Having to deal with a single capacitor hooked up
to a battery isn't all that difficult,
but when you have multiple capacitors,
people typically get much, much more confused.
There's all kinds of different ways
to hook up multiple capacitors.
But if capacitors are connected one
after the other in this way, we call
them capacitors hooked up in series.
So say you were taking a test, and on the test
it asked you to find the charge on the leftmost capacitor.
What some people might try to do is this.
Since capacitance is the charge divided by the voltage,
they might plug in the capacitance
of the leftmost capacitor, which is 4 farads, plug
in the voltage of the battery, which is 9 volts.
Solving for the charge, they'd get that the leftmost capacitor
stores 36 coulombs, which is totally the wrong answer.
To try and figure out why and to figure out
how to properly deal with this type of scenario,
let's look at what's actually going on in this example.
When the battery's hooked up, a negative charge
will start to flow from the right side of capacitor
3, which makes a negative charge get deposited
on the left side of capacitor 1.
This makes a negative charge flow
from the right side of capacitor 1
on to the left side of capacitor 2.
And that makes a negative charge flow
from the right side of capacitor 2
on to the left side of capacitor 3.
Charges will continue doing this.
And it's important to note something here.
Because of the way the charging process works,
all of the capacitors here must have the same amount
of charge stored on them.
It's got to be that way.
Looking at how these capacitors charge up,
there's just nowhere else for the charge
to go but on to the next capacitor in the line.
This is actually good news.
This means that for capacitors in series,
the charge stored on every capacitor
is going to be the same.
So if you find the charge on one of the capacitors,
you've found the charge on all of the capacitors.
But how do we figure out what that amount of charge
is going to be?
Well, there's a trick we can use when
dealing with situations like this.
We can imagine replacing our three capacitors
with just a single equivalent capacitor.
If we choose the right value for this single capacitor,
then it will store the same amount of charge
as each of the three capacitors in series will.
The reason this is useful is because we
know how to deal with a single capacitor.
We call this imaginary single capacitor
that's replacing multiple capacitors
the "equivalent capacitor."
It's called the equivalent capacitor
because its effect on the circuit
is, well, equivalent to the sum total effect
that the individual capacitors have on the circuit.
And it turns out that there's a handy formula that
lets you determine the equivalent capacitance.
The formula to find the equivalent capacitance
of capacitors hooked up in series looks like this.
1 over the equivalent capacitance
is going to equal 1 over the first capacitance
plus 1 over the second capacitance plus 1
over the third capacitance.
And if you had more capacitors that were in that same series,
you would just continue on this way
until you've included all of the contributions
from all of the capacitors.
We'll prove where this formula comes from in a minute,
but for now, let's just get used to using it
and see what we can figure out.
Using the values from our example,
we get that 1 over the equivalent capacitance
is going to be 1 over 4 farads plus 1 over 12 farads
plus 1 over 6 farads, which equals 0.5.
But be careful.
You're not done yet.
We want the equivalent capacitance, not
1 over the equivalent capacitance.
So we have to take 1 over this value of 0.5 that we found.
And if we do that, we get that the equivalent capacitance
for this series of capacitors is 2 farads.
Now that we've reduced our complicated multiple capacitor
problem into a single capacitor problem,
we can solve for the charge stored on this equivalent
capacitor.
We can use the formula capacitance equals charge
per voltage and plug in the value of the equivalent
capacitance.
And we can plug in the voltage of the battery
now because the voltage across a single charged-up capacitor
is going to be the same as the voltage of the battery that
charged it up.
Solving for the charge, we get that the charge stored
on this equivalent capacitor is 18 coulombs.
But we weren't trying to find the charge on the equivalent
capacitor.
We were trying to find the charge
on the leftmost capacitor.
But that's easy now because the charge
on each of the individual capacitors in series
is going to be the same as the charge on the equivalent
capacitor.
So since the charge on the equivalent capacitor
was 18 coulombs, the charge on each
of the individual capacitors in series
is going to be 18 coulombs.
This process can be confusing to people,
so let's try another example.
This time, let's say you had four capacitors hooked up
in series to a 24-volt battery.
The arrangement of these capacitors
looks a little different from the last example,
but all of these capacitors are still in series
because they're hooked up one right after the other.
In other words, the charge has no choice
but to flow directly from one capacitor
straight to the next capacitor.
So these capacitors are still considered to be in series.
Let's try to figure out the charge that's
going to be stored on the 16-farad capacitor.
We'll use the same process as before.
First we imagine replacing the four capacitors
with a single equivalent capacitor.
We'll use the formula to find the equivalent
capacitance of capacitors in series.
Plugging in our values, we find that 1 over the equivalent
capacitance is going to equal 0.125.
Be careful.
We still have to take 1 over this value
to get that the equivalent capacitance for this circuit
is going to be 8 farads.
Now that we know the equivalent capacitance,
we can use the formula capacitance
equals charge per voltage.
We can plug in the value of the equivalent capacitance, 8
farads.
And since we have a single capacitor now,
the voltage across that capacitor
is going to be the same as the voltage of the battery, which
is 24 volts.
So we find that our imaginary equivalent capacitor
would store a charge of 192 coulombs.
This means that the charge on each
of the individual capacitors is also going to be 192 coulombs.
And this gives us our answer, that the charge
on the 16-farad capacitor is going to be 192 coulombs.
In fact, we can go even further.
Now that we know the charge on each capacitor,
we can solve for the voltage that's
going to exist across each of the individual capacitors.
We'll again use the fact that capacitance
is the charge per voltage.
If we plug in the values for capacitor one,
we'll plug in a capacitance of 32 farads.
The charge that capacitor one stores is 192 coulombs.
So we can solve for the voltage across capacitor 1,
and we get 6 volts.
If we were to do the same calculation for each
of the other three capacitors, always being careful that we
use their particular values, we'll
get that the voltages across the capacitors
are 2 volts across the 96-farad capacitor,
12 volts across the 16-fard capacitor,
and 4 volts across the 48-farad capacitor.
Now, the real reason I had us go through this
is because I wanted to show you something neat.
If you add up the voltages that exist across each
of the capacitors, you'll get 24 volts,
the same as the value of the battery.
This is no coincidence.
If you add up the voltages across the components
in any single-loop circuit like this, the sum of the voltages
is always going to equal the voltage of the battery.
And this principle will actually let
us derive the formula we've been using for the equivalent
capacitance of series capacitors.
To derive this formula, let's say
we've got three capacitors with capacitances of C1, C2, and C3
hooked up in series to a battery of voltage V. We now
know that if we add up the voltage across each capacitor,
it's got to add up to the voltage of the battery.
Using the formula for capacitance,
we can see that the voltage across an individual capacitor
is going to be the charge on that capacitor divided
by its capacitance.
So the voltage across each capacitor
is going to be Q over C1, Q over C2, and Q over C3,
respectively.
I didn't write Q1, Q2, or Q3 because remember,
all the charges on capacitors in series
are going to be the same.
These voltages have to add up to the voltage of the battery.
I can pull out a common factor of Q
because it's in each term on the left.
And now I'm going to divide each side by Q.
I did that because look at what we've
got on the right-hand side of this equation.
The voltage across the battery divided by the charge stored
is just equal to 1 over the equivalent capacitance,
because Q over V is equal to the equivalent capacitance.
And here it is.
This is the formula we've been using,
and this is where it comes from.
It's derived from the fact that the voltages
across these capacitors in series
have to add up to the voltage of the battery.