Rsp and vi dividers 1 of 2

Okay, here's a problem that we
were working in class today, and

I would like to just go over
this example again for you.

We want to be able to find Vx.

We can see we have a lot of resistors here
we can combine in series and parallel.

Let's first color code all of our nodes.

Here is a red node,
that's a extraordinary node.

Here is an orange node,
that's another extraordinary node.

Here is a blue node, also extraordinary.

And we can see that at the bottom,
we have a black node that is also

an extraordinary node that
includes all of these elements.

Well let's begin by combining
our resistors in parallel.

We can see that,
these two 16 are in parallel,

because they are red on the top and
black on the bottom.

When we have resistors in parallel,
we combine them in this way and

if they happen to be
equal as the two 16 are.

Then we end up with half
of the original resistance.

So, when you combine these two in
parallel 16 and parallel with 16,

it's going to give us 8.

We can now see the 4 and
the 8 are in series,

because they have a single
ordinary node between them.

So, when we combine those in series
we just add them up, and the 8 and

4 together, right here,
is going to give me a resistance of 12.

Now, we have a 12 from yellow to black and
yellow to black,

those two are in parallel.

So 12 and 12 in parallel,
because of the same going to give me 6.

Now, something cool happens here.

I end up with a 6 on a resistance right
here, which I can bring to the other side

of my current source and
that's because, they have the same node

across the topic and rearrange anything
that's connected in the same nodes.

So, my 6 is going to come over there and
then the 6 and 6 in parallel,

is going to get me something that is 3.

So my final circuit,
is going to look like a 10 amp current

source in parallel with 3 ohms,
and then I have a 4 and

let's combine these in series 8,
and I want to find this voltage.

There's not a minus there.

I want to find Vx.

Now, I also can see that 8 and
4 go together in series.

But, because I want to find Vx, I need to
leave the element that it's across and

not combine it with other stuff or else
it will get all mixed up in my circuit.

Now, here's the circuit that we have, and

I'm going to show you two different
ways of being able to solve it.