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Question Video: Understanding the Design of Timing Circuits

The diagram shows a potential divider containing a resistor and a capacitor. Initially, no charge stored in the capacitor. What is the potential difference across the capacitor immediately after the switch is closed? What is the potential difference across the resistor immediately after the switch is closed? What is the output voltage immediately after the switch is closed? The capacitor takes 30 s to fully charge. What is the output voltage after 30 s?

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Video Transcript

The diagram shows a potential divider containing a resistor and a capacitor. Initially, no charge is stored in the capacitor. What is the potential difference across the capacitor immediately after the switch is closed?

Okay, so, first of all, let’s take a quick look at the diagram. We can see that, in this diagram, we’ve got a resistor with a resistance of 15 kiloohms. And we’ve got a capacitor which has been placed in series with the resistor. As well as this, we’ve got a switch which is on a branch of the circuit that is at 10 volts. And this end of the circuit is at zero volts.

Now, the reason that this circuit is called a potential divider is because of this branch over here because it allows us to measure the potential difference across this part of the circuit. And that potential difference is going to be some fraction of the potential difference across the entire circuit. And so, in essence, it’s dividing up this potential difference into two chunks. That’s this potential difference here and this potential difference here. Hence, it’s called a potential divider.

Now, we’ve been told that initially no charge is stored in the capacitor. And we’ve been asked to find the potential difference across the capacitor immediately after the switch is closed. So, initially, because this switch is open, there is no current running through the circuit. However, as soon as we close the switch, there is now going to be a current flowing through the circuit. So, as soon as we close the switch, the capacitor starts to charge up.

However, immediately after we closed the switch, the capacitor still has zero charge on its plates. And this is because it was uncharged before we closed the switch. Therefore, as soon as we close the switch, the capacitor is still uncharged for a very very small period of time. In other words, then, as soon as we close the switch, there is still no charge on either plate of the capacitor. And hence, there is no electric field between the plates of a capacitor. So, what we mean by this? Well, let’s zoom into our capacitor a little bit.

So, here is our capacitor in the circuit. Now, normally, in a circuit, the charge carriers are electrons. In other words, when there is a current flowing through the circuit, what that actually means is that it’s a flow of electrons around the circuit. And so, as soon as we close the switch, there’s a flow of electrons going around the circuit. So, if we say that electrons are moving this way around the circuit, and we can see on our close-up diagram of the capacitor that lots of electrons will be leaving this plate of the capacitor and lots of electrons will be deposited on this plate of the capacitor.

As a result of this, the first plate, which is losing electrons, ends up with a net positive charge. And the second plate, which is gaining electrons, ends up with a net negative charge. So, this is what happens when a capacitor charges up. And as a result of this, we’ve got lots of positive charges on one plate and lots of negative charges on the other plate. What this means is that there’s going to be an electric field between the two plates.

And this electric field which we’ve called 𝐸 happens to be directly proportional to the potential difference across this capacitor. In fact, it’s also related to the distance between the plates, which we’ll call 𝑑. But that’s not really relevant to us right now. All we need to know is that the electric field is directly proportional to the potential difference across the capacitor. And we’ve been asked to find the potential difference across the capacitor immediately after the switch is closed. But then, as we said earlier, immediately after the switch is closed, the capacitor is still uncharged. Therefore, there is zero electric field between the plates. And so, the potential difference across the capacitor is zero as well.

But this is true only immediately after the switch is closed. Because as time goes on after the switch is closed, the charge starts to build up on the capacitor plates. And hence, the potential difference across the capacitor plates increases as well. But anyway, the instant after we closed the switch, the potential difference across the capacitor is zero volts. Moving on, then, let’s now take a quick look at the resistor in the circuit.

What is the potential difference across the resistor immediately after the switch is closed?

Okay, so, to answer this question we need to realise that the voltage at this part of the circuit is 10 volts and the voltage at this part of the circuit is zero volts. This means that the potential difference between this part of the circuit and this part of the circuit is equal to 10 volts minus zero volts. And that is equal to 10 volts.

As well as this, we also know that the resistor and capacitor are in series. Therefore, this 10 volts of potential difference is going to be split between the resistor and the capacitor. However, we’ve just seen in the previous part of the question that the potential difference across the capacitor is zero volts immediately after we close the switch. Therefore, the remaining 10 volts must be across the resistor. And that is our answer here. The potential difference across the resistor immediately after the switch is closed is 10 volts.

Once we end up leaving the circuit for some time after the switch is closed though, we’ve realised that the charges on the capacitor will start to build up. Which means that the electric field between the plates will start to build up. And hence, the potential difference across the capacitor will no longer be zero volts but instead will be increasing.

So, for example, if sometime later the potential difference across the capacitor is one volt, then at that same time the potential difference across the resistor will be nine volts. Because that’s what’s left of these 10 volts. But anyway, we don’t need to worry about that bit. Let’s now move on to considering the whole point of a potential divider, the output voltage here.

What is the output voltage immediately after the switch is closed?

Okay, so, as we mentioned at the beginning, because we’ve got a potential divider, the output voltage is related to the potential difference across this part of the circuit. Specifically, the output voltage minus whatever this voltage is is equal to the potential difference across that part of the circuit. In this case, that’s the potential difference across this component here, which happens to be the capacitor.

But then, as we saw earlier, as soon as the switch is closed, the potential difference across the capacitor happens to be zero volts. And so, what we’ve got is that the output voltage, that’s the output voltage here, minus zero volts, that’s this voltage here, is equal to the potential difference across the capacitor. And this means that the output voltage must also be zero volts. So, that’s our answer to this part of the question. Now, up until this point we’ve only been considering what happens as soon as we close the switch. Let’s now think about what happens when we’ve left the circuit alone for some time after closing the switch.

The capacitor takes 30 seconds to fully charge. What is the output voltage after 30 seconds?

Alright, so, to answer this question, we need to think about what happens when a capacitor gets fully charged. Coming back to our zoomed-in picture of the capacitor from earlier, we can recall that we said that one of the plates becomes positively charged. And the other becomes negatively charged. And this occurs as electrons flow away from one plate and electrons are deposited onto the other plate.

Now, as current continues to flow throughout the circuit over a period of time, the top plate, in this case, will become more and more positively charged. And the bottom plate will become more and more negatively charged. This makes it harder and harder for electrons to flow through the circuit. Because now we’re trying to force even more electrons onto a plate which is highly negatively charged. And we’re trying to force even more electrons away from the positively charged plate.

Now, this is a problem because negative charges will repel negative charges. And so, all of these negative charges already on the lower plate will work towards repelling the new negatively charged electrons that are trying to be deposited on this plate. Similarly, negative charges are attracted to positive charges. So, all the electrons leaving the top plate will find it harder and harder to leave as this plate becomes more and more positive. The only thing that’s forcing these electrons to keep moving around the circuit in the same direction is this 10-volt potential difference.

But then, as it becomes harder and harder for electrons to keep moving, eventually current stops flowing in the circuit. And at that point, we say that the capacitor is fully charged. In other words, there is no more flow of electrons onto this plate or away from this plate. And that is our key to answering this question, the fact that the current at this point is zero. Because if there’s no more flow of electrons in the part of the circuit that has the capacitor in it, then because the resistor is in series with the capacitor, there is zero current to the resistor as well. And so, we can say that the current through the resistor, which we’ll call 𝐼 res, is equal to zero amps.

Now, at this point, we can recall that the resistor follows something known as Ohm’s law. Ohm’s law tells us that the potential difference across a component in a circuit is equal to the current through that component multiplied by the resistance of that component. Now, we know that the resistance of the resistor is 15 kiloohms. But that doesn’t matter if we’re trying to work out the potential difference across the resistor. Because we can say that the potential difference across the resistor, which we’ll call 𝑣 res, is equal to the current through the resistor, which is zero amps. Multiplied by the resistance of the resistor, which is 15 kiloohms, multiplied by 15 kiloohms, which is the resistance of the resistor.

But then, because the current for the resistor is zero, this means that the potential difference across the resistor is also zero. In other words, then, once the capacitor is fully charged, the current through the circuit is zero, and so the potential difference across the resistor is zero volts. But then, we said earlier that this 10 volts of potential difference gets split between the resistor and the capacitor. And so, the potential difference across the capacitor must be the remaining 10 volts.

And we can recall from earlier that the output voltage minus whatever this voltage is here is equal to the potential difference across this component of the circuit, which happens to be the capacitor. And so, we can say that the output voltage minus zero volts is equal to 10 volts. Which means that we can finally say that, after 30 seconds, when the capacitor is fully charged, the output voltage is equal to 10 volts.

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