Capacitors in parallel
A capacitator is a device that stores electrical energy in an electrical field. This video discusses the behavior of two capacitors connected in parallel. It compares two capacitators, and shows how to calculate the amount of charge each will receive. Finally, it discusses how to find the equivalent capacitance of the two capacitors combined. Created by David SantoPietro.
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- Ok, I'm slightly confused. Why the capacitors with the bigger capacitance have the lower Voltage? I understand the formula, I just would like a different interpretation to it.(25 votes)
- What happens if we hook up a resistor and a capacitor with the same battery in parallel and in series?(2 votes)
- In a series circuit a capacitor will initially allow current as if the resister is the only thing in the path but as the capacitor gains a charge it will limit the current until it is charged to the maximum voltage and then stop current flow. In a parallel circuit the capacitor will initially act like a short across the resister ans all of the current will flow through that branch until the capacitor is charged to the batteries voltage and the circuit will act like only the resister is in it.(5 votes)
- Help. How come for capacitors in series the total charge of the equivalent capacitor is the charge of each capacitor and for capacitors in parallel the total charge of the equivalent capacitor is equal to the sum of the charge stored by each capacitor ?(4 votes)
- Why is he showing the charges getting stripped off of the right sides of the Capacitors and not getting provided from the battery ?(3 votes)
- Charges are not provided by the battery. Energy is provided by the battery. The energy is used by the charged particles in the metal, as metals have delocalised electrons which gain this energy, the electrical potential energy to move from the positive terminal to the negative terminal, which you see on the left hand side. They do not flow back to the positive terminal due to the insulating material between the plates and hence, capacitors store energy and the material which the capacitors are made up of, the charged particles, use this energy to move back to the positive terminal when the circuit is complete. Hope I was able to answer your question and provide some depth to your understanding.(2 votes)
- What if one of the capacitors is different? As in, what if instead of their positive sides matching, one capacitor has it's respective positive side to the right and the other capacitor has its negative side pointing to the right?(1 vote)
- With one battery that is impossible because the electrons flow in one direction, from the negative to positive, which gives the capacitors its charges.(5 votes)
- In the last bit of the video, finding the individual charges of the capacitors in parallel, would it also be possible to simply multiply the total charge of the equivalent capacitor by the proportion of each capacitor in relation to the total capacitance? e.g. doing 54 * 3/9 to find the charge of the 3F capacitor and 54 * 6/9 to find the charge of the 6F capacitor. I know it works for this problem, but is this a generally acceptable way of doing things?(3 votes)
- In the second problem would it be wrong to solve for the capacitances of capacitors in series first then add capacitance of parallel? What is the reason to do so?(2 votes)
- What if I have two sources of voltage (V) and two capacitors in parallel? How can I solve the problem of that type?(1 vote)
- In ideal circuits like are dealt with in problems like this having 2 voltage sources of different voltages in parallel causes problem because you are applying differing potentials to the same wire and if the voltages are the same there is no difference than if you had a single voltage source of that voltage. If you have 2 voltages sources in series then you just add them together.(2 votes)
- I don't understand why we are dividing 1/Ceq. Don't we just use the numerical value determined for Ceq in 1/Ceq and stop there? Dividing Ceq into 1 would get you a value apart from Ceq...right? or wrong?(1 vote)
- If there were two capacitors in parallel and C1 was charged but C2 was not, but C2 had double the capacitance of C1, when you connect the two, how much charge would be transferred to C2?(1 vote)
- If they are in parallel, the potential difference V has to be the same for each capacitor. See if you can take it from there.(1 vote)
Look at the way that these six and three farad capacitors are connected to each other. What's going to happen if we hook them up to an eight volt battery? Well, like all capacitors, charge is going to get separated, so negatives are going to get stripped off of the right sides of these capacitors and pulled towards the positive terminal of the battery. But when they reach the other side, something interesting happens here. The charges reach this junction, or fork in the road. And now they have a choice in whether they're going to get deposited onto the three farad capacitor or the six farad capacitor. Each capacitor is going to get some of the charge, but since the six farad capacitor has twice the capacitance that the three farad capacitor does, the six farad capacitor's going to get twice as much charge stored on it as does the three farad capacitor. So twice as many negatives are going to get pulled off of the right side of the six farad capacitor, and twice as many negatives are going to get deposited on to the left side of the six farad capacitor. OK. So the six farad capacitor is going to get twice as much. But exactly how much charge are these capacitors going to get? Even though the circuit looks a little complicated, finding the charge in this case is actually really easy. The reason it's going to be easy is that both of these capacitors are hooked up directly to the terminals of the battery. In other words, the positive side of the six farad capacitor is hooked directly up to the positive terminal of the battery. And the negative side of the six farad capacitor is connected directly to the negative terminal of the battery. This means that the voltage across the six farad capacitor is going to be the same as the voltage of the battery, which is eight volts in this case. The same is also true for the three farad capacitor. So the voltage across the three farad capacitor is also eight volts. In fact, the way these capacitors are hooked up, it's as if they were connected to the eight volt battery all by themselves, because they both experience the entire voltage of the battery. Now that we know the voltage across these capacitors, we can use the definition of capacitance to solve for the charge. For the three farad capacitor, we can plug-in a capacitance of three farads and a voltage of eight volts. And we get that the charge stored on the three farad capacitor is 24 coulombs. We could do the same type of calculation for the six farad capacitor. We plug-in six farads an eight volts, and we get that the charge on the six farad capacitor is 48 coulombs. And see, just like we said, the charge on the six farad capacitor is twice as much as the charge on the three farad capacitor. We call capacitors hooked up in this way capacitors in parallel. You'll know that two capacitors are hooked up in parallel if their positive sides are directly connected to each other with a wire, and their negative sides are also directly connected to each other with a wire. We could ask ourselves now, what should the value be of a single capacitor whose effect on this circuit would be equivalent to that of the individual parallel capacitors? To find the equivalent capacitance of capacitors hooked up in parallel, all you need to do is add up the individual capacitances. And the reason is, just look at these capacitors. Since their positive sides are connected with a wire, you may as well have just merged all of the positive sides together to form one big positive side. And since their negative sides are all connected with a wire, you may as well have just merged the negative sides into one big negative side. So all you've really done by hooking up capacitors in parallel is to make one big capacitor out of smaller capacitors. Now, keep in mind that the capacitance of a capacitor is proportional to the area of the capacitor plates. So since we added the available areas of the capacitors together to get the total capacitance, all we need to do is to add up the individual capacitances. Even though the charge on the individual parallel capacitors might not be the same, they're charge has to add up to the total charge that would be stored on the equivalent capacitor. So if these parallel capacitors stored one coulomb, two coulombs, and three coulombs individually, their equivalent capacitor would store six coulombs. Let's try to apply these ideas to the circuit we just examined in the beginning of this video. The equivalent capacitance of these six farad and three farad capacitors would be a single nine farad capacitor. Now, let's solve for the amount of charge that this nine farad equivalent capacitor would store when hooked up to the eight volt battery. Using the definition of capacitance, we find that the charge on a nine farad capacitor would be 72 coulombs. And this makes sense, because remember the charge stored on the six farad capacitor was 48 coulombs, and the charge stored on the three farad capacitor was 24 coulombs. So the total charge on the six farad and three farad capacitors is 72 coulombs, which is the same charge that their equivalent capacitor stores. Let's try another problem that's a little more challenging. Say we introduce a 27 farad capacitor into this circuit. When the battery's connected, the capacitors will all store charge and have a certain voltage across them. So let's try to figure out the charge on and the voltage across all of these capacitors. Well, to start, we might notice that the three farad and six farad capacitors are still in parallel with each other, which means they have to have the same voltage as each other. But this time, the value of that voltage is not going to be the same as the voltage of the battery. Because even though their negative sides are connected directly to the negative terminal of the battery, their positive sides are not connected directly to the positive terminal of the battery. This 27 farad capacitor is getting in the way here. Similarly, the voltage across the 27 farad capacitor is also not going to be the same as the voltage of the battery. Because even though it's positive side is connected directly to the positive terminal of the battery, it's negative side is not connected directly to the negative terminal of the battery. So in summary, we don't know the voltage across any of these capacitors. And if we don't know the voltage across any of these individual capacitors, how are we ever going to solve for the charge on these capacitors? Well, the one thing that we do know is that the voltage across the whole circuit is eight volts. We just don't know the individual voltages across each capacitor. So what we're going to try to do is to replace these individual capacitors with a single equivalent capacitor. To do that, let's start with the six farad and three farad capacitors, because we know those are in parallel. We know they're in parallel because their positive sides are connected directly to each other and their negative sides are connected directly to each other. Using the rule to combine parallel capacitors, we get that the equivalent capacitance of the three and six farad capacitors is a single nine farad capacitor. So now we have a nine farad capacitor and a 27 farad capacitor. These are connected in series, because they're hooked up one right after the other. Or in other words, the positive side on one capacitor is connected to the negative side on the other capacitor. We can replace these two capacitors with a single equivalent capacitor by using the formula for adding capacitors in series, which is 1 over the equivalent capacitance equals 1 over C1 plus 1 over C2. So plugging in the values of nine farads and 27 farads, we get that 1 over the equivalent capacitance equals 0.148148. Don't forget to take 1 over this number to get that the equivalent capacitance is 6.75 farads. So we can replace the nine farad and 27 farad capacitors with a single 6.75 farad capacitor. Now, finally, we can solve for the charge on this 6.75 farad equivalent capacitor. Because it's positive side is connected directly to the positive terminal of the battery, and its negative side is connected directly to the negative terminal of the battery. That means the voltage across this 6.75 farad capacitor is going to be eight volts. We can use the definition of capacitance, and we get that the charge on this 6.75 farad capacitor is 54 coulombs. So since this was the equivalent capacitor for two series capacitors, both of these series capacitors must have the same charge as their equivalent capacitor. So both the 27 farad and nine farad capacitors have 54 coulombs each stored on them. At this point, we can figure out the voltage across these two capacitors. Using the definition of capacitance, we can plug-in 27 farads and 54 coulombs to get that the voltage across the 27 farad capacitor is two volts. Doing the same type of calculation for the nine farad capacitor, we get that the voltage across the nine farad capacitor is six volts. Notice that two volts and six volts adds up to the voltage of the battery, eight volts, just like they have to. And now we can find the charge stored on the individual three farad and six farad capacitors. We know now that the voltage across both the three farad and six farad capacitors is going to be six volts. Because the voltage across the individual capacitors in parallel has to be the same as the voltage across their equivalent capacitor. Now that we know the voltage, we can use the definition of capacitance. And for the three farad capacitor, we get that the charge stored is going to be 18 coulombs. And doing the same type of calculation for the six farad capacitor, we get that the charge is 36 coulombs. This makes sense, because 18 coulombs plus 36 coulombs adds up to 54 coulombs, which was the charge stored on their equivalent nine farad capacitor.