Tutorial 10 A - Mutual Inductance and Transformers

Learning Objectives

To learn about mutual inductance.

To use magnetic circuit theory to explain transformer equations.

To consider the structure of a transformer and different types of transformer.

To identify and quantify losses in a transformer.

To learn about the induction coil

Key Questions

What is mutual inductance?

How do we derive the transformer equations?

How are transformers made?

Why are transformers not 100 % efficient?

How can the transformer effect be made to work with DC?

Mutual Inductance

In previous tutorials, we have seen the effect of self inductance in a single coil.  We pass a current through the coil, we get a magnetic field.  If we make that magnetic field pass by a second coil, we get an induced voltage in that second coil.

We actually don't have to have any movement at all.   If we change the current, there is a change in the magnetic field, and that induces a voltage in a second unconnected coil. This is called the transformer effect or mutual induction.

If a changing current I1 A is passed through Coil 1, there will be a changing voltage E2  V in Coil 2.  This can be worked out with the formula:

It's the same pattern as the formula for self-inductance.  The term M is the mutual inductance which has units of Henry (H).

 When a coil of wire carries a current of 4.0 A that decays to zero in 0.15 s, an EMF of 10.0 V is measured for the duration.  What is the mutual inductance?

Mutual Inductance and Reluctance

Let us have both coils wound around a closed iron ring like this:

Coil 1 has N1 turns and has a supplied current of I1 A flowing in it.  Coil 2 with N2 turns is connected to a complete circuit which has an induced current I2 A flowing.  The iron ring has a reluctance S H-1.  Within the ring, there is a flux F Wb.  Since this is a magnetic circuit, we can say that F is the same throughout.  From magnetic circuits (Tutorial 7) we know that:

So we can write:

Since the flux is produced by the current in Coil 1, we can write an expression for the mutual inductance (which affects Coil 2):

We now untidy things by multiplying by (N1/N1).  Yes I know it is 1, but it's essential for the argument:

From the equation above:

So substituting in, we get:

And cancelling gives us:

 Worked Example Consider the device below: The current in coil 1 increases linearly from 1.0 A to 5.0 A in 150 ms.  As a result an EMF is induced in Coil 2 that is measured at 18.0 V.  Calculate: (a) the mutual inductance; (b) the reluctance of the material on which the two coils are wound; (c) the self-inductance of the primary coil. Answer (a)  The rate of change of the current = (5.0A - 1.0 A) ÷ 150 × 10-3 = 26.7 A s-1              Ignoring the minus sign:       M = 18.0 V ÷ 26.7 A s-1 = 0.675 H = 0.68 H (2 s.f.)   (b) Formula:    S = (520 turns × 1000 turns) ÷ 0.675 H = 7.7 × 105 H-1   (c) Formula (see Tutorial 8)   L = (1000 turns)2 ÷ 7.703 × 105 H-1 = 1.30 H = 1.3 H (2 s.f.)

 The mutual inductance between two coils is 240 mH.  The current in Coil 1 changes from 15 A to 6.0 A in 12 ms.  Calculate: (a) The average EMF induced in Coil 2; (b) The change of flux linked with Coil 2 which consists of 400 turns.

The Current Equation

We know from above that:

The flux is the same, so the F term cancels out to leave:

I1N1 = I2N2

This rearranges to:

This assumes that the transformer is perfect, i.e. there are no losses.

 The diagram shows a transformer:   Coil 1 has a current of 3.5 A flowing in it.  Calculate the current in Coil 2 if it's connected to a suitable load.

The Voltage Equation

We know that:

Coil 1 has N1 turns and Coil 2 has N2 turns.  The EMF for Coil 1 is E1 V and the EMF for Coil 2 is E2 V.  The rate of change in flux is the same.  So we can write:

Since EMF is a voltage we can rewrite the equation as:

We can then rearrange the equation to give:

 The diagram shows a transformer:   Coil 1 has a voltage of 28 V across it.  Calculate the voltage in Coil 2.

Transformers

Transformers use the principle of mutual induction to change the value of alternating voltages and currents, as we have seen in the example above.  The devices are generally very efficient with low losses and life expectancy is long.  The picture below shows a variety of different transformers that I have at home:

The transformer is a machine that is simplicity itself.  It consists of:

• A primary coil connected to the alternating power source.  This provides the changing magnetic field.

• A secondary coil connected to the load (in this case, the rectifier and control equipment in the locomotive).

• A laminated soft iron core.

The transformer in the picture above is referred to as a core transformer.  This is the sort that schools and colleges have for demonstration purposes, the demountable transformer.

Most practical transformers have the secondary wrapped around the primary.  This saves space and reduces flux leakage. The construction is referred to as a shell transformer.  The picture below shows the inside of the labpack, which has a shell transformer.

This transformer has several wires coming from its coils.  They are called taps and allow for a range of voltages  that are selected by switches.  The switches are sometimes called tap-changers.  The thermal switch switches off if excessive current is taken from the transformer, and prevents it getting too hot.  The rectifier turns the AC to DC (strictly speaking, rectified AC).  In this particular example, the rectifier does not work after a reverse voltage ruined it.

The two coils are electrically completely different circuits.  Either of the coils can act as a primary.  The laminated core is made up of layers of soft iron separated by an insulating layer of varnish or glue. The laminated soft iron core is shown in the picture below:

This reduces losses from eddy currents.  Eddy currents can have a significant heating effect.  If there aren't any laminations, the core can get very hot indeed.

Soft iron loses its magnetism immediately the current is turned off.  Therefore the magnetic field can change forwards to backwards as the current changes.

The ratio of the input voltage to the output voltage is the same as the ratio of the number of turns on the primary to the number of turns on the secondary.   We have seen the equation before:

If N1 is greater than N2, we have a step-down transformer, because the voltage is reduced.  A step-up transformer increases the voltage.

If a transformer is 100 % efficient (and it nearly is) we can say that:

power in = power out

Equation:

V1I1 = V2I2

Therefore we can say that when the voltage is lower, the current is bigger.  We have seen the transformer equation in terms of current:

In practice, the transformer is about 97 % efficient.  When a large transformer is transferring a lot of energy, even 3 % losses produce a fair amount of heat.  Therefore the transformer is cooled with oil, which is pumped to heat exchangers.

The giant transformers here step the output voltage of a power station from 25 000 V to 132 000 V for transmission.

Transformers can only work with alternating current; they cannot work with direct current.

 A transformer has a primary of 3600 turns and a secondary of 150 turns. It takes 1.5 amps from the 240 V mains. What is the output voltage and current?

Losses in a Transformer

In schools and colleges, we often teach that the power into a transformer is the same as the power out.  This is summed up in the equation:

P = V1I1 = V2I2

This is alright as a first approximation.  If it were completely correct, this tutorial would stop here.  There is no such thing as a 100 % efficient device.  Even in the best transformer, there are losses, which we will explore further.  What are the sources of inefficiency in a transformer?

• The coils will have a certain value of resistance.  If the currents are large, the energy loss is large, since P = I2R.  The losses are called copper losses.

• With the laminated soft iron core, there are still eddy currents, even though they are much reduced.  These will heat up the core.  These are called iron losses.

• Transformers are inductive devices.  Therefore there is a certain reactance as well as resistance.

• Work has to be done in building up the magnetic field in the core.  It takes energy to line up all the domains.  The energy recovered as the magnetic field falls is less than what is put in.

This is called hysteresis, shown on the graph below:

A few points to note:

• The area within the loop is the energy lost.

• All magnetic materials have a certain remanance, meaning that the material is magnetised, even when the current is zero.  In soft iron, the remanance is low.

• There is a value of current at which all the tiny molecular magnets are lined up.  The magnetic field cannot increase further.  This is called saturation

The efficiency is given by the equation:

The strange looking symbol, h, is "eta", a Greek letter long 'ē'.  It used as the physics code for efficiency.  The equation can be rewritten as:

Efficiency is a fraction, and is often expressed as a percentage.

Power Factor

Another important quantity for transformers is the power factor.  Since a transformer is reactive device connected to an alternating supply, some energy is lost in making the alternating magnetic field.  Therefore the power that can be usefully got out of the transformer is not simply the product of the voltage and the current.  For more details, see my Electrical Principles notes, Tutorial 5 and Tutorial 6.

The product of voltage and current is called the apparent power, and is given the physics code S.  Its units are volt-ampère (VA).  The equations for this is:

S = VI

The true power, physics code P, is the product of the apparent power and the power factor.  It is measured in watts (W) and is given by:

True Power (W) = Apparent Power (VA) × power factor

In Physics code, this is written:

The term cos f is the power factor and we can  see this in the power phase vector diagram below.

The reactive power, Q, is the power taken by the changing magnetic field.   The units are reactive volt-amp (var).  Work has to be done to make the magnetic field.  Some energy is got out as the magnetic field collapses as the voltage falls, but it's not the same as the work done to make the magnetic field.  Some is lost as shown on the hysteresis graph above.  The formula is:

For many transformers, the power factor works out at about 0.8.  Let's do a worked example:

 Worked Example A transformer has a power factor of 0.85 and is rated at 300 kVA.  The iron loss is 2.0 kW and the copper loss is 2.5 kW at full load.  What is the efficiency? Answer Useful power = 300 kVA × 0.85 = 255 kW Total losses = 2.0 kW + 2.5 kW = 4.5 kW  Efficiency = 1 - (4.5 kW ÷ 255 kW) = 0.982 Percentage efficiency = 98.2 %

Iron losses are constant, since they depend on the frequency of the alternating current.  Most transformers are connected to a 50 Hz AC supply.  Copper losses stem from the resistance of the wire.  So they will vary with the current.  If the load is reduced to half, the current in both the primary and secondary will be reduced to half.  Therefore the losses will be reduced to a quarter, since losses depend on the square of the current.

 The transformer in the worked example is operating at half power.  The power factor remains the same.  What is the efficiency now?

The maximum efficiency occurs when the copper losses are equal to the iron losses.  Working out the maximum efficiency needs to be approached carefully.  Follow through the worked example to see how.

 Worked Example A transformer has a power factor of 0.85 and is rated at 300 kVA.  The iron loss is 2.0 kW and the copper loss is 2.5 kW at full load.  What is the maximum efficiency? Answer 1. Let x be the fraction of the full load in kVA at which the efficiency is the maximum.   Corresponding copper loss =  2.5 kW × x2 Iron loss = copper loss 2.5 kW × x2 = 2.0 kW x2 = 0.8 x = 0.894   2. Work out the load to which this corresponds: Load = 0.894 × 300 = 268.3 kVA   3. Use the power factor to work out the useful power: Output power = 268.3 kVA × 0.85 = 228.1 kW   4. Work out total losses for the copper and the iron: Total losses = 2.0 kW + 2.0 kW = 4.0 kW   5. Work out the input power: Input power = output power + losses = 228.1 kW + 4.0 kW = 232.1 kW   6. Now you can work out the efficiency: Maximum efficiency = 1 - (4.0 kW ÷ 232.1 kW) = 0.9824   Percentage efficiency = 98.24 %

 A 250 kVA  transformer has a full load copper loss of 3.0 kW and an iron loss of 2.0 kW.  It has a power factor of 0.80.  Calculate: (a) The output apparent power at which the efficiency of the transformer is at a maximum; (b) The maximum percentage efficiency.

Different Transformer Types

We have seen above the core type transformer that is used in the demountable transformer in physics labs.  The coils are made of enamelled copper wire.  The enamel acts as an electrical insulator.  A picture is shown below:

There is a range of accessories that allow the tutor to demonstrate induction heating and welding.  Students like to see the demonstration that shows nails being welded together.

The more common type of transformer is the shell type in which the secondary is wrapped around the primary as in the picture below:

In power transformers the core is generally made of silicon steel, a material in which hysteresis and eddy currents are minimised.  Even so there can be a marked effect for very large power transformers, so cooling systems are needed.  In very large transformers, insulated aluminium wire used to keep the costs down.

Audio frequency transformers are rated from a few mVA to about 20 VA and operate at frequencies in the range of human hearing, 20 Hz to 20 kHz.  They are similar in construction to a shell transformer, but in small audio frequency transformers, the core is made of ferrite.  In larger ones the core is made of silicon steel. They are used for impedance matching, which we will look at in Tutorial 10 C.

In radio frequency transformers the core is of air, ferrite, or dustFerrite is a ceramic material that has a high resistivity, but similar magnetic properties to silicon steel.  Dust cores consist of fine particles of iron, each particle of which is insulated from its neighbour.

Radio transmission uses the transformer effect with an air (or even vacuum) core.  The transmitter acts as the primary of a very inefficient transformer, while the aerial on your radio set is the secondary.  Tuned filters consisting of inductors and capacitors can be made to resonate at the desired frequency to select the right frequency.

DC Transformers

The sorts of transformers we have seen above will not work with direct current.  A secondary voltage will only be induced when the transformer is turned on or off.  While the DC is flowing there will be a constant level of flux in the core.  There will be heat losses due to the resistance in the copper windings.  Otherwise all we have is a physics curiosity.

The earliest experiments with transformers used induction coils.  These devices use DC, and have a make and break circuit (See Tutorial 4B).  The diagram below shows the idea:

Image from Wikimedia Commons

By Original work: PieterJanR.  This version: Chetvorno

This formed the basis of induction coils in school and college physics labs:

Image from Wikimedia Commons

By Harry Winfield Secor - Downloaded from Harry Winfield Secor (1920) The How and Why of Radio Apparatus, 1st Ed., Experimenter Publishing Co., New York, p.5 on Google Books, Public Domain,

These devices could easily generate 100 kV at the terminals.  They were used in early radio transmission equipment and early X-ray tubes.

One of the earliest physicists who did work on the induction coil was Father Nicholas Callan (1799 - 1864) who was Professor of Natural Philosophy at Maynooth College in Ireland.  He found that when a secondary coil was wrapped around a primary, hefty electric shocks could be experienced if the primary circuit was broken.  His coil could easily generate 60 000 V.

Image from Wikimedia Commons - Auguste Blanqui

It is said that his voltmeter consisted of a number of Catholic seminary students in series.  Presumably precise calibration of such an instrument was not easy and there was always a high degree of uncertainty.  He measured the voltage on how high they jumped when he turned the contraption on.  This could be an explanation for the generally crusty nature of Catholic priests in Ireland at this time...

The secondary voltages were considered highly dangerous in later years, and the use of induction coils in schools and colleges was banned.

Induction coils are used in car ignition systems, although nowadays the switching is done electronically.