### Video Transcript

A diagram shows a circuit containing a capacitor, a resistor, and a switch connected
in series. The capacitor is initially charged, and the switch is initially open. What is the magnitude of the current through the circuit immediately after the switch
is closed?

Okay, so in this diagram, we’ve got the switch, the resistor, and the capacitor. And as the question says, they’ve been connected in series. Now, we’ve been told that the capacitor is initially charged. And we can see that this is true because there’s a potential difference of 24 volts
across the plates of the capacitor.

Now, what this means is that when this switch is closed, the circuit will be complete
and current will be allowed to flow. In other words, the charges on the plates of the capacitor will start flowing in
order to discharge the capacitor. And so, as soon as the switch is closed, this capacitor will essentially act like a
cell providing 24 volts of potential difference across this resistor.

But the difference between this capacitor and a normal cell is that as the capacitor
discharges, the value of the potential difference will drop. So right away as the switch is closed, there will be 24 volts. But then a few seconds later, it’ll drop to 23 volts, 22 volts, 21 volts, et cetera,
et cetera until it reaches zero volts, at which point the capacitor is completely
discharged. However, we only need to worry about when the switch is closed and what happens right
away after that, because we’ve been asked to find the magnitude or size of the
current through the circuit immediately after the switch is closed.

Now, in order to work out the current in the entire circuit, we can simply work out
the current through this resistor here, because the circuit is a series circuit. So any current flowing through the resistor must be the same as the entire current
flowing through the circuit. And then at this point, we can recall Ohm’s law. This law will allow us to calculate the current 𝐼 through a resistor when we know
the potential difference 𝑉 across it and the resistance 𝑅 of the resistor.

So to work out the current, we need to rearrange the equation. We can do this by dividing both sides of the equation by the resistance 𝑅. This way, 𝑅 cancels out on the right-hand side. And then at that point, what we’re left with is 𝑉 divided by 𝑅 is equal to 𝐼.

So now, we can plug in some values. We can say that as soon as the switch is closed, the potential difference across the
resistor is 24 volts. And the resistance of the resistor is 30 ohms. So dividing 24 volts by 30 ohms will give us the current 𝐼 through the resistor and
therefore the current through the entire circuit.

Now before we evaluate the fraction on the left-hand side of the equation, we should
first consider units very quickly. So we see that we’ve got volts in the numerator and ohms in the denominator. Now these are the standard units of potential difference and resistance,
respectively. Therefore, whatever we calculate to be the right-hand side of this equation will be
in its own standard units as well. And the standard unit of current is amperes or amps.

So we’ll remember that and then evaluate the left-hand side of this equation. The numerical value ends up being 0.8. And as we’ve just said, the unit is amps. And this is not only the current through the resistor, but it’s the current through
the circuit as well.

Therefore, our final answer is that the magnitude of the current through the circuit
immediately after the switch is closed is 0.8 amps.