## An ideal Transformer

Table of Contents

**An ideal Transformer**

**Ideal Transformer Construction**

**AnÂ ideal TransformerÂ is an imaginary Transformer. An ideal transformer is one which has the following Characteristics** âˆ’

- The primary and secondary windings have negligible (or zero) resistance.
- No leakage flux, i.e., whole of the flux is confined to the magnetic circuit.
- The magnetic core has infinite permeability, thus negligible mmf is require to establish flux in the core.
- There are no losses due winding resistances, hysteresis and eddy currents. Hence, the efficiency is 100 %.

### Working of Ideal Transformer

#### Ideal Transformer on No-Load

Consider an ideal transformer on no-load, i.e., its secondary winding is open circuited (see the figure). Thus, the primary winding is a coil of pure inductance.

A very small magnetising current Im is drawn to create the flux in the core, which lags the applied voltage by 90Â° when an alternating voltage V1 is applied to the primary winding. An alternating flux m that is proportional to and in phase with the magnetising current Im is created. The primary and secondary windings are magnetically connected by this alternating flux (m), which also induces EMF E1 in the primary winding and EMF E2 in the secondary winding.

The EMF induced in the primary winding E_{1}Â is equal to and in opposition to the applied voltage V_{1}Â (according to Lenzâ€™s law). The EMFs E_{1}Â and E_{2}Â lag behind the flux (Ï•_{m}) by 90Â°,. Although their magnitudes depend upon the number of turns in the primary and the secondary windings. From the phasor diagram of the ideal transformer on no-load, it is clear that the flux is common to both the windings, hence it can be taken as the reference phasor. Also, the EMFs E_{1}Â and E_{2}Â are in phase with each other, but E_{1}Â is equal to V_{1}Â and 180Â° out of phase with it.

#### Ideal Transformer On-Load

When load is connected across the terminals of secondary winding of the ideal transformer,. The transformer is said to be loaded and a load current flows through the secondary winding and the load.

Consider an inductive load of impedance Z_{L} is connected across the secondary winding of the ideal transformer (see the figure). Then, the secondary EMF E_{2} will cause a current I_{2} to flow through the secondary winding and the load, which is given by,

**I2=E2ZL=V2ZLI2=E2ZL=V2ZL**

Since, for an ideal Transformer, the EMF E_{2}Â is equal to Secondary terminal voltage V_{2}.

##### Here, the load is Inductive, therefore, the current I_{2}Â will lag behind the E_{2}Â or V_{2}Â by an angle Ï•_{2}. Also, the no-load current I_{0}Â being neglected because the Transformer is ideal one.

The Secondary winding’s current (I2) creates a mmf (N2I2). That generates a flux (2) that is in the opposite direction of the main flux (m). As a result, the core’s total flux deviates from its initial value, but it shouldn’t deviate from its original value. Therefore, the primary current must create an MMF. That can balance out the demagnetizing effect of the secondary mmf N2I2 in order to keep the flux in the core at its initial value. In order to achieve this, the primary current I1 must flow.

**N1I1=N2I2N1I1=N2I2**

**â‡’I1=N2N1Ã—I2=KI2â‡’I1=N2N1Ã—I2=KI2**

As a result, the primary winding needs to draw enough current to counteract the Demagnetizing effects of the Secondary current and maintain a constant main flux in the core. In order to maintain a constant mutual flux (m),. When the secondary current (I2) increases, the primary current (I1) also increases in the same way.

It is clear from the phasor diagram of the ideal Transformer on-load. That the Secondary current I_{2}Â lags behind the Secondary Terminal voltage V_{2}Â by an angle of Ï•_{2}.