Electrical Engineering ⇒ Topic : Ideal Transformer
When the primary and secondary winding resistance drops are neglected, the secondary induced voltage E2 will be equal to its terminal voltage V2.Likewise, on the primary side, the induced emf E1 will be equal to the applied voltage V1. This is an ideal transformer. However, V1 and E1 will oppose each other. The phasor diagram shown in Figure (a) shows these details along with the load currents
Consider a load on the transformer secondary such that it gives a lagging power factor Φ2. Then the load current in the secondary /2 will lag the secondary voltage V2 by an angle Φ2. /2 in the secondary causes a current I to flow in the primary such that neutralises the demagnetising effect of /2. In addition to /2', the no-load current /0 (discussed in the previous section) also flows in the primary (see Fig (A)). The primary current I1 will
FIGURE (A) Phasor diagram of an ideal transformer on load
now be the phasor sum of I0 and I2'. The power factor of the primary side will be cos Φ1 where Φ1 is the phase angle difference between V1 and I1.However, as far as the magnetisation of the core is considered,the magnetising current(I0) is same on load as on noload.
It is an imaginary transformer having the following properties:
The resistance of primary and secondary windings is negligible.
There is no leakage flux and no leakage inductance.
The core has infinite permeability so that the m.m.f. required to establish the flux in -is negligible.
The entire flux is confined to the core.
There are no losses, i.e., no ohmic loss, hysteresis, and eddy current losses. Therefore,the efficiency in this case is 100 percent.
The practical transformers do not satisfy the above conditions.
An ideal transformer is shown in Figure (a). The voltage applied to coil 1 is V1 and the secondary is connected to a load 4. The voltage across the load is V2. Since the ideal transformer has zero primary and zero secondary impedance, the voltage induced in the primary, E1 is equal to the applied voltage V1 and the output voltage V2 is equal to the secondary induced voltage E2. Here E1 is called counter e.m.f. or back e.m.f. of the primary and E2 is called the mutually induced e.m.f. in the secondary. Both are in anti-phase with V1. Figure (b) shows the phasor representation of an ideal transformer
The current in the primary is just sufficient to provide the required m.m.f. I1N1 to overcome the demagnetizing effect of secondary m.m.f. I2N2 which is a result of connected load
From Eqs. (1) and (2), it can be written as:
Thus, in an ideal transformer input kVA and output kVA are identical.
For ideal transformer, there is no copper loss and core loss. But for practical transformer when secondary is put on load, there is iron loss in core and copper loss in winding and these losses are not negligible.
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