### Video Transcript

Illustrate using vectors the phase difference between the voltage and the current in a circuit consisting of an AC supply and an inductive coil of negligible ohmic resistance.

Okay, so the circuit that we’re considering in this question consists of an AC supply and an inductive coil. So here’s a circuit diagram of a circuit showing the AC source and the inductive coil which we’ll say has a self-inductance 𝐿. Now, we’ve also been told that the inductive coil has negligible ohmic resistance. This means that we can ignore any resistive affecting the circuit. The coil doesn’t have any. In other words, it behaves like a perfect inductor. It only has a self-inductance 𝐿 and has no resistance 𝑅 is equal to zero.

So we’ve been asked to illustrate using vectors the phase difference between the voltage and the current in the circuit. To do this, we can first recall the relationship between voltage and current in an inductor. We can recall that the voltage across the terminals of an inductor is given by multiplying the self-inductance of the inductor by the rate of change of current through the inductor. Because Δ𝐼, that’s the change in current through the inductor, divided by Δ𝑡, that’s the time taken for that change in current to occur, altogether represents the rate of change of current, how fast current is changing through the inductor.

So because in the circuit we have an AC source, let’s assume that the current through the inductor looks something like this where on the horizontal axis, we’ve got time. And on the vertical axis, we’ve got current. So we see how the current through the inductor varies with time. Now, the reason that we’ve drawn a sinusoid is because we’ve got an AC source. And the AC source produces a sinusoidal potential difference. But anyway, so now that we’ve thought about the current through the inductor, let’s use this equation to give us the voltage across the terminals of the inductor.

We’re told that the voltage is equal to the inductance, that’s just some constant, multiplied by the rate of change of current. So we can find the rate of change of current at any point in time by finding the gradient of this curve here because the curve shows us the current through the inductor at any time. And so the rate of change of current at that time is given by the gradient of this graph. And then once we found that gradient, we multiply it by some constant 𝐿, that’s the inductance of the coil, to give us the voltage across the inductor at that point in time.

So for example, at the beginning, 𝑡 is equal to zero, we can draw a tangent to the sinusoid which looks something like this. And then calculate its gradient. Now, the gradient of this tangent is going to be the same as the gradient of the curve at that point. And we can see that this is the point on the sinusoid where the gradient is largest because at no other point can we draw a steeper tangent. The tangent is as steeper as it can be. And so the gradient is maximal. At this point, let’s say that we’re also plotting the voltage against time on the same graph. So in orange, we’ve got the current. And in pink, we’ve got the voltage.

But anyway, so we’ve said that the voltage at 𝑡 is equal to zero is maximum. And then we can multiply it by some constant 𝐿 to give us some maximum voltage value. Let’s say that this is that value. It doesn’t matter exactly what that voltage value is as long as we know that is the maximum value. In other words, the voltage doesn’t get any higher than this. And once we’ve done that, we can now move on to another part of the graph. Let’s say now, we move forward a quarter of a cycle which, in other words, is this point here on the current graph because remember this whole period of time is one cycle and so this distance represents a quarter of a circle.

So now, we’ve moved quarter of a cycle ahead in time. And we can once again draw a tangent to represent the gradient of the curve. This time, we see that it’s a flat line. in other words, its gradient is zero. And so if we take this gradient , that’s Δ𝐼 divided by Δ𝑡, and multiplied by the self-inductance 𝐿, it’s still going to be zero because zero multiplied by anything is still zero. And hence, at that point in time at a quarter of a cycle, we see that the voltage across the inductor is going to be zero at which point we can see that we’ve drawn two little dots here and here. And this is going to form the basis of our voltage curve.

Let’s continue doing this. Let’s move forward another quarter of a cycle to find the gradient of the current curve at this point here. Once again, we draw a tangent to the curve which looks something like this. And we see that, once again, this tangent is as steep as it can be but this time is pointing in the downward direction. In other words, this is the steepest negative gradient that we’ll find on this curve. In other words then, the rate of change of current is the largest possible negative value that it can be. And remember, because we’ve got a sine curve, the steepness of the gradient of the curve here is going to be the same as the steepness here, except this one is going to be negative.

Hence, when we multiply once again by the self-inductance of the inductor, the voltage is going to be at its maximum value but negative. So at this point, we’ve got three dots now for the voltage curve. Let’s move yet another quarter of a cycle ahead to this point. We can see that the current value is minimum. And when we draw a tangent to the curve, we see it’s a flat line once again. In other words, at that point in time, the rate of change of current is zero. And so the voltage is going to be zero as well. So we plot a point here. And then finally, completing the current cycle, we can see that that point is identical to the 𝑡 is equal to zero point which means that the tangent that we draw is going to have the same gradient as the tangent that we drew here.

Hence, the rate of change of current is maximum. And then multiplying it by the self-inductance, we’re going to get the maximum voltage once again. So we draw a point here at which point we can start to connect up the voltage graph which will look something like this. And we can actually see that the voltage curve is a sinusoid once again. But it’s been shifted along the 𝑡-axis, the time axis, relative to the orange curve. So let’s try and find the phase relationship between the pink curve and the orange curve.

Let’s start at 𝑡 is equal to zero. We can see that, on the pink curve, at 𝑡 is equal to zero, that’s the voltage curve, the voltage value is maximum and then decreasing. And then exactly a quarter of a cycle later, we can see that the current curve is at its maximum value and decreasing. Repeating this process with, for example, this point over here, we can see that the voltage value is minimal and increasing. And then exactly a quarter of a cycle later, we see that the current value is minimal and increasing. And in fact, this is true for the entire curve.

Whatever point we pick on the voltage curve, for example, if we pick this point over here, and at this point we can see that the voltage value is negative and decreasing, and then if we move a quarter of a cycle ahead, we can see that the current curve is doing exactly the same thing. We’ve got a negative value. And it’s also decreasing. In other words then, the current curve is lagging behind the voltage curve by a quarter of a cycle. And we can represent this relationship using vectors or other phases because phases are vectors that can be used to show phase relationships. In this case, we want to show the phase relationship or the phase difference between the voltage and the current.

So let’s use a phase to represent the current first. let’s say it’s pointing in this direction. Now, this is an arbitrary choice. It doesn’t matter which direction we choose the initial vector to point in. The important thing though is the relationship between this vector and the vector that we’ll draw to represent the voltage. So we said that the current is lagging behind the voltage by a quarter of a cycle. Or in other words, the voltage is ahead of the current by a quarter of a cycle. Now, if the voltage and the current were in phase, we would draw them pointing in the same direction. And in fact, the direction that this second vector points in is going to represent the phase relationship because it’s not actually the direction that matters but the angle between the two vectors.

In this case, they’re pointing in the same direction. So the angle between them is zero degrees. Now, if we were to draw our second vector pointing in this direction instead, then we can see that there’s a 90-degree angle between the current vector and the blue vector that we’ve drawn. This can be used to show that the blue vector is 90 degrees or a quarter of a cycle ahead of the orange vector. If instead we drew the blue vector pointing in this direction, then we can see that there’s a 180-degree angle between the current vector and the blue vector. This shows that the current vector and the blue vector are exactly out of phase. Or there is a 180-degree phase relationship between them. And if we’ve drawn the blue vector pointing in this direction, then we’d see that there’s either a 270 degrees’ relationship or a 90-degree relationship between the two depending on how we wanna see it.

In other words, in this situation, the blue vector is ahead of the orange vector by 270 degrees. Or the orange vector is ahead of the blue vector by 90 degrees. It doesn’t really matter which; they’re both equivalent. We could also choose to say that this blue vector is ahead of the orange vector by three-quarters of a cycle. Or we could say that the orange vector is ahead of the blue vector by a quarter of a cycle. So basically, the angle between the two as a fraction of a full 360 degrees represents the phase relationship between the two vectors.

And so because we know that the voltage vector is a quarter of a cycle ahead of the current vector, we’re going to draw it like this. We’ll label this one 𝑉 for voltage and see that there’s a 90-degree angle between the two because 90 degrees is a quarter of 360 degrees. And we want a quarter cycle relationship between the two where the voltage is leading. And hence at this point, we’ve arrived at our final answer. We’ve illustrated using vectors the phase difference between the voltage and the current in a circuit consisting of an AC supply and an inductive coil of negligible ohmic resistance.