,0:01:31.250
which we're gonna do in[br]each of the three methods.

0:01:31.250,0:01:35.300
Or at least consider whether each of those[br]three different methods will work for

0:01:35.300,0:01:36.650
this circuit.

0:01:36.650,0:01:41.780
Method one works when we have at least one[br]independent source, which we have here.

0:01:42.790,0:01:45.741
Method one involves finding[br]the short circuit current.

0:01:47.716,0:01:53.640
And then, forming the ratio, V,[br]open circuit divided by I, short circuit.

0:01:53.640,0:01:58.749
So if we short this out,[br]short the terminals,

0:01:58.749,0:02:05.590
the current flowing through[br]that short is I, short circuit.

0:02:05.590,0:02:10.251
Now, how do we find I,[br]short circuit in this circuit?

0:02:10.251,0:02:17.738
Well, by shorting that, out that brings[br]this, the voltage here, down to ground.

0:02:17.738,0:02:25.072
We have 0.5 amps flowing into now[br]a node which consists of two resistors,

0:02:25.072,0:02:29.783
this 30 ohm resistor and[br]this 5 ohm resistor.

0:02:29.783,0:02:33.296
Some of the current's gonna through the 5[br]ohm resistor, that's our short circuit

0:02:33.296,0:02:36.627
current, the rest of the current is[br]gonna come through this 30 ohm resistor.

0:02:36.627,0:02:39.090
We have a current divider here.

0:02:39.090,0:02:42.600
So using current divider,[br]we can calculate short circuit directly.

0:02:42.600,0:02:49.640
I short circuit is equal to 0.5 times,[br]and remember from the current divider,

0:02:49.640,0:02:55.740
the term that goes in the numerator is[br]the resistance in the other branch.

0:02:55.740,0:03:01.340
So that would be 30 ohms divided[br]by the sum of those two,

0:03:01.340,0:03:04.930
30 plus 5 is 35 ohms.

0:03:04.930,0:03:10.740
And we get then the short circuit current

0:03:10.740,0:03:16.970
is equal to 0.4286, 0.4286 amps.

0:03:16.970,0:03:22.098
And then RTh is equal to Voc

0:03:22.098,0:03:27.927
divided by Isc, which is 15

0:03:27.927,0:03:34.100
divided by 0.4286.

0:03:34.100,0:03:39.155
Which, when we get the calculator out on[br]that, you'll find that equals 35 ohms.

0:03:42.080,0:03:45.143
So, method 1.

0:03:45.143,0:03:50.166
Method 2,[br]let's clear this up just a little bit.

0:03:50.166,0:03:50.838
All righty,

0:03:50.838,0:03:54.768
now let's use method 2 to determine[br]the Thevenin equivalent resistance.

0:03:54.768,0:04:00.038
In method 2, which works if you have,

0:04:00.038,0:04:04.200
only, independent sources.

0:04:05.320,0:04:07.930
Method two, you deactivate the sources.

0:04:07.930,0:04:10.390
In this case,[br]we have an independent current source.

0:04:10.390,0:04:12.770
We turn the independent[br]current source to zero.

0:04:12.770,0:04:17.457
And, that is effectively,[br]open circuiting this branch,

0:04:17.457,0:04:23.094
which means that there will be no[br]current flowing through this branch.

0:04:23.094,0:04:27.362
Such that as we look[br]back in to the circuit,

0:04:27.362,0:04:33.661
this is just hanging unconnected and[br]does not enter into this.

0:04:33.661,0:04:36.352
Effectively, we have,[br]because this is open circuit here,

0:04:36.352,0:04:38.639
this is effectively[br]an infinite resistance here.

0:04:39.840,0:04:46.150
And we see coming back into this then,[br]the 5 ohms in series with the 35.

0:04:46.150,0:04:47.843
So method 2,

0:04:52.796,0:05:00.782
We get that R Thevenin is simply[br]the 30 + 5 ohms = 35 ohms,

0:05:00.782,0:05:04.864
same value we got for method 1.

0:05:04.864,0:05:06.592
Now, let's take a look at method 3.

0:05:06.592,0:05:12.062
Method 3 works regardless of[br]the types of sources that we have,

0:05:12.062,0:05:17.344
but in method 3,[br]we do deactivate the independent sources.

0:05:17.344,0:05:19.777
So once again,[br]we have this independent current source.

0:05:19.777,0:05:23.240
By deactivating it, we open the circuit.

0:05:23.240,0:05:28.185
And then, method 3 says that[br]we apply a test voltage,

0:05:28.185,0:05:33.238
which then forces a test[br]current into the AB terminals,

0:05:33.238,0:05:39.497
and we then algebraically work to[br]make the ratio of V test over I test.

0:05:39.497,0:05:44.476
So method 3 then Is form the ratio V

0:05:44.476,0:05:49.988
test over I test to give us R Thevenin.

0:05:49.988,0:05:55.286
All righty, with that open circuit,[br]this is a simple situation.

0:05:55.286,0:06:00.987
I, going in here,[br]we can write a KVL loop going

0:06:00.987,0:06:05.969
around this loop in the direction of IT.

0:06:05.969,0:06:12.450
So KVL, we have then, minus v test[br]going in the direction of current flow.

0:06:12.450,0:06:19.354
We have then a voltage drop of plus five[br]I sub t coming on down here plus 30.

0:06:21.180,0:06:27.325
That's a plus sign, 30 I sub t,[br]the sum of those terms equals zero.

0:06:27.325,0:06:30.349
Let's bring in V test over to[br]the other side of the positive V test,

0:06:30.349,0:06:31.378
factor out the I test.

0:06:31.378,0:06:38.500
We had I test x 5 + 30 is 35 = V test.

0:06:38.500,0:06:42.972
Divide both sides by I test,[br]that gives us our R Thevenin,

0:06:42.972,0:06:47.557
which is equal to the ratio of[br]V test over I test = 35 ohms.

0:06:47.557,0:06:52.837
So we get the same results in[br]any one of the three methods,

0:06:52.837,0:06:55.478
and we then, can or now can,

0:06:58.746,0:07:03.966
Draw our Thevenin equivalent circuit,[br]where V Thevenin

0:07:03.966,0:07:09.523
is equal to 15 volts, and[br]R Thevenin is equal to 35 ohms.

0:07:11.296,0:07:14.359
Notice that the original[br]circuit had a current source.

0:07:15.720,0:07:20.260
But the Thevenin Equivalent Circuit[br]replaces that source and

0:07:20.260,0:07:24.500
all of the resistance, the effects of[br]the resistance within this circuit,

0:07:25.600,0:07:30.840
are represented by a single voltage source[br]and the Thevenin Equivalent resistance.