# Ideal Transformer

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  T1 turns and is connected to a voltage source of V1 volts. The secondary has T2 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 I0 amps is passed through the primary winding of T1 turns it sets up an MMF of I0T1 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 I0 is enough to set up a flux which is finite. As I0 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,

v1 – e1  = 0    or v1 = e1

Where v1 = instantaneous value of the applied voltage

e1 = 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   v1 =  V1peak Cos ωt      where V1peak is the peak value  and ω = 2πf , f= frequency of supply

As v1 = e1 (From above equation) where e1 = E1peak Cosωt

we can say that E1 = V1

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 ZL 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  ZL 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 E1 and E2 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.