- [Voiceover] We've learned about series
and parallel resistors, we've learned how to
simplify series and parallel resistors
into an equivalent resistor, and just to review,
for the series resistor, R series, an equivalent R series
is equal to the sum of resistors in series.
R1 plus R2.
And we learned that if we have resistors in parallel,
meaning they share the same nodes, if they're in parallel,
we can get a parallel equivalent resistors
R parallel, and if there was two resistors,
if there's two resistors in parallel, the formula was
R1 times R2 over
R1 plus R2, plus R2.
And if there's three resistors or more in parallel
it's a little more complicated.
This formula look like this.
One over R parallel, this is for three or more resistors.
One over R1
plus one over R2,
plus one over R3, etcetera,
for as many resistors as you have.
These two parallel formulas do the same thing.
This one's a little more convenient to use
when you just have two resistors.
So now, this problem in this video,
we're going to look at a complicated resistor circuit.
There's a lot of resistors here.
The question we're going to ask is, if we plug in
this voltage source here into our resistor network
there's the two little plug points,
the question is what kind of current is going to be
pulled out of this voltage source?
That's what I want to know.
And if I stare at this, I go,
well how am I going to figure that out?
The important part of what we're doing here is to
stop for a minute and just see what this circuit looks like,
just be with it for a second.
Let your eyes wander over it and find
resistor patterns that you recognize.
And you just wait and do that.
All right, so we're looking for series in parallel patterns
and there's one, and this is not in series, no.
Okay, oh, there's a possibility, here's a possibility
of parallel resistors.
This, is there a series one over here.
I don't see a series head to tail connection here
and there's a series connection here that we can use.
Okay, so there's a couple of possibilities
here for simplification.
And one of the questions is, okay, where do we start?
We want to start this problem at the end farthest away
from the thing we care about, what we care about
is this right here, this is what we're looking for.
And so, a good way to start this problem
is to start at the very far end of the circuit
and work our way backwards.
And this is opposite the way you read sentences
and things like that, but in circuits
it's sometimes what we want to do.
So what I'm going to do is start over here on this side
and we're going to disassemble this circuit
and simplify it as we go.
So we've actually at this point, we've actually done
the hard work of this, and the rest of it's going to be
relatively straightforward application
of these two formulas.
We decided what our unknown is, and we decided
where to start and that strategy is the key
to making this a simple straightforward process.
All right, so let's go after it.
We're going to, we're basically now going to do
the work of simplifying this network.
So, what I see here is two resistors in series,
two ohms plus eight ohms equals 10 ohms.
So I can remove these resistors here
and we place them with
a 10 ohm resistor, that's step one.
What do we do next? Now we look at it again.
Right here we see we have two parallel resistors.
So we're actually going to use this form.
We're going to use this form of the parallel resistor
formula, and that says that 10 times 10
over 10 plus 10,
this is equal to 100 over 20
or five ohms
and I could've done this a little quicker
if I know that if two resistors in parallel
have the same value, 10 and 10.
I know that the parallel combination
of those is exactly half.
So now we have an equivalent resistor for those two guys.
We can take them out
and we replace that with
a five ohm resistor.
You can see where this is going,
we're just basically going to collapse this circuit
one step at a time.
Five ohms plus one ohm is this series combination here
and that is equal to six ohms, so I can replace this now
with six ohms, we're just consuming our circuit
and erasing and rewriting as we go.
And I said six ohms here, and what do we have now?
Now this is a little more challenging.
I have three resistors in parallel
and I'm going to use now this formula here
for resistors in parallel, so
one over the parallel combination is
one over 12 plus
one over four plus
one over six and the common denominator is 12
so that is
something over 12, which will be
one plus three over 12
plus two over 12, and that equals six over 12
which is one half
and one over Rp, that's one over Rp,
so Rp equals two over one or two ohms.
And now I can replace all three of these.
I get to erase all three.
We're almost done.
And that was, I said two ohms, two ohms.
The final step is one and two ohms in series.
One plus two equals three
and finally we get to the last step
which is
All right, so we're done.
What that means is, from the viewpoint,
as far as that voltage source can tell,
it's putting out enough current to drive three ohms.
And that whole rest of that complicated network,
looks like three ohms to that resistor,
to that voltage source.
And we have taken a complicated circuit
and turned it into a really simple one.
So that's the steps of simplification.
The key was figuring out what you want
and then going to the far end of the circuit
and working your way backwards
and simplifying as you go.