### Video Transcript

Consider the circuit shown. What is the value of the ohmic resistance 𝑅?

Alright, so we’ve got a circuit here, which consists of a resistor with the ohmic resistance 𝑅. And as well as this, we’ve got a capacitor that’s in series with the resistor. And we’ve got an AC source. Now, we’ve been told that the AC source produces 20 volts and that the current through it is two amps. As well as this, we’ve been told that there’re 12 volts of potential difference across the capacitor. Now, with this information, we need to find the value of the resistance 𝑅. Now, we can take a hint from the question.

We’ve been told that the resistance is an ohmic resistance. This means that it must follow Ohm’s law. And Ohm’s law tells us that the potential difference or voltage across an ohmic resistor is given by multiplying the current through that resistor by the resistance of that resistor. And since, in this situation, we’re trying to find the value of the resistance 𝑅, we can rearrange Ohm’s law by dividing both sides of the equation by the current 𝐼, which results in 𝐼 cancelling out on the right-hand side. And so we’re left with the potential difference divided by the current is equal to the resistance. So basically, what this means is that if we can work out the potential difference across the resistor and the current through the resistor, then we’ll be able to work out the resistance of the resistor.

Now, at this point, let’s also add a subscript 𝑅 to 𝑉, so that we’re talking about the potential difference across the resistor specifically. So why aren’t we doing this to the current value as well? Why aren’t we adding a subscript 𝑅? Well, it’s because we’ve been told in the diagram that the current through the AC source is two amps. And it’s important for us to realise that the resistor and capacitor are both in series with the AC source. This means that the current through the entire circuit is the same as the current through any one of the components because it’s a series circuit. In other words, there’re two amps of current through the resistor and two amps of current through the capacitor. And hence, we don’t need to add a subscript to the current value because the current value is the same for the resistor, the capacitor, and the AC source.

However, because we have a series circuit, we can see that the potential difference is going to be shared across different components in the circuit. And hence, we do need to differentiate between the potential difference across the resistor and the potential difference across the capacitor, which we’ll call 𝑉 subscript 𝐶, which also happens to be given to us as 12 volts in the question. Now, it’s at this point that we need to think about the fact that we’ve got an AC source. This means that we can’t simply split up the 20 volts of AC into 12 volts across the capacitor and the remaining potential difference across the resistor. Because with an alternating current source, we need to consider the fact that the current through a component in the circuit and the potential difference across it might not be in phase.

To help us understand what this means, we can use phasor diagrams. Now, for a resistor, as we said earlier, it follows Ohm’s law. 𝑉 is equal to 𝐼𝑅. And since the resistance of a resistor is just a constant value, this means that the voltage is directly proportional to the current through it. Therefore, if we had to plot a graph where the horizontal axis was time and we had a sinusoidally oscillating current going through the resistor, then the voltage across that resistor would have the same phase. It would also be sinusoidally oscillating. But it most likely will have a different amplitude, where this value, the amplitude of the potential difference, is equal to the amplitude of the current multiplied by the resistance of the resistor.

But anyway, so the main point here is that the voltage across a resistor and the current through it are in phase with each other. Therefore, if we had to represent the current and the voltage with a phasor diagram, starting with the current, which we can arbitrarily say is pointing in this direction, then because the voltage is in phase with the current, this means that the phasor representing the voltage will be pointing in the same direction. So here, we can label the current through the resistor, which we said was 𝐼, and the voltage through the resistor, which is 𝑉 subscript 𝑅.

Now, if we instead think about the capacitor, we have a slightly different story. We can recall that the current through a capacitor, which we’ll call 𝐼, is equal to the capacitance of that capacitor multiplied by the rate of change of potential difference. In other words, the current varies as the rate of change of potential difference, where 𝐶, the capacitance, is the constant of proportionality. So what does this mean in our situation here? Well, if we draw an arbitrary sign curve representing the voltage across the capacitor, then we can work out what the current should look like based on this equation.

We can see that the current is directly proportional to the rate of change of potential difference as we’ve already said, which can be found by finding the tangent to this curve at any point. So if we once again label the 𝑡-axis, representing time, that is, we can look at 𝑡 is equal to zero and see that the tangent to the curve has its maximum possible gradient, because the curve doesn’t get any steeper than this. And so this gradient, that’s the maximum gradient, multiplied by the capacitance is going to give us the current value at that point. Then, we can do the same thing at this point where the voltage is maximum. But then, where the voltage is maximum, the tangent to the curve is going to be a flat line. And a flat line has a gradient of zero. Therefore, the rate of change of potential difference, that’s d𝑉 over d𝑡, is zero. And so when we multiply that by the capacitance, it’s still zero. Therefore, the current at that point is zero.

Repeating the process once more for the point at which the voltage curve crosses the axis, we can see that the tangent to the curve is at its maximal gradient, but in the negative direction. In other words, this is the steepest that the curve gets. But it’s in the negative direction. Hence, the current is going to be at its minimum value or maximum value in the negative direction, if you like. Similarly, looking at this point here, that’s the minimum value of the voltage or maximum value in the negative direction, the tangent is going to be flat once again. So the current is going to be zero. And then, finally, at this point over here, the tangent to the curve is at its maximum value because this point is identical to this point. And so we reach the maximum value of the current once again.

So now, we’ve plotted enough points to be able to reasonably trace out the current curve, which looks something like this. And then, at this point, we can see that, for example, here the current is maximum. And then, a quarter of a cycle later, the voltage is maximum. Now we know that that distance is a quarter of a cycle because the entire cycle is this distance here. And the time that it takes for the current to go from maximum to zero is a quarter of a cycle. Similarly, the current here is zero and decreasing. And then, a quarter of a cycle later, the voltage is zero and decreasing. So what this is telling us is that the voltage across the capacitor is lagging behind the current through the capacitor by a quarter of a cycle.

So if we say that the current through the capacitor is pointing in this direction as our phasor, remember this current 𝐼 needs to be the same as this current 𝐼, because the current through the resistor and the capacitor have to be the same, because they’re in series. And then, we can show the potential difference across the capacitor as a phasor that’s pointing in this direction, because the voltage across the capacitor is lagging behind the current through the capacitor by 90 degrees or by a quarter of a cycle. And so the angle between this phasor over here, representing the current, and this phasor over here, representing the potential difference across the capacitor, is 90 degrees or quarter of a cycle.

And so simply, knowing the fact that the current through the resistor and the current through the capacitor are identical, we now have a phase relationship between the potential difference across the resistor and the potential difference across the capacitor. We know that the phase relationship is 90 degrees as well, because we’ve used our current phases as references. And the potential difference across the resistor points in the same direction whereas the potential difference across the capacitor is lagging by 90 degrees. Now, this is useful because we can work out the total voltage or potential difference across the resistor and capacitor combined.

The way to do this is to add the voltage across the resistor and the voltage across the capacitor vectorially, which looks something like this. Let’s call this vector 𝑉 subscript total, for the total potential difference across the resistor and the capacitor combined. In other words then, we’ve worked out the total potential difference across this part of the circuit or all of the components in the circuit that are not the voltage source. Now, what this essentially means is that the voltage source is supplying 20 volts. And all of those 20 volts must be dropped across the resistor and capacitor in some way. So we know that the magnitude or size of the total potential difference must be 20 volts. And we’ve been told in the diagram that the potential difference across the capacitor 𝑉 𝐶 is equal to 12 volts. So we can use this information to work out the potential difference across the resistor.

We can simply rearrange one of the vectors, so that we’ve got a right angle triangle and then apply Pythagoras’s theorem. We can say that the size or magnitude of 𝑉 subscript 𝑅, that’s the potential difference across the resistor squared, plus 𝑉 subscript 𝐶 squared, that’s the potential difference across the capacitor squared, is equal to 𝑉 total squared. Because 𝑉 total is the hypotenuse of the right angle triangle. And so because we know the values of 𝑉 𝐶 and 𝑉 total, we can substitute those in. We then have 𝑉 𝑅 squared plus 12 volts squared is equal to 20 volts squared. Then, we can subtract 12 volts squared from both sides of the equation, which means that 12 volts squared cancels on the left-hand side. Then, we can actually square out these two quantities, which ends up being 400 volts squared minus 144 volts squared. And 400 minus 144 gives us 256 volts squared.

Now we can take the square root of both sides of the equation. And so on the left, we’re simply left with 𝑉 subscript 𝑅. And on the right-hand side, we’re left with the square root of 256 multiplied by the square root of 𝑉 squared, which just happens to be volt. Now, the square root of 256 is equal to 16. So we’ve just found out that the potential difference across the resistor is 16 volts. We’ve just found 𝑉 𝑅. And because from the question, we already knew what the current through the resistor is, we know that the current is two amps, we also know at this point the value of 𝐼. So let’s sub in those values. So now, what we have is that 16 volts divided by two amps is equal to the resistance of the resistor 𝑅. And at this point, we can remember that one volt per amp is the same thing as one ohm. Therefore, the unit of resistance that our answer is going to be in is ohms. And 16 divided by two is eight. Therefore, at this point, we found our final answer.

The value of the ohmic resistance 𝑅 in the circuit is eight ohms.