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## Physics library

### Course: Physics library > Unit 12

Lesson 2: Circuits with capacitors# Capacitors in series

When capacitors are connected one after another, they are said to be in series. For capacitors in series, the total capacitance can be found by adding the reciprocals of the individual capacitances, and taking the reciprocal of the sum. Therefore, the total capacitance will be lower than the capacitance of any single capacitor in the circuit. . Created by David SantoPietro.

## Want to join the conversation?

- How can charges move between the two sides of a capacitor if they are separated?(12 votes)
- They don't move between the two sides.

A build up of negative charges on one side of the capacitor causes the negative charges on the other side to be pushed away or repelled from the edge of the capacitor. As more and more negative charges build up, more and more negative charges get repelled away from the other side and soon that side is very positive with little electrons. The electrons or negative charges keep on going in a circle.(25 votes)

- Why is the sum of the voltages of the capacitors equal to the voltage of the battery?(14 votes)
- Because the total voltage supplied must be equal to the total voltage across the circuit. Voltage gets divided among the capacitors when they are connected in series.(11 votes)

- At6:32, why is the charge stored on each of the individual capacitors equal to the charge stored on the equivalent capacitor? Why aren't the charges divided between the four- like each one has 192/4 C of charge?(11 votes)
- NO, remember that the Capacitance unit is F, not C, So basically you messed up, you should NOT sum like this, they have the same amount of Charge NOT Capacitance.

So you add (1/48F) + (1/16F) + (1/96F) + (1/32F) = 0.125F, Then taking the reciprocal you get 8F which is the equivalent of CAPACITANCE.

When you try to find the Voltage you do this ( 192/48 ) + ( 192/16 ) + ( 192/96 ) + ( 192/32 ) = 24v which is the same voltage of the battery.

I Hope that helped!(1 vote)

- If the voltage increases as charge increases, which increases as time goes on. How does the loop rule apply when the battery is just connected to the circuit and the capacitors haven't had enough time to build up charge yet (i.e. the sum of voltages across the capacitors wouldn't equal the batter voltage)?(7 votes)
- Right, the voltage is not initially equal, and that's why some current flows to charge the capacitors. The current flows until the voltage does equalize, and then it stops.(4 votes)

- I have a slightly off topic question, about Resistors being in series with a capacitor.

If the source Pd = the resistor Pd + the capacitor Pd, can it be said that the voltage across the resistor decreases as the charge increases on the capacitor (since Q is proportional to V) and as this is for charging, will discharging be : source Pd =resistor Pd - Capacitor Pd ?(4 votes)- nice question.

Lets take a step back a bit. The capacitor consists (as you know) of two plates separated by a dielectric...insultor. So, in a DC circuit (which you have here..) no current will flow once the capacitor is charged.

This means there will be zero voltage drop across the resistor when the capacitor is fully charged. and the the voltage across the charged resistor = source voltage.

during the charging process, the voltage drop across the resistor will be equal to the current at any time t multiplied by the resistance.

I would say that the The voltage across the capacitor will be source voltage - voltage drop across resistor.

Hope that helps

IM(5 votes)

- Does this mean that the higher the capacitance, the lower the voltage of a capacitor hooked up in a series?(5 votes)
- Yes, that is basically correct. Let's say you have two capacitors connected in series to a voltage of 3V. One has a capacitance of 1F and the other has a capacitance of 2F. The larger capacitor (the 2F one) has a voltage across it of 1V while the smaller capacitor (the 1F one) has a voltage across it of 2V.(1 vote)

- At4:51, why is the charge on the equivalent capacitor equal to the charge on EACH of the 3 capacitors? Why is it not divided by 3, so each capacitor holds 18/3= 6 C of charge?(3 votes)
- Because charge is conserved. If you connect one capacitor to a battery, what happens? You get some positive charge on one plate and some negative charge on the other. The sum of the + and the - is 0. Now if you add another capacitor in series with the first one, the net charge is still going to be zero. You are going to have + charge on top plate of top capacitor, and - charge on bottom plate of bottom capacitor. What happens to the plates "in the middle". They have to have net charge of zero, too.

Note that it is a bit of a misconception to say that the capacitor "stores charge". It doesn't. The net charge on the capacitor is zero. What it does is hold separated charges separate.(1 vote)

- why 1/Cequ = 1/c+1/c+1/c? why we do not use Cequ=c+c+c?(1 vote)
- Because when you put them in series, it is like the inner plates don't matter anymore and the outermost ones are further away from each other, so the Ceq is lower.(4 votes)

- So are capacitors what are used in backup generators, like when the power goes out? But obviously much larger so they can store more energy.(1 vote)
- Hello Bilbeisiomar,

On the power grid the capacitor is good for second to second smoothing (tremendous power for a short period of time). As Andrew stated, even the largest capacitors cannot store significant amounts of energy. Many would argue the same for batteries...

To appreciate the size of problem I made this silly video to introduce pumped hydro - the king of modern energy storage.

https://www.youtube.com/watch?v=hDmvNoGxXcs

Ref: https://en.wikipedia.org/wiki/Pumped-storage_hydroelectricity

Please leave a comment below if you would like to continue the conversation.

Regards,

APD(4 votes)

- I get mathematically why the charge on each of the capacitors is 18 but why wouldn't it conceptually be 18/3=6C?(2 votes)

## Video transcript

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.