Tutorial 5 B  Inductor Reactance 

Learning Objective 
To understand the reactance of an inductor; To recognise and use the equation X_{L} = 2pfL To measure the reactance with different frequencies
Understand the phase relationship between voltages; 

Key Questions 
What is the reactance of an inductor? How do we demonstrate the reactance of an inductor?
What is the phase relationship with an
inductive circuit? 

Reactance of an InductorIn the last tutorial, we saw that if an inductor was put in series with a bulb:
It was as if the inductor had a kind of “resistance”.
Electrically, the inductor is simply a wire coiled up. A perfect inductor has zero resistance. In reality there is resistance, because copper wire has resistance, but it is very low. In this section we will assume that the inductor is perfect. The equation for reactance is:
If we plot reactance against frequency, the graph is a straight line of positive gradient going through the origin. This means that the reactance is directly proportional to the frequency for a perfect inductor.
The graph is like this:
You can determine the value of the inductance by taking the gradient, and dividing by 2p. Note that the line, taken from an actual experiment, is not quite a perfect straight line, but a line of best fit has been added.
The derivation of the reactance is given in the extension page.

Discuss whether an inductor would allow a constantly changing unidirectional current to pass. 

Question 7 
A 10 mH inductor has a reactance of 320 W. a) Show that the frequency of the AC supply is about 5100 Hz; b) Calculate the reactance at 1000 Hz. 
Phasor diagrams and inductor reactanceWe know that the current in an inductor is dependent on the frequency of the alternations. We use this equation for the reactance of an inductor:
In the derivation for this equation, we started off with the relationship for instantaneous voltage:
In the derivation we used this expression in the argument (Click HERE to see the derivation):
There is something that is much more significant about the result above. The voltage is related to time through the sine function, while the current is related through the negative cosine function. So what? Let’s keep w the same and see how the voltage and current vary across the same time period. If we plot the graphs we will see the significance:
The current graph is lagging the voltage graph by 90^{o} or p/2 rad. So the voltage phase vector is leading the current phasor by p/2 rad.
We can show this in a phasor diagram as this:

Simple LR circuitsA purely inductive circuit is simply an electrical curiosity. With resistive elements in the circuit, it becomes more interesting. In reality there are resistive elements, such as the internal resistance of the source, and the resistance of the wires.
Let’s use the same circuit as above, but this time we add a resistor, R.
This is a simple series LR circuit.
We will measure the voltage across the resistor as well as the inductor.
We need to draw the current phasor first. By convention we always draw the quantity which is the same in a circuit first, i.e. at the zero position.
So our current vector goes from left to right at the 3 o’clock position. Parallel to that is the voltage across the resistor.
Draw the phasor diagram here:
The voltage across the inductor is at 90^{o} and is leading the current, so its phase vector points vertically upwards. The resultant voltage is shown by the phasor V_{res}. We can work out V_{res} by simply using Pythagoras.
The sum of the two voltage is a vector sum, not an arithmetical sum. This is because the vectors are 90^{o} apart.
Now add in the phase angle:
We can easily work out the phase by which the current leads the resultant voltage:
