WEBVTT

00:00:00.960 --> 00:00:03.500
Okay, here's a problem that we
were working in class today, and

00:00:03.500 --> 00:00:06.620
I would like to just go over
this example again for you.

00:00:06.620 --> 00:00:08.400
We want to be able to find Vx.

00:00:08.400 --> 00:00:12.080
We can see we have a lot of resistors here
we can combine in series and parallel.

00:00:12.080 --> 00:00:14.310
Let's first color code all of our nodes.

00:00:14.310 --> 00:00:18.630
Here is a red node,
that's a extraordinary node.

00:00:18.630 --> 00:00:22.694
Here is an orange node,
that's another extraordinary node.

00:00:25.197 --> 00:00:28.550
Here is a blue node, also extraordinary.

00:00:28.550 --> 00:00:32.790
And we can see that at the bottom,
we have a black node that is also

00:00:32.790 --> 00:00:36.540
an extraordinary node that
includes all of these elements.

00:00:36.540 --> 00:00:39.560
Well let's begin by combining
our resistors in parallel.

00:00:39.560 --> 00:00:41.856
We can see that,
these two 16 are in parallel,

00:00:41.856 --> 00:00:44.960
because they are red on the top and
black on the bottom.

00:00:44.960 --> 00:00:49.940
When we have resistors in parallel,
we combine them in this way and

00:00:49.940 --> 00:00:52.840
if they happen to be
equal as the two 16 are.

00:00:52.840 --> 00:00:56.610
Then we end up with half
of the original resistance.

00:00:56.610 --> 00:01:00.647
So, when you combine these two in
parallel 16 and parallel with 16,

00:01:00.647 --> 00:01:02.112
it's going to give us 8.

00:01:02.112 --> 00:01:04.680
We can now see the 4 and
the 8 are in series,

00:01:04.680 --> 00:01:08.680
because they have a single
ordinary node between them.

00:01:08.680 --> 00:01:13.381
So, when we combine those in series
we just add them up, and the 8 and

00:01:13.381 --> 00:01:18.019
4 together, right here,
is going to give me a resistance of 12.

00:01:18.019 --> 00:01:21.829
Now, we have a 12 from yellow to black and
yellow to black,

00:01:21.829 --> 00:01:23.900
those two are in parallel.

00:01:23.900 --> 00:01:28.670
So 12 and 12 in parallel,
because of the same going to give me 6.

00:01:28.670 --> 00:01:30.510
Now, something cool happens here.

00:01:30.510 --> 00:01:35.130
I end up with a 6 on a resistance right
here, which I can bring to the other side

00:01:35.130 --> 00:01:40.040
of my current source and
that's because, they have the same node

00:01:40.040 --> 00:01:43.780
across the topic and rearrange anything
that's connected in the same nodes.

00:01:43.780 --> 00:01:47.734
So, my 6 is going to come over there and
then the 6 and 6 in parallel,

00:01:47.734 --> 00:01:50.460
is going to get me something that is 3.

00:01:50.460 --> 00:01:56.096
So my final circuit,
is going to look like a 10 amp current

00:01:56.096 --> 00:02:01.280
source in parallel with 3 ohms,
and then I have a 4 and

00:02:01.280 --> 00:02:07.738
let's combine these in series 8,
and I want to find this voltage.

00:02:07.738 --> 00:02:08.810
There's not a minus there.

00:02:08.810 --> 00:02:10.900
I want to find Vx.

00:02:10.900 --> 00:02:14.040
Now, I also can see that 8 and
4 go together in series.

00:02:14.040 --> 00:02:18.270
But, because I want to find Vx, I need to
leave the element that it's across and

00:02:18.270 --> 00:02:23.200
not combine it with other stuff or else
it will get all mixed up in my circuit.

00:02:23.200 --> 00:02:24.620
Now, here's the circuit that we have, and

00:02:24.620 --> 00:02:27.300
I'm going to show you two different
ways of being able to solve it.