Electrical Engineering ⇒ Topic : Ideal Transformer

Luis
 
When the primary and secondary winding resistance drops are neglected, the secondary induced voltage E_{2} will be equal to its terminal voltage V_{2}.Likewise, on the primary side, the induced emf E_{1} will be equal to the applied voltage V_{1}. This is an ideal transformer. However, V_{1} and E_{1} 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 Φ_{2}. Then the load current in the secondary /_{2} will lag the secondary voltage V_{2} by an angle Φ_{2}. /_{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 noload current /_{0} (discussed in the previous section) also flows in the primary (see Fig (A)). The primary current I_{1} will
FIGURE (A) Phasor diagram of an ideal transformer on load now be the phasor sum of I_{0} and I_{2}^{'}. The power factor of the primary side will be cos Φ_{1} where Φ_{1} is the phase angle difference between V_{1} and I_{1}.However, as far as the magnetisation of the core is considered,the magnetising current(I_{0}) is same on load as on noload.  
 
Gaurav
 
IDEAL TRANSFORMER 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 V_{1} and the secondary is connected to a load 4. The voltage across the load is V_{2}. Since the ideal transformer has zero primary and zero secondary impedance, the voltage induced in the primary, E_{1} is equal to the applied voltage V_{1} and the output voltage V_{2} is equal to the secondary induced voltage E_{2}. Here E_{1} is called counter e.m.f. or back e.m.f. of the primary and E_{2} is called the mutually induced e.m.f. in the secondary. Both are in antiphase with V_{1}. Figure (b) shows the phasor representation of an ideal transformer .............. (1) The current in the primary is just sufficient to provide the required m.m.f. I_{1}N_{1} to overcome the demagnetizing effect of secondary m.m.f. I_{2}N_{2 }which is a result of connected load ................ (2) From Eqs. (1) and (2), it can be written as: ............... (3) 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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