Tutorial 10 B -  Transformer Action

Learning Objectives

To learn how transformers behave when off load and on load.

To learn how transformers are used to match a source with a load.

Key Questions

How do transformers behave off load?

How do transformers behave on load?

How do we match a source and a load using a transformer?

Transformers off Load

The transformer in the labpack in the picture in the previous tutorial has a resistance of 11.0 ohms (as measured with an ohm-meter).  If this were connected across a 230 V dc supply, it would take a current of nearly 21 A and certainly blow the fuse in the plug.  When connected to the mains with no load, the current is actually very small.  Why is this?


Consider a transformer of N1 turns on the primary connected to a voltage source of V1 volts.  There is a secondary of N2 turns which supplies a no-load voltage of V2 volts.  The two coils are linked by a flux of F Weber which is common to both.  The flux is set up by a small current I0 amps in the primary, which is in phase with the flux.  Lenz's Law applies so that V1 is opposed by and induced EMF of E1 volts and V2 is opposed by an induced EMF of E2 volts.


We know that the voltage leads the current by 90 degrees.  We will assume that the inductor is perfect.  The phasor diagram looks like this:



The induced EMF E1 is in antiphase to V1 (as is the EMF E2 with V2).  In a perfect transformer there would zero resultant voltage.  There would be no copper losses either.  The current  I0 is used to magnetise the core i.e., to provide the flux.


In a real transformer there are iron losses.  We will continue to ignore the copper losses.  There is a magnetising component, IM, which is in phase with the flux.  Then there is a core loss component, IC, which is due to eddy currents and hysteresis.  So our phasor diagram looks like this:

The current lags the voltage by an angle of f0.


The relationships between the currents I0 , IC, and IM are as follows.  For the vector sum:


The magnetising current:

The core loss component:


We can work out the power factor at no load:

The iron losses are given by:


None of these are easy to measure.  But we can use easily measured quantities to work these out.  Let's do a worked example see how:


Worked Example

A single phase transformer has a primary voltage of 230 V and a secondary voltage of 12 V.  It takes a no-load current of 0.05 A and the iron loss is known to be 4.0 W  What are the core loss and magnetising components?


Iron loss = 4.0 W = 230 V × 0.050 A × cos f0


cos f0 = 4W ÷ 11.5 W = 0.3478

f0 = cos-1 0.3478 = 69.64o


Magnetising current:

IM = 0.05 A × sin 69.64 = 0.0469 A (= 46.9 mA)


Core loss component:

IC = 0.05  A × cos 69.64 = 0.0174 A (=17.4 mA)


Question 8

A transformer has an input voltage of 3300 V and an output voltage of 400 V.  The iron loss is known to be 500 W and the no load current is measured at 0.80 A.  Determine the values of the magnetising and the core loss currents.



Loading a Transformer

Consider a transformer that has negligible resistance in both the primary and secondary.  The number of primary turns is the same as the number of secondary turns.  It is a 1:1 transformer.  You may be thinking what is the point of that, but they are used as isolation transformers, for example for a razor socket in the bathroom.  We can say that there is negligible voltage drop:


V1 = E1


V2 = E2


The secondary has a current, I2 amps flowing in it.  The resulting secondary EMF will try to reduce the core flux, as it opposes the change (Lenz's Law).  The result of this is that the primary EMF is reduced, which allows a bigger current to flow through the primary.  The increased current leads to an increased magnetomotive force to balance out the change in the flux.  This restores the flux to what was before.


The key point is that the flux remains constant.


Let's use a phasor diagram to show this rather complex idea:


The current I2 lags the voltage V2 by f2.  There is a reflected increase in current in the primary which we will call I1' ("I-one-prime").  This also lags the voltage V1 by phase angle f2.  When there was no load, we saw that there was the current vector I0, which was the resultant of the magnetising current IM and the core loss current IC.   It lags the voltage V1 by f0


Therefore the primary current I1 is the vector sum of the currents I2 and I0.  It lags the voltage V1 by the phase angle f1.


In this argument, the number of turns was equal for the primary and secondary.  If they are not equal, we need to work out the values of V1, V2, I1', and I2 using the transformer equations.  We will use phasor diagrams to help us in the calculations.  Let us look at a worked example:


Worked Example

A transformer has 2000 turns on the primary and 800 turns on the secondary.  The power factor is 0.2 lagging, and the no-load current is 5.0 A.  The voltage drop is negligible.  On load, the secondary current is 100 A with a power factor of 0.85 lagging.  Determine:

(a) The primary current;

(b) The primary power factor.


The reflected current I1' can be worked out using the equation:


N1I1' = N2I2


I1' × 2000 turns = 800 turns × 100 A

I1' = 40 A


This is NOT the primary current, because we have to take into account the current vector I0.


The secondary power factor is 0.85.  Therefore the phase angle (lagging) is:

f2 = cos-1 0.85 = 31.79o


The phase vector diagram (NOT to scale) so far looks like this:


We know that the no-load current I0 = 5.0 A and the power factor is 0.20:

f0 = cos-1 0.20 = 78.46o


We can add that to the phasor diagram as shown:

Now we need to draw the measured current I1 and this is shown in the phase vector diagram:



We could work out the resultant current I1 and the phase angle f1 by accurate drawing.  However if your "accurate" drawing is anything like mine, it is far from accurate.  A better way is by adding the vector components.  This is shown below with much of the clutter removed:


The horizontal components add up:


IH = 40 A × sin 31.79 + 5.0 A sin 78.46 = 21.07 A + 4.899 A = 25.97 A


The vertical components add up as well.  Cos f is the power factor, so we can write:


IV = 40 A × 0.850 + 5.0 A × 0.20 = 34 A + 1.0 A = 35 A


Now we can do the vector sum:


I12 = (25.97 A)2 + (35 A)2 = 1899 A2


I1 = 43.6 A = 44 A (to 2 s.f.)


Now we can work out the power factor for I1:


Cos f1 = 35 A ÷ 43.6 A = 0.8031 = 0.80 (to 2 s.f.)


(Just for the sake of completeness f1 = 36.58o, lagging the voltage.)


Question 9

A single phase transformer has 2400 turns on the primary and 600 turns on the secondary.  Its no load current is 4.0 A with a power factor of 0.25 lagging.  Assuming that the voltage drop in the windings is negligible, calculate the primary current and power factor when the secondary current is 80 A with a lagging power factor of 0.80.




Transformer Matching

Transformers are normally associated with stepping up and stepping down commercial mains electricity.  However they also have an important use in impedance matching.  In my notes on Electrical Principles, we saw that for maximum power transfer from a Thévenin source that the load matched the internal resistance.  Click HERE if you want to revise it.


For an alternating current, reactance is important as a result of the inductance of coils of wire.  The load may have a certain impedance.


For maximum power transfer in an AC circuit, the impedance of the load must match the internal impedance of the supply.   We can achieve this with a transformer, and the turns ratio (N1 ÷ N2) is given by:

With stereo radio receivers, it is essential to use a decent aerial.  If there isn’t one, the radio won’t pick up anything, or at best it will be very hissy, which is not pleasant to listen to.  There are two kinds of aerial available:

·         A balanced ribbon aerial;

·         An unbalanced antenna connected by a coaxial cable.


The input impedance for a balanced aerial is 300 ohms, while that for an unbalanced aerial is 75 ohms.  If you connect a ribbon aerial to a 75 ohm socket, the signal will be weak.  You need a matching transformer called a balun (from balanced-unbalanced).


We can work out the turns ratio, knowing that the source resistance is 300 ohms and the load (input impedance) is 75 ohms:

The schematic is like this:


We know that for an ideal transformer, the voltage is related by:


We also know that current is related by:

Use Ohm’s Law to get the effective resistance, R', in the primary coil:

If we have a load resistance, R¸ we know that it is given by:

So we can write:

In words, we can rewrite this as:

So we can make R' equal to the internal resistance of the source.


Impedance matching in an audio system

There are many incidences of where a source might not be properly matched with its load.  Consider an audio amplifier that can drive a 2 ohm load to 50 W.  We can work out the current easily enough.


Question 10

(a) Work out the current in the amplifier above.
(b) Show that the voltage is 10 V
(c) Show that the power transfer is 32 W for an 8 ohm load.



Most loudspeakers are 8 ohms.  So if we connect the supply directly to the loudspeaker, we get only 32 W.  You can see that the possible power transfer is a lot less.


So we need to increase the voltage to the loudspeaker so that it transfers 50 W of power.  We know that:

So rearranging:

V2 = PR = 50 W × 8 W = 400 V2


V = 20 V


So we need to increase the voltage from 10 V to 20 V using a 1:2 turns ratio.  So if the primary has 1000 turns, the secondary has 2000 turns.  We can show the consistency of the result using the equation:

In other words we can say that the primary has half the number of turns as the secondary.


Question 11

A transformer is connected to a 240 V AC supply to supply 12 V to a low voltage lighting system for a village on a model railway. The total resistance of the system is 2.0 .

(a) What is the ratio of turns N1:N2?
(b) What is the current in the secondary?
(c) What power is delivered to the load?
(d) What resistance, when connected to the 240 V supply will give the same power as the transformer?
(e) Show that the ratio of the source resistance to the load is the same as the square of the turns ratio.





Tutorial 10 A

Tutorial 10 C

Tutorial 10 D