To understanding of the behavior of the transformer, initially certain idealizations are made and the resulting **ideal transformer** is studied. The magnetic core to carry the flux introduced losses. The losses due to hysteresis and eddy current for the available grades of steel are very small at power frequencies. Also the copper losses in the winding due to magnetization current is reduced to an almost insignificant fraction of the full load losses.

**Property of Ideal Transformer**

- The magnetic circuit is linear and has infinite permeability.
- Hysteresis loss is negligible.
- There is no leakage flux.
- Windings do not have resistance.
- There are no copper losses, nor there is any ohmic drop in the electric circuit.

## Working of Ideal Transformer

The primary winding has T_{1} turns and is connected to a voltage source of V_{1} volts. The secondary has T_{2} turns. Secondary can be connected to a load impedance for loading the transformer. The primary and secondary are shown on the same limb and separately.

As a current I_{0} amps is passed through the primary winding of T_{1} turns it sets up an MMF of I_{0}T_{1} ampere which is in turn sets up a flux Φ through the core. Since the** reluctance **of the iron path given by **R = l/μ **is zero as** μ = infinity. ** A small value of current I_{0} is enough to set up a flux which is finite. As I_{0} establishes the field inside the transformer it is called the magnetizing current of the transformer.

Current is produced due to the sinusoidal voltage V applied to the primary. As the current through the loop is zero, at every instant of time, the sum of the voltages must be zero inside the same. Writing this in terms of instantaneous values we have,

v_{1} – e_{1 } = 0 or v_{1 }= e_{1}

Where v_{1} = instantaneous value of the applied voltage

e_{1} = induced emf due to Faradays principle

Negative sign is due to the application of the **Lenz’s law **and shows that it is in the form of a voltage drop.

Let v_{1 }= V_{1peak }Cos ωt where V_{1peak} is the peak value and ω = 2πf , f= frequency of supply

As v1 = e1 (From above equation) where e_{1 }= E_{1peak} Cosωt

we can say that E_{1} = V_{1}

The RMS primary induced emf is

The same mutual flux links the secondary winding.The induced emf in the secondary can be similarly obtained as

**Voltage ratio** as

Let the **transformer** is connected to a load impedance Z_{L} is connected across the terminals of the secondary winding a load current flows. This load current produces a demagnetizing MMF and the flux tends to collapse.

However this is detected by the primary immediately as both E1 and E2 tend to collapse. The current drawn from supply increases up to a point the flux in the core is restored back to its original value. The demagnetizing MMF produced by the secondary is neutralized by additional magnetizing MMF produces by the primary leaving the MMF and flux in the core as in the case of no-load. Thus the transformer operates under constant induced emf mode.

If the reference directions for the two currents are chosen as in the Fig then the above equation can be written in phasor form as,

Thus voltage and current transformation ratio are inverse of one another. If an impedance of Z_{L} is connected across the secondary,

The input impedance under such conditions is

Transformer thus acts as an impedance converter. The transformer can be interposed in between a source and a load to match the impedance.

## Phasor Diagram of Ideal Transformer

The voltages E_{1} and E_{2} are obtained by the same mutual flux and hence they are in phase. If the winding sense is opposite i.e., if the primary is wound in clockwise sense and the secondary counter-clockwise sense then if the top terminal of the first winding is at maximum potential the bottom terminal of the second winding would be at the peak potential. A similar problem arises even when the sense of winding is kept the same, but the two windings are on opposite limbs.