autotransformer cours and solved exercise

autotransformer cours

Consider a single autotransformer winding having N1 turns, mounted on an iron core  The winding is connected to a fixed-voltage AC source E1, and the resulting exciting current I0 creates an ac flux Φm in the core. As in any transformer, the peak value of the flux is fixed so long as E1 is fixed .
Suppose a tap C is taken off the winding, so that there are N2 turns between terminals A and C. Because the induced voltage between these terminals is proportional to the number of turns, E2 is given by

E2 = (N2/N1) . E1

Clearly, this simple coil resembles a transformer having a primary voltage Ex and a secondary voltage E2. However, the primary terminals B, A and the secondary terminals C, A are no longer isolated from each other, because of the common terminal A.
If we connect a load to secondary terminals CA, the resulting current I2 immediately causes a primary current I2 to flow .

The BC portion of the winding obviously carries current I1. Therefore, according to Kirchhoff’s current law, the CA portion carries a current (I2 - I1). Furthermore, the mmf due to I1 must be equal and opposite to the mmf produced by (I2 - I1). As a result, we have:

I1(N1 - N2) = (I2- I1).N2

which reduces to :

I1N1=I2N2

Finally, assuming that both the transformer losses and exciting current are negligible, the apparent power drawn by the load must equal the apparent power supplied by the source. Consequently,

E1I1=E2I2

These Equations are identical to those of a standard transformer having a turns ratio Nt/N2. However, in this autotransformer the secondary winding is actually part of the primary winding. In effect, an autotransformer eliminates the need for a separate secondary winding. As a result, autotrans-formers are always smaller, lighter, and cheaper than standard transformers of equal power output. The difference in size becomes particularly important when the ratio of transformation E1E2 lies between 0.5 and 2. On the other hand, the absence of electri¬cal isolation between the primary and secondary windings is a serious drawback in some applications.
Autotransformers are used to start induction motors, to regulate the voltage of transmission lines, and, in general, to transform voltages when the primary to secondary ratio is close to 1.

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Solved exercise : Autotransformer

3.6 kW load is connected across the secondary of an autotransformer , calculate :

1. The secondary voltage and current
2. The currents that flow in the winding
3. The relative size of the conductors on windings BC and CA

Solution

1. The secondary voltage is :
   
   E2 = 80% X 300 = 240 V 
   The secondary current is
   
   I2 = P/E2 = 3600/240 = 15 A 
2. The current supplied by the source is:
   
   I1 = P/E1 = 3600/300 = 12 A
   
   
   
   

   the current in winding   BC = 12 A
   the current in winding   CA = 15 - 12 = 3 A 
3. The conductors in the secondary winding CA can be one-quarter the size of those in winding BC because the current is 4 times smaller . However, the voltage across winding BC is equal to the difference be¬tween the primary and secondary voltages,
   namely (300-240) = 60 V. Consequently,
   winding CA has four times as many turns as winding BC. Thus, the two windings require essentially the same amount of copper.

verification en cours . . .