Last video, we saw what happens when we have
resistors in series.
Now let's see what happens when we have
resistors in parallel.
All right, let me pick a new color.
New color will be magenta.
There's my battery: positive, negative.
There's my ideal conducting wire.
Here's my ideal conducting wire, but then-- and this is
new-- it branches off, and I have two resistances.
I have one here.
And that's another resistance.
And let's say that this has a resistance R1, this has a
resistance R2.
And, of course, the non-intuitive convention is
that the current flows from the positive to the negative
terminal, but we know that the electrons are actually flowing
in the other direction.
And I want to keep saying that, because I think it's so
important to understand what's actually happening as opposed
to the convention.
Well, anyway, in the previous video, we said, well, when we
have devices or components in series, that the current
through the entire circuit is constant.
But let's think about what happens here.
So we have these electrons-- actually, let's think about it
from the electron flow.
The electrons are flowing, flowing, flowing at a given
rate, and then here they have a choice.
Some of them can take this top path, some of them can take
this bottom path.
So if you think about it, the flow of electrons in this
branch plus the flow of electrons in that branch have
to add up to the flow of the electrons in
this branch, right?
And then they're going to meet back up, and then the flow of
electrons here-- so if we think of it this way, and now
I'm going to go back to the convention, this is I1.
So you have these electrons flowing at a given rate.
This is the current right here.
They're going to branch off, and maybe half of them go--
we'll see if the resistances are equal, if both of these
branches have an equal amount of capacity in terms of how
fast the electrons can flow through.
If they're equal, or since we're going to current in this
direction, let's talk about positrons or positive charges.
If the positive charges-- although I just want to keep
saying that it is not the positive
charges that are moving.
It's the electrons.
But if you say that the lack of electrons can flow equally
easily between both paths, that's if the resistances were
the same, we could imagine that the current, the flow,
would split itself up, and then over here
would meet back up.
And then we would say that the current here would also be I1.
But let's figure out where the current's going.
Let's call this current I2 and let's call this current I3.
So I think it is reasonable, and you can imagine with water
pipes or anything, that the current going into the branch
is equal to the current exiting the branch.
Or you could even think of it that the current entering--
when the currents I2 and I1 merge, that they combine and
they become Current 1, right?
I mean, think about it.
In a given second, if this is 5 coulombs per second-- I'm
just making up numbers-- and this is 6 coulombs per second,
in a given second right here, you're going to have 5
coulombs coming from this branch and 6 coulombs coming
from this branch, so you're going to have 11 coulombs per
second coming out once they've merged, so this would be 11
coulombs per second.
So I think hopefully that makes sense to you that this
current is equal to the combination of this current
and that current.
Now, what do we also know?
We also know the voltage along this entire ideal wire is
constant, so then voltage-- let me draw that in another
color, in blue.
So, for example, the voltage anywhere along this blue that
I'm filling in is going to be the same, because this wire is
an ideal conductor, and you can almost view this blue part
as an extension of the positive
terminal of the battery.
And very similarly-- I'll do it in yellow-- we could draw
this wire as an extension of the negative
terminal of the battery.
This is an extension of the negative
terminal of the battery.
So the voltage difference between here and here-- so
let's call that the total voltage, or let's just call it
the voltage, right?
The voltage difference between that point and that point is
the exact same thing as the voltage difference between
this point and this point, which is the exact same thing
as the voltage difference between this
point and this point.
So what can we say?
What is the total current in the system?
If we just viewed this as a black box, that this is some
type of total resistance, well, the total current in the
system would be the total voltage, the voltage divided
by-- let's call this our total resistance, right?
Let's say we couldn't see this and we just said, oh, that's
just some total resistance, and that is equal to the
current going through R1.
This is I1.
This is a 1 right here.
This is current I1.
What's current I1?
Well, it's going to be the voltage across this resistor
divided by the resistance, right?
That's what Ohm's law tells us: V is equal to IR, or
another way we could say it is V over R is equal to I, right?
So I1 is equal to the voltage across this resistor, but we
just said that voltage is the same thing as
this voltage, right?
The voltage here is the same thing as the voltage here.
So the voltage across that resistor is still V, and so
the current flowing across that resistor is V over R1.
And the same logic: what is I2?
I2 is this current.
What is the voltage across this device?
Well, that's just V again, right?
It's the same thing as the voltage across this device, so
it's V over R2 by Ohm's law.
Well, all these V's are the same, so we can divide both
sides of that equation by V, and we get 1 over the total
resistance is equal to 1 over R1 plus 1 over R2.
And you could make that argument if we had an R3 here.
Let's say that we had another device, and that is R3.
You could use the exact same argument, and you would have a
plus 1 over R3.
And if you had Rn or 10 of them, you'd just keep that 1
over R4, R5, et cetera.
So let's see if we can use this information we have
learned to actually solve a problem, and I actually find
that it's always easier to solve a problem than to
explain the theory behind a problem.
You'll see that with most of these circuit problems, it's
actually very basic mathematics.
So let's say I have a 16 volt battery plus,
minus, it's 16 volts.
And just to hit the point home that you always don't have to
draw circuits the same, although it is nice if you're
actually drawing complicated circuits, I could
draw it like this.
I could draw the circuit like this, and let's say that
there's a resistor here.
And then let's say there's a wire and then there's another
resistor here, and that this decides to do some random
loopy thing here and that they connect here, and that they
come back here.
This strange thing that I have drawn, which you will never
see in any textbook, because most people are more
reasonable than me, is the exact same-- you can almost
view it topologically as the exact same circuit as what I
drew in the previous diagram, although now I will assign
numbers to it.
Let's say that this resistance is 20 ohms and let's say that
this resistance is 5 ohms. What I want to know is, what
is the current through the system?
First, we'll have to figure out what the equivalent
resistance is, and then we could just use Ohm's law to
figure out the current in the system.
So we want to know what the current is, and we know that
the convention is that current flows from the positive
terminal to the negative terminal.
So how do we figure out the equivalent resistance?
Well, we know that we just hopefully proved to you that
the total resistance is equal to 1 over this resistor plus 1
over this resistor.
So 1 over-- I won't keep writing it.
What's 1 over 20?
Well, actually, let's just make it a fraction.
So it's 1/20 plus-- 1/5 is what over 20?
That's 4/20, right?
So 1 over our total resistance is equal to 5/20, which is
equal to what?
1/4.
So if 1/R is equal to 1/4, R must be equal to 4.
R is equal to 4 ohms.
So we could redraw this crazy circuit as this.
I'll try to draw it small down here.
We could redraw this where this resistance is 4 ohms and
this is 16 volts.
We could say that this whole thing combined is really just
a resistor that is 4 ohms. Well, if we have a 16-volt
potential difference, current is flowing that way, even
though that's not what the electrons are doing.
And that's what our resistance is, 4
ohms. What is the current?
V equals IR, Ohms law.
The voltage is 16 volts.
It equals the current times 4 ohms. So current is equal to
16 divided by 4, is equal to 4 amps.
So let's do something interesting.
Let's figure out what the current is flowing through.
What's this?
What's the current I1 and what's this current I2?
Well, we know that the potential difference from here
to here is also 16 volts, right?
Because this whole thing is essentially at the same
potential and this whole thing is essentially at the same
potential, so you have 16 volts across there.
16 volts divided by 20 ohms, so let's call this I1.
So I1 is equal to 16 volts divided by 20 ohms, which is
equal to what?
4/5.
So it equals 4/5 of an ampere, or 0.8 amperes.
And similarly, what is the amount of current flowing
through here?
I2?
I'm going to do this in a different color.
It's getting confusing.
I'll do it in the vibrant yellow.
So the current flowing through here, once again, the
potential difference from here-- that's not different
enough-- the potential difference from here to here
is also 16 volts, right?
So the current is going to be I2, is going to be equal to
16/5, which is equal to 3 1/5 amps.
So most of the current is actually flowing through this,
and that makes sense because the resistance is less, right?
So that should hopefully give you a little bit of intuition
of what's going on.
And less current is flowing through here, so I2 through
the 20-ohm resistor is 0.8 amps is I1, and I2 through the
5-ohm resistor is equal to 3.2 amps.
And it makes sense that when you add these two currents
together, the 3.2 amperes flowing through here and the
0.8 amperes flowing through here, that when they merge,
they merge and then you have 4 amperes flowing through there.
Anyway, hopefully, I have given you some intuition on
what happens when we put resistors in parallel.
I will see you in the next video.