### Video Transcript

In this video, weโre going to learn how to find characteristics of capacitors including equivalent capacitance for capacitors that are only in parallel with one another, as well as only in series with one another. We will begin by looking at capacitors that are hooked up in parallel with one another. In the diagram, we have a battery that provides a potential difference of ๐ total or ๐ subscript ๐. The battery is hooked up in parallel to capacitor ๐ถ one and capacitor ๐ถ two.

Weโre going to see a lot of connections between capacitors in parallel and resistors in parallel. With that in mind, before we go over capacitors in parallel, letโs do a quick recap of resistors in parallel. We should remember that when we have resistors in parallel, we find equivalent resistance by doing one over the equivalent resistance is equal to one over ๐
one plus one over ๐
two and so on and so forth for as many resistors as we have in parallel. We also need to recall a couple of laws known as Kirchhoffโs laws that we can apply to our circuit.

In the first law, the current into a junction equals the current out of a junction. We can turn this rule into an equation by looking at the junction between resistor ๐
one and resistor ๐
two. With respect to the diagram, we can say that the total current from the battery ๐ผ T Is equal to the current that splits to go through the first resistor ๐ผ one plus the current that splits to go through the second resistor ๐ผ two and so on and so forth for as many branches as we have. Kirchhoffโs second law states that the sum of the voltages around a closed loop equals zero. Each branch in a parallel circuit is considered a closed loop. Therefore, we can say the potential difference of the battery ๐ T is equal to the potential difference across resistor one, ๐ one, is equal to the potential difference across resistor two, ๐ two, and so on.

Now letโs compare what we know about resistors in parallel to capacitors in parallel. For capacitors in parallel, we have the same relationship for potential difference across each branch as we did with the resistors. Once again, we apply Kirchhoffโs second law, and we have single capacitors in parallel in our circuit. This means that the potential difference of the battery ๐ total is equal to the potential difference across capacitor one, ๐ one, is equal to the potential difference across capacitor two, ๐ two, and so on and so forth for as many capacitors as we have in parallel.

In terms of current, we want to study capacitors that are fully charged after having been connected into the circuit for a while. But when capacitors are fully charged, thereโs no more charge flow in the circuit and so thereโs no current in the circuit. However, we should remember that current ๐ผ is equal to charge ๐ divided by time ๐ก. So instead of using current, we can use charge stored in the capacitor. For resistors, the total current is split among subbranches. For capacitors, itโs the total charge such that the total charge ๐ T is equal to the charge stored in capacitor one, ๐ one, plus the charge stored in capacitor two, ๐ two, dot dot dot. For this reason, the charge of the battery will be greater than any of the individual charges stored in the capacitors in parallel.

Finding equivalent capacitance will be different than finding equivalent resistance. For resistors in parallel, the more resistors we add in parallel, the smaller the equivalent resistance becomes. It turns out that the capacitances actually add together, so the total capacitance is equal to the sum of all of the capacitances. More specifically, the total capacitance of the circuit ๐ถ T is equal to the capacitance of capacitor one, ๐ถ one, plus the capacitance of capacitor two, ๐ถ two, and so on. Therefore, the more capacitors we add in parallel, the larger our total capacitance becomes. And we can say the total capacitance will always be greater than the value of the individual capacitors in parallel.

Letโs apply our equations for capacitors in parallel by giving values to our diagram such that our potential difference of the battery is 12 volts, the capacitance of capacitor one is 25 microfarads, and the capacitance of capacitor two is 11 microfarads. Letโs start by determining the total capacitance of our circuit. As discussed earlier, the equation to find the total capacitance of a parallel circuit is ๐ถ T equals ๐ถ one plus ๐ถ two for as many capacitors as we have. Based on the values that we chose for our diagram, we can say that the total capacitance is equal to 25 microfarads for capacitor one plus 11 microfarads for capacitor two. When we add 25 microfarads plus 11 microfarads, we get 36 microfarads.

Next, letโs figure out what the potential difference is across each of our two capacitors, ๐ถ one and ๐ถ two. For potential difference, we can use the equation ๐ T equals ๐ one equals ๐ two, so on and so forth for as many capacitors as we have in parallel. Plugging in the values from our diagram, we know that the potential difference of the battery is 12 volts. Therefore, we can say the potential difference across capacitor one, ๐ one, is 12 volts and that the potential difference across capacitor two, ๐ two, is also 12 volts.

Letโs also look at how much charge is stored not only in each of the capacitors, but also in the total charge of the battery. To determine these values, we first need to remember the equation that relates capacitance ๐ถ, charge ๐, and potential difference ๐. We need to rearrange the formula ๐ถ equals ๐ over ๐ to solve for charge ๐. To do this, we multiply both sides of the equation by ๐. This will cancel out the ๐ on the right side of the equation, leaving us with the equation ๐ times ๐ถ is equal to ๐.

We can now apply this equation to each of the capacitors individually. Letโs start with capacitor ๐ถ one. We found out earlier that the potential difference across ๐ถ one was 12 volts. And from the diagram, we know that the capacitance is 25 microfarads. When we multiply 12 volts by 25 microfarads, we get a charge ๐ one of 300 microcoulombs. Letโs do the same thing, but this time for ๐ถ two. We determined earlier that the potential difference across ๐ถ two was 12 volts, and we know from the diagram that the capacitance is 11 microfarads. When we multiply 12 volts by 11 microfarads, we get a charge stored on capacitor two, ๐ two, of 132 microcoulombs.

Now, we need to determine the total charge. To do this, we use the equation that total charge is equal to the charge stored on capacitor one plus the charge stored on capacitor two and so on. Plugging in the values that we just determined, we get that the total charge ๐ T is equal to 300 microcoulombs, the charge stored on capacitor one, plus 132 microcoulombs, the charge stored on capacitor two. When we add 300 microcoulombs plus 132 microcoulombs, we get a total charge of 432 microcoulombs. Now that we know how to find total capacitance, total charge, and total potential difference for capacitors in parallel, letโs move on and look at capacitors in series.

Before we go over capacitors in series, letโs first refresh our memory on resistors in series. Letโs recall that the equivalent resistance is equal to the sum of the individual resistors, ๐
one plus ๐
two and so on and so forth for as many resistors as we have in series. Once again, we apply Kirchhoffโs laws. With respect to Kirchhoffโs first law, the current into a junction equals the current out of a junction. In a series circuit, the current doesnโt split because there is no junction. This means that the total current is equal to the current through the first resistor, ๐ผ one, is equal to the current through the second resister, ๐ผ two, and on and on.

In the second law, sum of voltages around a closed loop equals zero. In a series circuit, all the resistors are on the same closed loop. This means that the potential difference of the battery, ๐ total, is equal to the potential difference across resister one, ๐ one, plus the potential difference across resister two, ๐ two, dot dot dot. For our capacitors in series, we have a connection with the resistors in series in terms of potential difference being split amongst the individual components. We can say the total potential difference of the battery ๐ T is equal to the potential difference across capacitor one, ๐ one, plus the potential difference across capacitor two, ๐ two, so on and so forth for as many capacitors as we have in series.

For resistors in series, weโre able to compare the currents. However, for capacitors in series just as in parallel, we studied the capacitors that are fully charged after having been connected into the circuit for a while. But when capacitors are fully charged, there is no more charge flow in the circuit, and so there is no current in the circuit. Once again, weโre gonna have to look at the charge, just as we did for capacitors in parallel. In terms of charge, we can say the total charge ๐ T is equal to the charge stored on capacitor one, ๐ one, is equal to the charge stored capacitor two, ๐ two, and on and on.

To find the total capacitance in series, we use the relationship that looks like the relationship that we used for resistors in parallel. One over the total capacitance is equal to one over ๐ถ one plus one over ๐ถ two and so on. For resistors, the more resistors we add in series, the greater the total resistance. However, for capacitors, the more capacitors we add in series, the smaller the total capacitance.

Looking back at the equation ๐ถ equals ๐ over ๐, when we add more capacitors in series, the potential difference of the battery stays the same. But we just saw that the total capacitance decreases. This means that the total charge stored in our circuit will also decrease. So as we add more capacitors, the total charge of the circuit will actually decrease and will be less than if we had just one capacitor hooked up to the battery by itself, whereas in parallel we get more charge stored for each capacitor that we hook up to the battery.

We have chosen values for our variables in our diagram such that the potential difference of the battery is 10 volts, the capacitance of capacitor one is three microfarads, and the capacitance of capacitor two is six microfarads. We can now apply the equations for capacitors in series to our diagram. Letโs begin by determining the total capacitance of our circuit using the equation one over total capacitance is equal to one over ๐ถ one plus one over ๐ถ two and so on and so forth for as many capacitors as we have in series.

Using the values we chose for our diagram, one over the total capacitance is equal to one over ๐ถ one, three microfarads, plus one over ๐ถ two, six microfarads. When adding fractions, we need to use the least common denominator. In this case, that would be six microfarads. This means that we need to multiply one over three microfarads by two over two, giving us two over six microfarads. When we add two over six microfarads plus one over six microfarads, we get three over six microfarads, which can be reduced to one over two microfarads.

To solve for total capacitance, we multiply both sides of the equation by two microfarads and total capacitance. This cancels out the two microfarads on the right side of the equation and the total capacitance on the left side of the equation, leaving us with two microfarads is equal to the total capacitance. Next, we can find the total charge along with the charge stored on each of the capacitors as we know that theyโre all equal. We know that all the charges are equal, and we know the total capacitance as well as the total potential difference. Therefore, we can use the equation capacitance is equal to charge divided by potential difference.

We need to rear in to the formula to solve for charge. To do this, we multiply both sides by ๐. This will cancel out the ๐ on the right side of the equation. Now that we have ๐ times ๐ถ is equal to ๐, we can plug in our known variables. Our total potential difference was 10 volts. We found that our total capacitance was two microfarads, and weโre solving for our total charge. When we multiply 10 volts by two microfarads, we get 20 microcoulombs. So we can say the total charge, the charge stored in capacitor one, and the charge stored in capacitor two are all equal to 20 microcoulombs.

We know that the total potential difference of the battery is 10 volts. To find the individual potential difference across each capacitor, we can use the equation capacitance is equal to charge over potential difference. To rearrange the formula to solve for the potential difference, we multiply both sides of the equation by ๐. This will cancel out the ๐ on the right side of the equation. Next, we need to divide ๐ถ from both sides of the equation. This will cancel out the ๐ถ on the left side of the equation. We now have that the potential difference across the capacitor is equal to the charge stored in the capacitor divided by the capacitance.

We can now apply this to each individual capacitor. Substituting in the values for capacitor one, we have a charge stored on the capacitor of 20 microfarads and a capacitance of three microfarads. When we divide 20 microcoulombs by three microfarads, we get 6.7 volts. Now, weโll apply the same equation to the second capacitor. The potential difference across capacitor two is equal to 20 microcoulombs divided by six microfarads.

When we divide 20 microcoulombs by six microfarads, we get 3.3 volts. Even though we know the potential difference of the battery is 10 volts, we can validate that using the potential difference formula. The total potential difference will be equal to the potential difference across capacitor one, 6.7 volts, plus the potential difference across capacitor two, 3.3 volts. When we add 6.7 volts plus 3.3 volts, we get 10 volts.

Weโre now ready to apply our understanding of capacitors to an example problem.

The circuit in the diagram contains two capacitors connected in parallel. What is the total capacitance of the circuit?

In the diagram, we can see that we have a 65-microfarad capacitor and a 35-microfarad capacitor hooked up in parallel with our battery. To determine the total capacitance of our circuit, we need to remember the equation for capacitors connected in parallel. Total capacitance in parallel is equal to capacitor one plus capacitor two and so on and so forth for as many capacitors as we have in parallel. In our circuit, we have a 35-microfarad capacitor and a 65-microfarad capacitor. When we add 35 microfarads and 65 microfarads, we get a total capacitance of 100 microfarads. Therefore, we can say that the total capacitance of our circuit is 100 microfarads.

Key Points

To find the total capacitance for capacitors in parallel, use ๐ถ T equals ๐ถ one plus ๐ถ two plus dot dot dot. To find the total capacitance of capacitors in series, use one over ๐ถ T equals one over ๐ถ one plus one over ๐ถ two and so on and so forth. Capacitors in series store equal charge. The voltage across capacitors in series is an inverse ratio to the ratio of their capacitances. Capacitors in parallel have equal voltages across their branches.