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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.

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  • leaf green style avatar for user Lemuel Taeza
    How can charges move between the two sides of a capacitor if they are separated?
    (12 votes)
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    • mr pants teal style avatar for user not.too.late
      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.
      (24 votes)
  • marcimus pink style avatar for user Rheann Erica Sequeira
    Why is the sum of the voltages of the capacitors equal to the voltage of the battery?
    (14 votes)
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  • leafers ultimate style avatar for user Shreyas Pai
    At , 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)
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    • aqualine ultimate style avatar for user razoq1999
      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)
  • male robot donald style avatar for user clayrjackson26
    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)
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  • blobby green style avatar for user lucybuckwell1
    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)
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    • blobby green style avatar for user Teacher Mackenzie (UK)
      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)
  • blobby green style avatar for user sonictif2
    Does this mean that the higher the capacitance, the lower the voltage of a capacitor hooked up in a series?
    (6 votes)
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    • spunky sam blue style avatar for user Michael Kines
      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)
  • blobby green style avatar for user Divya Talesra
    At , 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)
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    • male robot hal style avatar for user Andrew M
      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)
  • leaf green style avatar for user s.m.shahid hasan ridhoy
    why 1/Cequ = 1/c+1/c+1/c? why we do not use Cequ=c+c+c?
    (1 vote)
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  • piceratops ultimate style avatar for user bilbeisiomar
    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)
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  • mr pants teal style avatar for user SanFranGiants
    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)
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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.