Tutorial 12 - Inductors and AC (A-level Extension)
The content of this tutorial is not on the AQA syllabus (or, for that matter, the OCR or EDEXCEL). It is on the SQA Advanced Higher syllabus. It is part of the option A (Alternating Currents) on the Welsh Board (WJEC). Other Option A content can be found in Tutorials 9 and 10.
I have included it here because students studying the AQA Electronics Option need to know something about the concepts of AC Theory to understand the ideas of filters. Also students studying a syllabus that is not AQA will find these notes helpful.
Before you study this tutorial, make sure that you understand inductors.
Reactance of an Inductor
Consider this circuit:
When the bulb is connected to DC, it glows with full brightness.
When it is connected to a signal generator giving out the RMS voltage that corresponds with the DC voltage, we observe the following:
At low frequencies, the bulb in series with the inductor is bright;
As we increase the frequency, the bulb becomes dimmer.
It allowed DC to flow freely, but opposed the flow of AC.
It is as if the inductor has a kind of resistance to AC, which is the reactance.
Electrically, an 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:
What is the reactance of a 3.5 mH inductor connected to a 12 V AC supply that has a frequency of 500 Hz?
A 10 mH inductor has a
reactance of 320 .
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:
These data were taken from a real experiment, and a line of best fit was added. The points are slightly away from the ideal line predicted by the equation. This is because the inductor will have a certain resistance.
You can determine the value of the inductance by taking the gradient, and dividing by 2.
Phase relationship in an inductor
If we plot the current and voltage in a pure inductor against the time, we get a graph like this:
The current graph is lagging the voltage graph by 90o 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:
Note that a purely reactive circuit based on a perfect inductor does not cause energy to be used. Energy is merely shuttled backwards and forwards between the inductor and the source. In reality there are resistive elements in any inductor circuit, not least in the inductor itself. Measuring current with a milliammeter introduces a resistive element to the circuit as well.
Simple LR Circuit
A purely inductive circuit is simply an electrical
curiosity (and doesn't exist in reality). With resistive elements in the
circuit, it becomes more interesting. In reality there are resistive elements,
such as the internal resistance of the inductor, and the resistance of the
Consider this circuit:
This is a simple series
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.
The voltage across the inductor is at 90o and is leading the current, so its phase vector points vertically upwards. The resultant voltage is shown by the phasor Vres. We can work out Vres by simply using Pythagoras.
At a certain frequency, the
voltage across a inductor is found to be 3.5 V while the voltage across
the resistor is found to be 4.0 V.
Impedance of the LR Circuit
When we studied capacitors, we introduced a new quantity, impedance, which was given the Physics code Z and had the units Ohms (W). Impedance takes into account the resistive and reactive elements in a circuit. The same applies to reactance of an inductor.
The formal definition of impedance is:
The ratio between resultant potential difference and the current in a reactive AC circuit
We can write this as:
We know that for the resistive elements:
We also know that for the reactance of an inductor:
Since the current is the same, we can redraw our phasor diagram as:
So we can say that the impedance is the vector sum of the resistance and the reactance. So by using Pythagoras again, we can write:
In a purely inductive circuit (i.e. with no
resistor) the average power dissipated is zero. However during each cycle,
energy is transferred forwards and backwards.
At a certain frequency, an inductor has a reactance of 20 ohms. It is in series with a resistor of 10 ohms. What is the impedance?
A fluorescent tube for use in a 50 Hz mains light fitting is marked 240 V, 60 W. In operation it has a resistance of 50 ohms and has an inductor (sometimes called a choke) to limit the current.
a. What power would the tube use if it were directly connected to the mains?
b. Work out the current in a 50 ohm resistor if the correct power of 60 W is flowing through it.
c. Show that the voltage across the resistor is about 55 V.
d. Using a phasor diagram, or by calculation, calculate the voltage across the coil.
e. Work out the reactance of the coil.
f. Work out the inductance of the coil. Give your answer to an appropriate number of significant figures.
g. Calculate the impedance of the light fitting.
h. Comment whether the value you worked out in (g) is consistent with the resultant voltage of 240 V.