Tutorial 9 - Magnetic Components and Alternating Currents

Learning Objectives

To learn about the response of inductive circuits to alternating currents.

To see how the ideas are applied to a simple low-pass filter.

To learn about the skin effect in wires carrying an alternating current.

Key Questions

How do inductive circuits behave with AC?

What is a low-pass filter?

What is the skin effect?

Demonstrating Reactance with AC

Note: Quite a lot of the material at the start of this tutorial is revision.  If you are happy with it, click HERE.

 

As far as electric currents are concerned, an inductor is simply a piece of coiled wire and should not affect the flow of charge at all.  Suppose we connect an inductor in series with a bulb.  The circuit is below:

The inductor has zero resistance.  If we connect the circuit to a DC battery, the bulb lights up to full brightness, just as we would expect.

 

Now letís connect it to an AC supply of exactly the same voltage.  We would expect the bulb to light up to exactly the same brightness.  But what we see is the bulb is slightly dimmer.  Somehow there seems to have been some extra resistance.  We call this extra resistance reactance.

 

Another demonstration is to connect a motor like that of a vacuum cleaner to a low voltage DC power supply (e.g. 12 V).  You need to increase and reduce the voltage slowly, otherwise the back emf could give you a shock.  You will find that the motor runs at quite a high speed.   Connect it to 12 V AC and nothing happens.  This is because of the reactance in the motor.

 

 

The motors have to be universal motors.  An induction motor will not work with DC at all.

 

Why does an inductor have a reactance?

Electric currents always produce a magnetic field.  That is an unchangeable law of Physics.  A direct current produces a steady magnetic field, while an alternating current produces a magnetic field that is changing all the time.

 

You can make a voltage across the ends of a wire by moving a magnet past a wire, or by moving a wire past a magnet.  You can even make a voltage by having a magnet sitting next to a wire that is stationary, but you have got to change the magnetic field.  If the magnetic field remains the same, no voltage will be induced, however strong the magnetic field.

 

A current-carrying coil of wire will act as an electromagnet, even though the coil itself is made of a non-magnetic material like copper.  Letís think about what would happen when we switch on a current that passes through a coil of wire.  When we switch on the current, a magnetic field is made as the current flows.  As the field being made, a reverse voltage is made to oppose the increase in voltage across the inductor. 

 

We can use Faradayís and Lenzís laws to help us to model the situation.  There are some terms that we need to know to help us think it through:

 

Quantity

Physics Code

Units

Flux density

B

Tesla (T)

Magnetic flux

F

Weber (Wb)

Electromotive Force (voltage)

E (Curly ĎEí)

Volts (V)

 

Flux is simply the total number of field lines.   The magnetic flux is the product between the field strength and the area.  In physics code it is written:

 

F = BA

 

Flux density is high when the field lines are close together.  That means that the magnetic field is strong.  So the magnetic field strength is shown by the concentration of field lines.  If the field line a spread out, the field is weaker. 

 

Faradayís and Lenzís Laws tell us that a change in a magnetic field will produce an electromotive force (EMF) that will act to oppose the change.  This can be summed up by the equation:

 

 

The term N is the number of turns in the coil, and the dF/dt is the rate of change of flux, i.e. how much the flux changes in a small time interval.  The minus sign tells us it's a reverse EMF, opposing the change.  If the dt term is very small, then the reverse EMF is very large.  This has important implications for electronic circuits with inductive components.

 

If we plot a graph of the voltage across a conductor when it is switched on, we get an exponential rise like this:

 

When the voltage is not changing when DC flows, then there is zero reverse EMF, and the current flows normally as if the inductor were simply a wire.  However, some work per coulomb is done to build up the magnetic field when the inductor is switched on.

 

However, in AC, as the current is changing direction all the time, there is a reverse EMF being induced all the time to prevent the change from happening (Lenz's Law).  The higher the frequency, the bigger the reverse EMF.

 

Inductance and Current

The self-inductance of a coil can be worked out from the equation:

 

Units for dI/dt are amps per second.  The minus sign tells us that the EMF is opposing the applied voltage.  The dI/dt term is the rate of change of the current:

We can determine this by measuring the gradient at any point, as shown in the diagram above.  We need to take a tangent and work out its rise and run.   But if we know L, we can easily work out the reverse EMF.

 

Question 1

A solenoid of 2.3 ◊ 10-2 H and negligible resistance is connected across a 12 V battery.  What is the rate of increase of current in the solenoid as itís turned on?

Answer

 

This is not a particularly easy way of measuring L.  An easier way is to consider the properties of an inductor.  The inductance of a solenoid is worked out using the following equation:

 

The terms involved are:

Quantity

Physics Code

Units

Inductance

L

Henry (H)

Permeability of free space

m0 (= 4p ◊ 10-7)

H m-1

Number of turns

N

 

Area of the solenoid

A

m2

Length

l

m

 

If there is a core of magnetic material, then m0 is multiplied by the relative permeability mr, which is just a number with no units.

 

Question 2

An inductor is made of a solenoid of 1200 turns of copper wire around a square former 2.0 cm ◊ 2.0 cm. The length of the solenoid is 5.0 cm. Calculate the inductance.

Answer

 

Reactance of an Inductor

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.  With an alternating current, there is an effective resistance called reactance.  The Physics code is XL and units are ohms (W). The equation for reactance is:

 

Quantity

Physics Code

Units

Reactance

XL

Ohm (W)

Frequency

f

Hz

Inductance

L

 Henry (H)

 

Question 3

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.

Answer

 

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:

The graph is derived from an actual experiment, so that the line is not quite straight.  The line of best fit has a small circle at each end.

 

You can determine the value of the inductance by taking the gradient, and dividing by 2p.

 

 

Phasor diagrams and inductor reactance

If you have done A-level Physics, you will have covered circular motion and simple harmonic motion.  You will remember that the two are very closely related.  Wave motion and SHM are also very closely related.  Since alternating current is generated using a rotary generator, we can use many of the ideas of SHM and circular motion in the following arguments.

 

If we use a data-logger to measure the voltage and current of  a resistor through which an alternating current is flowing, we find that the two are in step with each other.  They are in phase.   If we do the same with an inductor , we get a rather unexpected result.  The voltage and current are 90 degrees out of phase.

If you want to find out more about the reasons for this, click HERE to go to a tutorial on AC Theory.

 

The current graph is lagging the voltage graph by 90o or p/2 rad.  Drawing graphs is a bit of a pain, so we use a phase vector (or phasor) diagram which looks like this.  For more about this, click HERE to go to Tutorial 3 in AC Theory. 

 

The voltage and current are alternating at frequency f Hz.  This means that there are f cycles every second.  One cycle in the alternation can be regarded as a single revolution of the phase vector.  In one revolution the angle in radians is 2p rad. The rate of turning is described as the angular velocity, which is given the physics code w (omega, a Greek letter long 'o').  By convention, the phase vector turns anticlockwise, and the zero point is at the three o'clock position.

 

The phasor diagram for the graph is shown below:

 

So the voltage phase vector is leading the current phasor by p/2 rad.

 

Series LR circuits

A 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. 

 

In this circuit there is an inductor, L H, in series with a resistor, R W.   The voltage alternates at a frequency f Hz.

Now the voltage and current in the resistor are always in phase.  The voltage in the inductor leads the current by p/2 rad.  This is a series circuit so the current is the same all the way round.  Therefore our phase vector diagram looks like this:

 

Notice that we haven't shown the entire circle.  This is what it looks like at time zero, and that's what we are interested in.

 

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.

 

 

Question 4

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.
a. Why is the total voltage not 7.5 V?
b. What is the resultant voltage?

Answer

 

 

 

 

The sum of the two voltages is a vector sum, not an arithmetical sum.  This is because the vectors are 90o apart.

 

 

Now add in the phase angle, f (phi - Greek lower case letter 'ph' or 'f'):

We can easily work out the phase angle by which the current leads the resultant voltage:

Question 5

Using the data from question 3, calculate the phase angle between the current (the voltage across the resistor) and the resulting voltage.

Answer

 

Impedance

In a reactive circuit, we cannot talk about resistance as such.  We need to introduce a new quantity, impedance, which has 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:

We can work out the phase angle between the resistance and impedance quite easily, just like we measured it between the voltage across the resistor and the voltage across the inductor.

We can write other equations for the phase angle:

and:

 

The reactance and resistance do NOT add up arithmetically; you have to do a vector sum

 

The phase angle is in radians.  If you want to convert to degrees, you have to multiply by p and divide by 180

 

 

 

Question 6

At a certain frequency, a inductor has a reactance of 20 ohms. It is in series with a resistor of 10 ohms. What is the impedance? What is the phase angle between the resistance and the impedance?

Answer

 

Parallel RL circuits

Consider this circuit that consists of an inductor of inductance L H, and resistor of resistance R W.  It is connected to an alternating voltage V V that has a frequency of f Hz.

We know that:

         In a parallel circuit, the voltage is the same across each branch.

         In a capacitor circuit, the current leads the capacitor voltage by 90o.

         In an inductor circuit, the current lags the inductor voltage by 90o.

         In a parallel reactive circuit, the currents add up as a vector sum.

         The current is always in phase with the voltage across the resistor.

 

Since VR is in phase with I, we can draw a phasor diagram to show the phase relationship.  We will show IR leading, and IL lagging. 

 

We can show the resultant current, I.

We can easily see that I is the vector sum of IL and IR.  So we write:

where:

and :

We can write an expression for the phase angle:

We also know that:

So we can get an expression for Z by simple substitution into the first equation:

Since the voltage across a parallel circuit is the same, we can rewrite this as:

This reminds us of the equation for parallel resistors:

We work out the reactance of the inductor simply by using:

 

Question 7

A parallel circuit consists of a perfect inductor of 400 mH and a 200 ohm resistor. The components are connected to a supply of 6 V at a frequency of 300 Hz.
a. What is the reactance of the inductance?
b. What is the impedance of the circuit?
c. What is the resultant current?

Answer

 

Parallel Inductor which has a Resistance

Now we know that no inductor is perfect; it has a definite value for the resistance.  Letís look at this further.  It does complicate things, but itís not impossible.

 

We model the real inductor as a perfect inductor in series with a resistor, r.

 

The first thing we need to do is to work out the series impedance between L and r.  We will call this z.

 

z2 = r2 + XL2

 

Question 8

Use the data from Question 7. The frequency is now 50 Hz.
What is the impedance of the series part of the circuit, if the resistance of the inductor is 15 ohms?

Answer

 

As the frequency goes up with an inductor, the reactance also increases.  Therefore the ohmic resistance of the inductor becomes much less significant.  Using the numbers that we did in Question 3, we found that the reactance was 754 W.  When you add 152 to 7542, the square root hardly changes from 754 W.  (Try it for yourselves.  The answer is 754.15 W.) 

 

At much lower frequencies the change becomes significant.

 

If you want to see how inductors and capacitors interact in AC circuits, see my Electrical Principles tutorials.

 

Low Pass Filter

A loudspeaker is a transducer that converts and electrical waveform into a sound wave for us to listen to.  The radio in the picture below has a single loudspeaker.  The radio is good quality and gives a quality sound, but the bass notes do not come through as well since the speaker is rather small.

 

 

The Hi-Fi loudspeaker has two drive units.  The larger unit is called a woofer, while the smaller unit is the tweeter.

 

 

The woofer can handle the low frequencies well, but it cannot respond to high frequencies due to the inertia of the coil.  Therefore there is a low pass filter to allow only the low frequencies to be fed to the woofer.  In this case, the low pass filter is passive, which means that it depends on the reactance of an inductor.  (An active filter has an amplifier associated with it.)   The high pass filter for the tweeter is based on a capacitor.  We will look at the low pass filter for the woofer.

 Consider a woofer that has a resistance of R W.  Since it is a magnetic component, it will have an inductance, L2  H which is given by:

 

The woofer can be modelled as a resistance and inductance  in series.   The low pass filter will be an air-cored inductor of inductance L1 H and it will also have a resistance of R1 W.  The inductance is given by:

 

 

We have to be careful about our problem solving strategy.  In a series circuit, the resistances add up, and the inductances add up.  So we need then to so the vector addition to get the final impedance.  For the resistances:

 

And the series inductances add up as well:

 Then we need to work out the reactance, XL:

 

Then we work out the impedance, Z:

 

Do not be tempted to work out the impedances of the components and then add them together.  This is wrong:

and

 This will give you a false answer.  Look at the  vector diagram:

 

We can see how the resistance and inductance vectors make up the impedance vectors for Z1 and Z2.  Since the directions for Z1 and Z2 are different, we cannot simply add their lengths together to get the overall length that would give rise to Ztot.   Instead we need to write:

 

Example

A loudspeaker coil is made of 175 turns of fine copper wire mounted on a circular former of diameter 2.0 cm and length 2.0 cm.  The resistance of the coil is 8.0 W.  It is connected in series with a low-pass filter consisting of an inductor which has 50 turns of wire on a circular diameter 3.0 cm and length 3.0 cm.  The low pass filter has a resistance of 1.0 W  The loudspeaker is connected to alternating waveform of 250 Hz.

(a) Calculate the inductance of each component;

(b) Work out the reactance of each component;

(c) Hence work out the impedance of the combined speaker and low pass filter.

Answer

(a)  Work out the area of the speaker: A = pr2 = p ◊ (0.010 m)2 = 3.14 ◊ 10-4 m2

      Work out the area of the inductor: A = p ◊ (0.015 m)2 = 7.069 ◊ 10-4 m2

 

      For the speaker, L = (4p ◊ 10-7 H m-1 ◊ 1752 ◊ 3.14 ◊ 10-4 m2) ų 0.020 m = 6.04 ◊ 10-4 H

      For the inductor L = (4p ◊ 10-7 H m-1 ◊ 502 ◊ 7.069 ◊ 10-4 m2 ) ų 0.030 m = 7.40 ◊ 10-5 H

 

(b)

     For the loudspeaker, the reactance XL = 2 ◊ p ◊ 250 Hz ◊ 6.04 ◊ 10-4 H = 0.949 W

     For the inductor, XL = 2 ◊ p ◊ 250 Hz ◊ 7.40 ◊ 10-5 H = 0.116 W

 

(c) Combined resistance = 8.0 W + 1.0 W = 9.0 W

     Combined reactance = 0.949 W + 0.116 W = 1.065 W

 

     Z2 = (9.0 W)2 + (1.065 W)2 = 82.13 W2

 

     Z = 9.1 W (2 s.f.)

 

 

Question 9

A loudspeaker has an inductance of 1.25 ◊ 10-4 H and a resistance of 4.0 W.  It is connected in series with an inductor of inductance  3.0 ◊ 10-5 H and resistance 1.0 W.  A frequency of 1400 Hz is passed through the loudspeaker.   Calculate:

(a) the reactance of each of the components;

(b) the combined impedance at this frequency.

Answer

 

Inductance in a Cable

When a direct current flows in a cable, the distribution of charge can be considered to be even across the cross section of the conductor.  However, when the current is alternating, the charge tends not to be uniform across the area, but more concentrated on the outside surface.  This is called the skin effect.  At low frequencies, this effect is only marginal and can be ignored.  At very high frequency, it is significant and can cause the system no longer to follow the predicted behaviour.

 

Since flux is linked with current and inductance is linked with flux, there is a greater inductance towards the centre of the wire.  At high frequencies, the reactance becomes higher as:

 

Therefore the current can pass more easily around the outside of the wire.

 

Explanation

We know that the magnetic field around a single isolated wire is a series of concentric circles and follows the screwdriver rule:

The same thing happens within the wire.  We could model the system as an infinitely thin wire, so we get these concentric cylinders of flux.  Since the wire is usually made of copper, a completely non magnetic material, it is reasonable to assume that the pattern flux within a copper wire of a definite thickness will be the same as the the infinitely thin wire with flux around the outside.  The flux has a higher value around the centre of the circle.  If there is an alternating current, the magnetic field is alternating as well.  The changing magnetic field is causing a back EMF which is given by our friend:

The result of this is that there are eddy currents in the wire that look like this:

 

You can see that around the edges, the eddy currents flow in the same direction as the current going through the wire.  Towards the centre, the eddy currents oppose the flow of the current.  The result of this is that the majority of the current is carried at the edge of the wire.  The idea is shown in the diagram below:

The skin depth, d, is the depth in which 63 % of the current is carried.  The physics code d is delta, a Greek lower case letter 'd'.  Within the dotted circle only 37 % (1/e) of the current is carried.  The size of the skin depth depends on the frequency.  The lower the frequency, the wider the skin depth.  If the frequency is zero, the current is carried uniformly.

 

The current density, J A m-2 , varies according to this relationship:

 

Quantity

Physics Code

Units

Current Density

J

Amp per square metre (A m-2)

Current density at the surface

JS

Amp per square metre (A m-2)

Exponential number

e

 2.718...

Imaginary number

j

j2 = -1

Depth

d

Metre (m)

Skin depth

d

Metre (m)

 

This indicates that the current density decreases exponentially from the maximum value which occurs at the surface.  The use of the imaginary number j indicates that there is a phase change of 1 radian for every skin depth.  The j factor is determined by angular velocity of the alternating waveform.  Further discussion is beyond the scope of these notes.

 

The general formula for calculating the skin depth is this:

Looks horrible, doesn't it?  The second bit is a correction factor that is used at higher frequencies.

 

Quantity

Physics Code

Units

Resistivity

r

Resistivity (W m)

Angular velocity

w

Radians per second (rad s-1)

Absolute permeability

m

 Henry per metre (H m-1)

Absolute permittivity

e

Farads per metre (F m-1)

Skin depth

d

Metre (m)

 

The absolute permeability of any material is the product of the permeability of free space and the relative permeability:

 

m = m0mr

 

The value for permeability of free space m0 = 4p ◊ 10-7 H m-1

 

The absolute permittivity of any material is the product of the permittivity of free space and the relative permittivity:

 

e = e0er

 

The value for permittivity of free space e0 = 8.854 ◊ 10-12 F m-1.  The symbol e is epsilon, a Greek lower case letter 'e'.

 

The good news is that for frequencies of less than 1/re, the square root after the multiplication sign becomes very close to 1, and the formula looks a lot simpler:

 

Let's do a worked example:

 

Worked Example

What is the skin depth in a copper wire if the frequency of the alternating current is 400 Hz?  Resistivity of copper is 1.68 ◊ 10-8 W m

Answer

m for copper = 4p ◊ 10-7 H m-1

 

Work out the angular velocity:

w = 2pf = 2p ◊ 400 = 800 p rad s-1

 

d = [(2 ◊ 1.68 ◊ 10-8 W m) ų (800 p rad s-1 ◊ 4p ◊ 10-7 H m-1)0.5 = 3.26 ◊ 10-3 m

 

The implication of this is that AC currents do not need solid conductors.  Bus bars that carry heavy currents in power station generators are actually hollow, and carry coiling oil.

 

Question 10

A transmitter is connected to the aerial by a thick cable that is 5.0 mm in diameter.  The transmitter operates at a frequency of 200 kHz.  The cable is made from aluminium, which has a resistivity of 2.82 ◊ 10-8 W m.  Show that the depth of the layer that carries 63 % of the current is about 0.2 mm.

Answer

 

The answer to Question 10 shows that the skin effect can be very marked at radio frequencies.  In some cases the conductors are made of glass rod on which there is a thin layer of conducting material.  However there is an important implication. 

 

Consider a wire of area 1 mm2 that carries a current of 15 A.  That is the maximum safe rating, above which there is the risk of it heating up.  The current density is given by:

 

Therefore in this case J = 15 ◊ 106 A m-2

 

Now let's look at the current density using our result from Question 10.  Let's assume that the skin effect shows that the vast majority of the current is carried in the outer 0.4 mm of the 5 mm bus-bar.  We work out the area that carries the current by subtracting the region that carries little or no current from the total area of the circle:

Area of ring = [p ◊ (2.5 ◊ 10-3 m)2] -  [p ◊ (2.1 ◊ 10-3 m)2] = 1.963 ◊ 10-5 m2 - 1.385 ◊ 10-5 m2 = 5.78 ◊ 10-6 m2

 

Let's suppose a current of 100 A is flowing, as we might expect in a powerful transmitter. 

 

J = 100 A ų 5.78 ◊ 10-6 m2 = 17 ◊ 106 A m-2

 

While a 5 mm bus-bar will easily carry 100 A DC, this calculation shows that the current density at 200 kHz is excessive, leading to a significant heating effect.  This can be mitigated by using bus-bars of wider diameter.  To save material, a pipe would be used.

 

 

Self inductance happens in straight wires carrying DC or low frequency AC.  We will explore this more in Tutorial 11.

 

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