5.8 Series Connection of Capacitors from Office of Academic Technologies on Vimeo.

**5.08 Series Connection of Capacitors**

All right. Now let’s study the series connection of capacitors. In this case, again, let’s consider three capacitors with capacitances of C1, C2, and C3. And in order to connect them in series, we connect them one after each other. For the capacitors to be set in series, the sum of the potential differences across each capacitor should be equal to the potential difference applied to the whole combination. Therefore, we say capacitors are said to be in series if the sum of the potential differences across each capacitor is equal to the potential difference applied to the combination.

So, as I mentioned earlier, in this case we connect the capacitors, C1, C2, and C3, one after each other, like this. Similar to the couplings of train cars along the same rail or track. And then we apply a potential difference across the combination by connecting those two ends to the terminals of a power supply, let’s say a battery, which is generating V volts of potential difference, and introduce a switch over here. Here, we have the capacitor with a capacitance of C1, Capacitor 2 with a capacitance of C2, and C3 for the third capacitor.

As soon as we close the switch here, again as in the previous case, since these charges are continuously repelling each other at the terminals of the power supply of the battery, let’s say, and the positive charges will move through along this available path in order to get away from each other as much as possible. And they will get collected to the left plate of Capacitor C1 as q1, plus q1. Similarly, the negative ones will continue along this path and get collected to the terminals of right-hand plate of the capacitor C3 as minus, let’s say, q.

But since they’re connected, these plates are connected to the terminals of the power supply, therefore these charges, the magnitude of charge q1 and if you call this one as q3, they will all be equal to one another and they will all be equal to charge q, let’s say. Therefore, let’s denote this one over here as plus q and the other one as minus q. Again, they’re directly connected to the terminals of this power supply.

So as we remember from the structure of a capacitor, we said that it is a device, which consists of two conducting plates separated by an insulating medium. So these mediums, between the plates of each one of these capacitors, are insulating mediums. In other words, they do not net a medium for the charges to easily move around. They are insulators. So when we look at this circuit overall, it’s actually an open circuit. In other words, we do not have a complete closed path for the charges to move.

Then we can easily ask the question, okay, we can understand why this plate of capacitor C1 is getting charged positively, and why this plate of capacitor C3 is getting charge negatively, because they are directly connected to the terminals of the power supply But then how this plate, the other plate of C3 and then other plate of C1, as well as capacitor C2, will get charged during this process because they do not have direct conducting linkage to the terminals of the power supply.

Well, when we look at–let’s consider this unit over here. As we can see, that unit here is the plate of capacitor C2 and this plate of capacitor C3 and why this whole region over here is a conducting medium. It is separated by these insulating points. Now this conducting medium, a piece of wire and let’s say the metal plates of these capacitors, have an abundance of free electrons. So once this other plate here is charged to minus q value, then these negative charges will repel these free electrons, in this medium, away from themselves. So these free electrons are going to move as far as possible for them, and that is the other boundary of this region, and they will come and be collected therefore on the right-hand side plate of the capacitor C2.

Therefore, since we’re going to have this excess amount of negative charge, free electrons, repelled by this minus q, we will end up with minus q of charge to be collected at this plate, at the right-hand side plate of this capacitor C2. Since these charges are going to be moving from this end toward this region, then the other end here will lack that much of negative charge. Therefore, this plate will get charged with positive q.

And of course, similar type of charging will take place for the other unit over here. This negative charge will repel the same amount of free electrons as far away as possible from this region. So this plate will get charged minus q, and since, therefore, they will leave the other region lacking that much of negative charge, this plate will therefore get charged positive q. Therefore, the other plates and the capacitors which are not connected directly to the power supply will get charged as a result of induction.

And so as a first property of this connection or combination, we can say that the charge stored on each capacitor in series combination will be equal to one another. In other words, q1 will be equaled q2, which will be equal to q3, and they will all be equal to the amount of charge drawn from the power supply, which is q. Again, this is directly associated with the conservation of charge principle.

And if you look at the second property, and this directly coming from the general feature of the series combination, as we stated above here, the potential difference across the whole combination will be equal to sum of the potential differences across each capacitor. In other words, if you just take our voltmeter and measure the potential difference across the whole combination by connecting our voltmeter to these two points, across the combination, we are going to read V volts, whatever the voltage supplied by the power supply. So that will read us V volts.

And then if we measure the potential difference across the first capacitor will read V1 volts. Across C2, we will read V2 volts, and across C3 we’re going to read V3 volts. And we will see that the potential difference across the whole combination, which is V volts, will be equal to V1 plus V2 plus V3. And that is the general property of series connections. The potential difference across the whole combination is equal to sum of the potential differences across each component within the series connection.

Now, as we did in the case of parallel connection, we are going to simplify this circuit by replacing all these 3 capacitors in series connections with a single capacitor. And let’s call it as C equivalent, so that that single capacitor will do the same job in the circuit that these three were doing in series combination. Again, let’s introduce our switch over here. The same battery is providing the same potential difference of V volts as the previous case, and once we turn the switch on position, once we close it, again these positive charges will move along this path and get collected along the left-hand side plate of the equivalent capacitor. And the positive ones will move through the other path and get collected on the right-hand side plate of the C equivalent. And, of course, the charging will continue until we reach high charge density so that they will generate strong enough repulsive force to the incoming charges. And, at that time, the capacitor will be fully charged.

If we write down the capacitor C equivalent, capacitance from this C equivalent capacitor, from it’s definition, it is going to be equal to the total charge stored in the place of the capacitor, which is q divided by the potential difference between the plates of this capacitor. And that is going to be equal to whatever the potential difference generated by this battery. And that is V.

From here, if you solve for the potential difference, we can write down this expression as q over C equivalent, the amount of charge stored in the capacitor, divided by the capacitance of the capacitor. Of course, we can write down similar type of expressions for capacitors C1, C2, and C3. The potential difference across from C1, which was V1, will be equal to then q1 over C1. But since in series combination the amount of charge stored in each capacitor is the same, q1 is equal to q. Therefore we will have q over C1 for V1. And, similarly, V2 will be equal to q2 over C2, and that too will be equal to q over C2, since again q2 is equal to q. Moving on, V3 will be equal to q3 over C3. And again, since q3 is equal to q from property 1, we will have q over C3 for this capacitor.

By using property 2, from property 2, since V is equal to V1 plus V2 plus V3, and in terms of the charge and capacitance, we can write down V as q over C equivalent. That will be equal to for V1. We will have q over V1, plus for V2 we will have q over C2, plus for V3 we will have q over C3. Since charge is common along each one of these terms dividing both sides of the equation by q, we can eliminate qs and end up with a final expression such that 1 over C equivalent is equal to 1 over C1 plus 1 over C2 plus 1 over C3.

We can now easily see the trend. If we connect the capacitors in series combination, then we see that the inverse of the equivalent capacitors become sum of the inverses of the capacitors or capacitances in series combination. We can generalize this for N number of capacitors in series, one over C equivalent, the equivalent capacitance of the whole combination, becomes equal to sum of the inverses of each capacitance in the combination or series combination. Here, also, we can easily see that once we connect the capacitors in series connection, then the resulting capacitance becomes smaller than the smallest capacitance in the combination. In other words, the total capacitance of the circuit decreases once we connect the capacitors in series form.

One thing that you should be always very careful by using this equation in order to calculate the equivalent capacitance of a circuit, this expression gives you the inverse of the equivalent capacitance. So to be able to get the C equivalent, or the equivalent capacitance, once we calculate the right-hand side of this equation, you have to take the inverse of that to be able to get the equivalent capacitance of those capacitors.