Look at the way that these six and three farad capacitors
are connected to each other.
What's going to happen if we hook them up
to an eight volt battery?
Well, like all capacitors, charge
is going to get separated, so negatives
are going to get stripped off of the right sides
of these capacitors and pulled towards the positive terminal
of the battery.
But when they reach the other side,
something interesting happens here.
The charges reach this junction, or fork in the road.
And now they have a choice in whether they're
going to get deposited onto the three
farad capacitor or the six farad capacitor.
Each capacitor is going to get some of the charge,
but since the six farad capacitor
has twice the capacitance that the three farad capacitor does,
the six farad capacitor's going to get twice
as much charge stored on it as does the three farad capacitor.
So twice as many negatives are going
to get pulled off of the right side of the six
farad capacitor, and twice as many
negatives are going to get deposited
on to the left side of the six farad capacitor.
OK.
So the six farad capacitor is going to get twice as much.
But exactly how much charge are these capacitors going to get?
Even though the circuit looks a little complicated,
finding the charge in this case is actually really easy.
The reason it's going to be easy is
that both of these capacitors are hooked up directly
to the terminals of the battery.
In other words, the positive side of the six farad capacitor
is hooked directly up to the positive terminal
of the battery.
And the negative side of the six farad capacitor
is connected directly to the negative terminal
of the battery.
This means that the voltage across the six farad capacitor
is going to be the same as the voltage of the battery, which
is eight volts in this case.
The same is also true for the three farad capacitor.
So the voltage across the three farad capacitor
is also eight volts.
In fact, the way these capacitors
are hooked up, it's as if they were connected to the eight
volt battery all by themselves, because they both experience
the entire voltage of the battery.
Now that we know the voltage across these capacitors,
we can use the definition of capacitance
to solve for the charge.
For the three farad capacitor, we
can plug-in a capacitance of three farads
and a voltage of eight volts.
And we get that the charge stored on the three
farad capacitor is 24 coulombs.
We could do the same type of calculation
for the six farad capacitor.
We plug-in six farads an eight volts,
and we get that the charge on the six farad capacitor
is 48 coulombs.
And see, just like we said, the charge
on the six farad capacitor is twice as much
as the charge on the three farad capacitor.
We call capacitors hooked up in this way capacitors
in parallel.
You'll know that two capacitors are hooked up in parallel
if their positive sides are directly connected
to each other with a wire, and their negative sides
are also directly connected to each other with a wire.
We could ask ourselves now, what should the value
be of a single capacitor whose effect on this circuit
would be equivalent to that of the individual parallel
capacitors?
To find the equivalent capacitance of capacitors
hooked up in parallel, all you need to do
is add up the individual capacitances.
And the reason is, just look at these capacitors.
Since their positive sides are connected with a wire,
you may as well have just merged all of the positive sides
together to form one big positive side.
And since their negative sides are all connected with a wire,
you may as well have just merged the negative sides
into one big negative side.
So all you've really done by hooking up
capacitors in parallel is to make one big capacitor out
of smaller capacitors.
Now, keep in mind that the capacitance of a capacitor
is proportional to the area of the capacitor plates.
So since we added the available areas of the capacitors
together to get the total capacitance, all we need to do
is to add up the individual capacitances.
Even though the charge on the individual parallel capacitors
might not be the same, they're charge
has to add up to the total charge that
would be stored on the equivalent capacitor.
So if these parallel capacitors stored one coulomb, two
coulombs, and three coulombs individually,
their equivalent capacitor would store six coulombs.
Let's try to apply these ideas to the circuit we just
examined in the beginning of this video.
The equivalent capacitance of these six farad and three
farad capacitors would be a single nine farad capacitor.
Now, let's solve for the amount of charge
that this nine farad equivalent capacitor would
store when hooked up to the eight volt battery.
Using the definition of capacitance,
we find that the charge on a nine farad capacitor
would be 72 coulombs.
And this makes sense, because remember the charge stored
on the six farad capacitor was 48 coulombs,
and the charge stored on the three farad capacitor
was 24 coulombs.
So the total charge on the six farad and three
farad capacitors is 72 coulombs, which
is the same charge that their equivalent capacitor stores.
Let's try another problem that's a little more challenging.
Say we introduce a 27 farad capacitor into this circuit.
When the battery's connected, the capacitors
will all store charge and have a certain voltage across them.
So let's try to figure out the charge on
and the voltage across all of these capacitors.
Well, to start, we might notice that the three farad and six
farad capacitors are still in parallel with each other, which
means they have to have the same voltage as each other.
But this time, the value of that voltage
is not going to be the same as the voltage of the battery.
Because even though their negative sides are connected
directly to the negative terminal of the battery,
their positive sides are not connected directly
to the positive terminal of the battery.
This 27 farad capacitor is getting in the way here.
Similarly, the voltage across the 27 farad capacitor
is also not going to be the same as the voltage of the battery.
Because even though it's positive
side is connected directly to the positive terminal
of the battery, it's negative side
is not connected directly to the negative terminal
of the battery.
So in summary, we don't know the voltage
across any of these capacitors.
And if we don't know the voltage across any
of these individual capacitors, how
are we ever going to solve for the charge on these capacitors?
Well, the one thing that we do know
is that the voltage across the whole circuit is eight volts.
We just don't know the individual voltages
across each capacitor.
So what we're going to try to do is
to replace these individual capacitors
with a single equivalent capacitor.
To do that, let's start with the six farad and three
farad capacitors, because we know those are in parallel.
We know they're in parallel because their positive sides
are connected directly to each other
and their negative sides are connected directly
to each other.
Using the rule to combine parallel capacitors,
we get that the equivalent capacitance of the three
and six farad capacitors is a single nine farad capacitor.
So now we have a nine farad capacitor
and a 27 farad capacitor.
These are connected in series, because they're
hooked up one right after the other.
Or in other words, the positive side on one capacitor
is connected to the negative side on the other capacitor.
We can replace these two capacitors
with a single equivalent capacitor
by using the formula for adding capacitors
in series, which is 1 over the equivalent capacitance equals
1 over C1 plus 1 over C2.
So plugging in the values of nine farads and 27 farads,
we get that 1 over the equivalent capacitance
equals 0.148148.
Don't forget to take 1 over this number
to get that the equivalent capacitance is 6.75 farads.
So we can replace the nine farad and 27 farad capacitors
with a single 6.75 farad capacitor.
Now, finally, we can solve for the charge on this 6.75
farad equivalent capacitor.
Because it's positive side is connected directly
to the positive terminal of the battery,
and its negative side is connected directly
to the negative terminal of the battery.
That means the voltage across this 6.75 farad capacitor
is going to be eight volts.
We can use the definition of capacitance,
and we get that the charge on this 6.75 farad capacitor
is 54 coulombs.
So since this was the equivalent capacitor for two series
capacitors, both of these series capacitors
must have the same charge as their equivalent capacitor.
So both the 27 farad and nine farad capacitors
have 54 coulombs each stored on them.
At this point, we can figure out the voltage
across these two capacitors.
Using the definition of capacitance,
we can plug-in 27 farads and 54 coulombs
to get that the voltage across the 27 farad capacitor
is two volts.
Doing the same type of calculation
for the nine farad capacitor, we get
that the voltage across the nine farad capacitor is six volts.
Notice that two volts and six volts
adds up to the voltage of the battery, eight volts,
just like they have to.
And now we can find the charge stored on the individual three
farad and six farad capacitors.
We know now that the voltage across both the three
farad and six farad capacitors is going to be six volts.
Because the voltage across the individual
capacitors in parallel has to be the same
as the voltage across their equivalent capacitor.
Now that we know the voltage, we can
use the definition of capacitance.
And for the three farad capacitor,
we get that the charge stored is going to be 18 coulombs.
And doing the same type of calculation
for the six farad capacitor, we get
that the charge is 36 coulombs.
This makes sense, because 18 coulombs plus 36 coulombs
adds up to 54 coulombs, which was the charge stored
on their equivalent nine farad capacitor.