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.