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Look at the way that these
six and three farad capacitors

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are connected to each other.

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What's going to happen
if we hook them up

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to an eight volt battery?

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Well, like all
capacitors, charge

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is going to get
separated, so negatives

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are going to get stripped
off of the right sides

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of these capacitors and pulled
towards the positive terminal

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of the battery.

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But when they reach
the other side,

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something interesting
happens here.

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The charges reach this
junction, or fork in the road.

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And now they have a
choice in whether they're

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going to get deposited
onto the three

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farad capacitor or the
six farad capacitor.

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Each capacitor is going
to get some of the charge,

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but since the six
farad capacitor

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has twice the capacitance that
the three farad capacitor does,

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the six farad capacitor's
going to get twice

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as much charge stored on it as
does the three farad capacitor.

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So twice as many
negatives are going

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to get pulled off of the
right side of the six

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farad capacitor,
and twice as many

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negatives are going
to get deposited

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on to the left side of
the six farad capacitor.

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OK.

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So the six farad capacitor is
going to get twice as much.

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But exactly how much charge are
these capacitors going to get?

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Even though the circuit
looks a little complicated,

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finding the charge in this
case is actually really easy.

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The reason it's
going to be easy is

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that both of these capacitors
are hooked up directly

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to the terminals of the battery.

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In other words, the positive
side of the six farad capacitor

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is hooked directly up
to the positive terminal

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of the battery.

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And the negative side of
the six farad capacitor

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is connected directly
to the negative terminal

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of the battery.

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This means that the voltage
across the six farad capacitor

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is going to be the same as the
voltage of the battery, which

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is eight volts in this case.

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The same is also true for
the three farad capacitor.

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So the voltage across
the three farad capacitor

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is also eight volts.

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In fact, the way
these capacitors

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are hooked up, it's as if they
were connected to the eight

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volt battery all by themselves,
because they both experience

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the entire voltage
of the battery.

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Now that we know the voltage
across these capacitors,

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we can use the
definition of capacitance

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to solve for the charge.

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For the three
farad capacitor, we

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can plug-in a capacitance
of three farads

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and a voltage of eight volts.

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And we get that the
charge stored on the three

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farad capacitor is 24 coulombs.

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We could do the same
type of calculation

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for the six farad capacitor.

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We plug-in six farads
an eight volts,

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and we get that the charge
on the six farad capacitor

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is 48 coulombs.

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And see, just like
we said, the charge

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on the six farad
capacitor is twice as much

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as the charge on the
three farad capacitor.

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We call capacitors hooked
up in this way capacitors

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in parallel.

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You'll know that two capacitors
are hooked up in parallel

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if their positive sides
are directly connected

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to each other with a wire,
and their negative sides

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are also directly connected
to each other with a wire.

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We could ask ourselves
now, what should the value

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be of a single capacitor
whose effect on this circuit

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would be equivalent to that
of the individual parallel

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capacitors?

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To find the equivalent
capacitance of capacitors

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hooked up in parallel,
all you need to do

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is add up the
individual capacitances.

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And the reason is, just
look at these capacitors.

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Since their positive sides
are connected with a wire,

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you may as well have just
merged all of the positive sides

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together to form one
big positive side.

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And since their negative sides
are all connected with a wire,

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you may as well have just
merged the negative sides

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into one big negative side.

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So all you've really
done by hooking up

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capacitors in parallel is to
make one big capacitor out

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of smaller capacitors.

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Now, keep in mind that the
capacitance of a capacitor

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is proportional to the area
of the capacitor plates.

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So since we added the available
areas of the capacitors

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together to get the total
capacitance, all we need to do

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is to add up the
individual capacitances.

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Even though the charge on the
individual parallel capacitors

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might not be the
same, they're charge

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has to add up to the
total charge that

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would be stored on the
equivalent capacitor.

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So if these parallel capacitors
stored one coulomb, two

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coulombs, and three
coulombs individually,

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their equivalent capacitor
would store six coulombs.

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Let's try to apply these
ideas to the circuit we just

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examined in the
beginning of this video.

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The equivalent capacitance
of these six farad and three

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farad capacitors would be a
single nine farad capacitor.

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Now, let's solve for
the amount of charge

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that this nine farad
equivalent capacitor would

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store when hooked up to
the eight volt battery.

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Using the definition
of capacitance,

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we find that the charge
on a nine farad capacitor

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would be 72 coulombs.

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And this makes sense, because
remember the charge stored

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on the six farad
capacitor was 48 coulombs,

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and the charge stored on
the three farad capacitor

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was 24 coulombs.

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So the total charge on
the six farad and three

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farad capacitors is
72 coulombs, which

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is the same charge that their
equivalent capacitor stores.

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Let's try another problem that's
a little more challenging.

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Say we introduce a 27 farad
capacitor into this circuit.

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When the battery's
connected, the capacitors

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will all store charge and have
a certain voltage across them.

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So let's try to figure
out the charge on

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and the voltage across
all of these capacitors.

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Well, to start, we might notice
that the three farad and six

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farad capacitors are still in
parallel with each other, which

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means they have to have the
same voltage as each other.

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But this time, the
value of that voltage

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is not going to be the same
as the voltage of the battery.

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Because even though their
negative sides are connected

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directly to the negative
terminal of the battery,

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their positive sides are
not connected directly

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to the positive
terminal of the battery.

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This 27 farad capacitor is
getting in the way here.

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Similarly, the voltage
across the 27 farad capacitor

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is also not going to be the same
as the voltage of the battery.

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Because even though
it's positive

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side is connected directly
to the positive terminal

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of the battery,
it's negative side

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is not connected directly
to the negative terminal

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of the battery.

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So in summary, we
don't know the voltage

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across any of these capacitors.

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And if we don't know
the voltage across any

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of these individual
capacitors, how

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are we ever going to solve for
the charge on these capacitors?

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Well, the one thing
that we do know

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is that the voltage across the
whole circuit is eight volts.

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We just don't know the
individual voltages

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across each capacitor.

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So what we're going
to try to do is

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to replace these
individual capacitors

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with a single
equivalent capacitor.

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To do that, let's start
with the six farad and three

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farad capacitors, because we
know those are in parallel.

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We know they're in parallel
because their positive sides

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are connected
directly to each other

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and their negative sides
are connected directly

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to each other.

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Using the rule to combine
parallel capacitors,

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we get that the equivalent
capacitance of the three

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and six farad capacitors is a
single nine farad capacitor.

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So now we have a
nine farad capacitor

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and a 27 farad capacitor.

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These are connected in
series, because they're

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hooked up one right
after the other.

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Or in other words, the
positive side on one capacitor

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is connected to the negative
side on the other capacitor.

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We can replace
these two capacitors

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with a single
equivalent capacitor

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by using the formula
for adding capacitors

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in series, which is 1 over the
equivalent capacitance equals

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1 over C1 plus 1 over C2.

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So plugging in the values of
nine farads and 27 farads,

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we get that 1 over the
equivalent capacitance

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equals 0.148148.

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Don't forget to take
1 over this number

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to get that the equivalent
capacitance is 6.75 farads.

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So we can replace the nine
farad and 27 farad capacitors

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with a single 6.75
farad capacitor.

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Now, finally, we can solve
for the charge on this 6.75

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farad equivalent capacitor.

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Because it's positive
side is connected directly

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to the positive
terminal of the battery,

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and its negative side
is connected directly

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to the negative
terminal of the battery.

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That means the voltage across
this 6.75 farad capacitor

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is going to be eight volts.

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We can use the definition
of capacitance,

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and we get that the charge
on this 6.75 farad capacitor

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is 54 coulombs.

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So since this was the equivalent
capacitor for two series

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00:08:25,390 --> 00:08:28,310
capacitors, both of
these series capacitors

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must have the same charge as
their equivalent capacitor.

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00:08:31,770 --> 00:08:35,940
So both the 27 farad and
nine farad capacitors

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have 54 coulombs
each stored on them.

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At this point, we can
figure out the voltage

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00:08:41,360 --> 00:08:43,380
across these two capacitors.

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00:08:43,380 --> 00:08:45,550
Using the definition
of capacitance,

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00:08:45,550 --> 00:08:49,570
we can plug-in 27
farads and 54 coulombs

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00:08:49,570 --> 00:08:53,090
to get that the voltage
across the 27 farad capacitor

202
00:08:53,090 --> 00:08:54,680
is two volts.

203
00:08:54,680 --> 00:08:56,500
Doing the same
type of calculation

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00:08:56,500 --> 00:08:58,840
for the nine farad
capacitor, we get

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00:08:58,840 --> 00:09:02,960
that the voltage across the nine
farad capacitor is six volts.

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00:09:02,960 --> 00:09:05,620
Notice that two
volts and six volts

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00:09:05,620 --> 00:09:08,520
adds up to the voltage of
the battery, eight volts,

208
00:09:08,520 --> 00:09:10,220
just like they have to.

209
00:09:10,220 --> 00:09:13,480
And now we can find the charge
stored on the individual three

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00:09:13,480 --> 00:09:16,040
farad and six farad capacitors.

211
00:09:16,040 --> 00:09:18,640
We know now that the voltage
across both the three

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00:09:18,640 --> 00:09:22,980
farad and six farad capacitors
is going to be six volts.

213
00:09:22,980 --> 00:09:25,310
Because the voltage
across the individual

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00:09:25,310 --> 00:09:28,130
capacitors in parallel
has to be the same

215
00:09:28,130 --> 00:09:31,410
as the voltage across
their equivalent capacitor.

216
00:09:31,410 --> 00:09:32,910
Now that we know
the voltage, we can

217
00:09:32,910 --> 00:09:34,890
use the definition
of capacitance.

218
00:09:34,890 --> 00:09:37,150
And for the three
farad capacitor,

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00:09:37,150 --> 00:09:40,820
we get that the charge stored
is going to be 18 coulombs.

220
00:09:40,820 --> 00:09:42,710
And doing the same
type of calculation

221
00:09:42,710 --> 00:09:44,720
for the six farad
capacitor, we get

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00:09:44,720 --> 00:09:47,390
that the charge is 36 coulombs.

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This makes sense, because
18 coulombs plus 36 coulombs

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adds up to 54 coulombs,
which was the charge stored

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00:09:55,140 --> 00:09:58,460
on their equivalent
nine farad capacitor.

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