Video: Understanding Capacitors Connected in Parallel

What are the voltages across the capacitors after equilibrium has been reached?

08:15

Video Transcript

What are the voltages across the capacitors after equilibrium has been reached? a) C one, five volts; C two, five volts; C three, 10 volts. b) C one, 10 volts; C two, 10 volts; C three, 10 volts. c) C one, 10 volts; C two, five volts; C three, five volts. d) C one, five volts; C two, 10 volts; C three, five volts. e) C one, 10 volts; C two, 20 volts; C three, 20 volts.

The question refers to the circuit on the right, which consists of a single-celled battery connected to three capacitors. And the question asks us for the voltage across each capacitor after equilibrium has been reached. Equilibrium means that no voltages or currents are changing. Let’s recall what this means in the context of a capacitor. A capacitor stores energy by separating charges. So, let’s start with an initially uncharged capacitor. We’ve now added a battery of voltage 𝑉 across the capacitor. Current now flows around the circuit. Conventional current flows out of the positive terminal and into the negative terminal. Recall, though, that conventional current is the flow of positive charge. So, positive charge flowing into the top plate of the capacitor result in a net positive charge on the top plate.

Similarly, conventional current flowing away from the bottom plate leaves a void of positive charge, which is the same thing as saying it leaves behind a negative charge. The capacitor now has a charge of +𝑄 on the top plate and ‒𝑄 on the bottom plate. And these charges keep growing as the conventional current keeps flowing. This separation of charges results in a voltage across the capacitor. As the charge on each plate of the capacitor grows, so does the voltage across it. And as this happens, the conventional current flowing in the circuit begins to slow.

This is because the greater the voltage across the capacitor, the harder it is to continue increasing the charge. And eventually, the electromotive force of the battery is insufficient to push more charge onto the capacitor. This occurs when the voltage across the capacitor is equal to the voltage of the battery. And when this occurs, the conventional current stops flowing through the capacitor. This is what it means for the capacitor to be in equilibrium. No more conventional current is flowing through it, so no more charges building up, and the voltage across it is constant.

If there are more than one capacitor, as there are in this circuit diagram, the voltage across each capacitor is not necessarily the voltage of the battery. Nevertheless, the voltage at equilibrium is constant. We’ve labelled these voltages on the diagram with the symbols 𝑉 one, 𝑉 two, and 𝑉 three for voltage across capacitor one, voltage across capacitor two, and voltage across capacitor three. To approach finding these values, we’ll use a property of voltage between two points. The voltage between any two points is the same regardless of the path taken between those two points. Let’s consider the path between point A and point B that travels directly across capacitor one. To find the total voltage between the end points of a path, we simply add up the voltages encountered along that path.

In this case, the only voltage we encounter is 𝑉 one, the voltage across capacitor one. Calling the voltage between point A and point B 𝑉 AB, we can write that 𝑉 AB is equal to 𝑉 one. If we can independently find a value for 𝑉 AB, then we will have found the value for 𝑉 one. To find 𝑉 AB independent of 𝑉 one, we need a path between A and B with a known voltage. Note that we have a battery with a given voltage. So, if we can find a path from A to B that passes through this battery, we’ll know the voltage between A and B. To find that path, simply trace the circuit from A to the battery through the battery and onto point B. The only voltage present on this path is the voltage of the battery. So, the voltage of this path is 10 volts. But this is also a path between A and B, so it must have the same voltage as every other path between A and B or 𝑉 AB. So, we found that 𝑉 AB is equal to 10 volts.

Note that it’s positive 10 volts because we traverse the battery from plus to minus side. If we traverse the other direction, in other words, a path from B to A, we would’ve found that the voltage is minus 10 volts. Replacing 𝑉 AB with 10 volts, we find 𝑉 one, the voltage across capacitor one is also 10 volts. Since we know the voltage between points A and B, let’s find a path from A to B that includes capacitor two and capacitor three. This time, the path includes two voltages, the voltage across capacitor two and the voltage across capacitor three. Again, the total voltage is 𝑉 AB, since the path goes from A to B. And this is equal to the sum of the voltages encountered, 𝑉 two plus 𝑉 three.

Just like before, we replace 𝑉 AB with 10 volts. But the right-hand side contains two unknown variables. So, to solve for them, we’ll need another equation, some relation between 𝑉 two and 𝑉 three. We’ll first present an intuitive argument for this relationship and then back it up rigorously. C two and C three are the only two components on this right-hand branch of the circuit. And since they have the same capacitance, it ought to be that they behave identically. In other words, since there are no other components in the right-hand side of the circuit and C two and C three are effectively identical, there’s no reason for the circuit to prefer one over the other when it comes to voltage. So, because they’re the same component and ought to behave the same, they should have the same voltage across them. In other words, 𝑉 two equals 𝑉 three.

Substituting in for 𝑉 three, we find that 10 volts is equal to two times 𝑉 two. Dividing both sides by two, two divided by two is one and 10 divided by two is five. So, 𝑉 two equals five volts. And since 𝑉 two equals 𝑉 three, 𝑉 three is also equal to five volts. We now know the voltages across each of three capacitors. In order, they are 10 volts, five volts, and five volts. This is answer choice c, C one, 10 volts; C two, five volts; C three, five volts. To get this answer, we made an intuitive assumption based on the symmetry of C two and C three that 𝑉 two is equal to 𝑉 three. We can take a longer but more rigorous route to the same conclusion by understanding how the charges build up on these capacitors.

Recall that before equilibrium, charge builds up on the capacitor by conventional current flowing into one plate and out of the other. Recall also that if current is flowing in a circuit, then any areas without junctions have the same current everywhere in that area. Before equilibrium, capacitors two and three are charging up due to current flowing between points A and B along this path. This path has no junctions because the junctions are A and B at the beginning and end. Therefore, the current at all points along this path must be the same. So, the current flowing into capacitor two, out of capacitor two and into capacitor three, and out of capacitor three and back to point B must all have the same value.

But recall that current is the rate of transfer of charge. So, the current into capacitor two is depositing charge on that first plate. But that same value of current is also flowing out of capacitor two, leaving behind a charge of ‒𝑄 on the other plate. By the same logic, the currents flowing into and out of capacitor three are also the same as those into capacitor two, mean they result in the same charge. The last thing that we need is the relationship between the charge on a capacitor, the voltage across it, and its capacitance.

That relationship is voltage across the capacitor is equal to charge on the capacitor divided by the capacitance. Using this relationship, let’s write the voltage across capacitor two. 𝑉 two, the voltage across capacitor two, is equal to 𝑄, the charge on it, divided by 10 picofarads, the capacitance. As we saw before, capacitor three has the same charge. And it also is given to us that it has the same capacitance. So, 𝑉 three is also equal to 𝑄 over 10 picofarads. Since both 𝑉 two and 𝑉 three are equal to 𝑄 over 10 picofarads, it follows that our original assumption that 𝑉 two is equal to 𝑉 three is indeed correct and justified in this problem.

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