Page 110 - Electric Machinery Fundamentals

15() bLl:.Cll<IC MACHINbRY FUNUAMbNTALS

3. The leakage flux in the core must be zero, implying that all the flux in the
core couples both windings.
4. The resistance of the transformer windings must be zero.
While these conditions are never exactly met, well-designed power transformers
can come quite close.

2.5 THE EQUIVALENT CIRCUIT OF
A TRANSFORMER
The losses that occur in real transformers have to be accounted for in any accurate
model of transformer behavior. The major items to be considered in the construc-
tion of such a model are

1. Copper (PR) losses. Copper losses are the resistive heating losses in the pri-
mary and secondary windings of the transformer. They are proportional to the
square of the current in the windings.
2. Eddy current losses. Eddy current losses are resistive heating losses in the
core of the transfOlmer. They are proportional to the square of the voltage ap-
plied to the transformer.
3. Hysteresis Losses. Hysteresis losses are associated with the rearrangement of the
magnetic domains in the core during each half-cycle, as explained in Chapter I.
They are a complex, nonlinear function of the voltage applied to the transformer.
4. Leakage flux. The fluxes 4>LP and 4>LS which escape the core and pass through
only one of the transformer windings are leakage fluxes. These escaped
fluxes produce a leakage inductance in the primary and secondary coils, and
the effects of this inductance must be accounted for.
The Exact Equivalent Circuit of a
Real Transformer
It is possible to construct an equivalent circuit that takes into account all the ma-
jor imperfections in real transformers. Each major impeifection is considered in
turn, and its effect is included in the transformer model.
The easiest effect to model is the copper losses. Copper losses are resistive
losses in the primary and secondary windings of the transformer core. They are
modeled by placing a resistor Rp in the primary circuit of the transformer and a re-
sistor Rs in the secondary circuit.
As explained in Section 2.4, the leakage flux in the primary windings CPLP
produces a voltage 4>LP given by
d4>cp
eLP(t) = Np dt (2- 36a)

and the leakage flux in the secondat)' windings CPLS produces a voltage eLS given by
d4>cs
eLs(t) = NSdt (2-36b)