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Practical transformer

Practical transformer is different from ideal transformer in many respects. Ideal transformer has no losses but practical transformer have

• Iron losses
• Magnetic leakage
• Winding resistances

Iron losses in transformer

Alternating flux Φ passes through iron core. It creates eddy current and hysteresis loss in it. Two of these losses called iron loss or core loss. Iron loss depends upon the core volume, supply frequency, maximum flux density etc. Magnitude of iron loss is very small in practical transformer.

Winding resistances in transformer

Primary and secondary windings made with copper wire. Every conductor has own resistance. So primary and secondary side have resistance. Primary resistance R1 and secondary resistance R2 are in series with the respective windings. Figure 1 shows the windings resistances both sides. Due to winding resistance when current passes through the windings there will voltage drop IR and occurs power loss. It effects in power factor. Hence E1 < V1 and V2 < E2.

Leakage reactance in transformer

Common flux Φ links both primary and secondary side. Primary current I1 produces individual flux Φ1 in primary side and secondary current I2 produces flux Φ2 in secondary side, those two fluxes is not common in both sides. Here flux Φ1 and flux Φ2 are known as leakage flux in transformer. Figure 2 shows leakage fluxes.

The leakage flux path is through the air mainly. The effect of primary leakage flux Φ1 makes an inductive reactance X1 series in primary winding and secondary leakage flux Φ2 introduces an inductive reactance X­2 in series with the secondary winding shown in figure 1. Primary leakage flux Φ1 induces back e.m.f. e in the primary winding and it is alternating one.

Primary leakage inductance, L1 = primary leakage flux linkages / primary current = N1 Φ1 / I1

Primary leakage reactance, X1 = 2πf L1

In the same way secondary leakage inductance, L2 = N2 Φ2 / I2

Secondary leakage reactance, X2 = 2πf L2

No power loss occurs due to leakage reactance. But it changes the power factor as well as there is voltage loss due to IX drop. Flux leakage is quite small about 5% of mutual flux Φ in a transformer. Yet it can not be avoided or ignored.

A practical transformer diagram is shown in figure 3, there is no load in secondary side it is open circuited. When ac source is connected in primary a small current I0 flows through the primary. It occurs a very small amount of copper loss and iron loss in the primary. So that the primary no load current I0 is not 90˚ behind the applied voltage V1 but lags it by angle Φ0 < 90˚.

No load input power, W0 = V1 I0 cos Φ0

In primary side for I0 we get two components Iw and Im.

• The component Im is known as magnetizing component. This component produces mutual flux Φ in the core. Im lagging behind V1 by 90˚.

Im = I0 sin Φ0

• The component Iw is known as iron loss or active or working component. It is in phase with the applied voltage V1. It supplies a very small primary copper loss and iron loss.

Iw = I0 cos Φ0

It is clear that I0 is the phasor sum of Im and Iw.

I0 = √(Im2 + Iw2)

No load power factor, cos Φ0 = Iw / I0

At no load practical transformer primary copper loss I02R is very small and this loss may be neglected. Hence, primary no load input power of practical transformer is equal to the iron loss.

No load input power, W0 = Iron loss

As primary loss in practical transformer is quite small so it can be written at no load, V1 = E1.

There is no load in secondary so E2 = V2.

Practical transformer on no load phasor diagram

Figure 4 is the phasor of practical transformer on no load. Primary small current I0 is phasor sum of Im and Iw.

Primary no load current I0 lags by V1 an angle Φ0 < 90˚.

We can find the magnetizing and iron loss current using above equations now we will solve a math.

Math

A transformer takes a current 1.2 A and absorbs 128 W when primary is connected to its normal supply of 400V, 50 Hz, the secondary being on open circuit. Find the magnetizing and iron loss currents.

No load primary power, W = V1 I0 cosΦ0

Iron loss component, Iw = I0 cosΦ0 = W0 / V1 = 128/400 = 0.32 A

No load current, I0 = √(Im2 + Iw2)

Magnetizing current, Im =  √(I02 – Iw2) = √{(1.2)2 – (0.32)2} = 1.16 A

Md Ebrahim Shah

Electrical Engineer, teacher, professional article writer and founder of engineersblogsite.com. His passion is to describe electrical technology at easy way as people can get clear conception and upgrade the technology for the better human life.

rahul kumar - 28/11/2014

Thanks for help me in my study