# Practical transformer on no load

## 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 R_{1} and secondary resistance R_{2} 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 E_{1} < V_{1} and V_{2} < E_{2}.

## Leakage reactance in transformer

Common flux Φ links both primary and secondary side. Primary current I_{1} produces individual flux Φ_{1} in primary side and secondary current I_{2} produces flux Φ_{2} in secondary side, those two fluxes is not common in both sides. Here flux Φ_{1} and flux Φ_{2} 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 Φ_{1 }makes an inductive reactance X_{1} series in primary winding and secondary leakage flux Φ_{2} introduces an inductive reactance X_{2} in series with the secondary winding shown in figure 1. Primary leakage flux Φ_{1} induces back e.m.f. e in the primary winding and it is alternating one.

Primary leakage inductance, L_{1} = primary leakage flux linkages / primary current = N_{1} Φ_{1} / I_{1}

Primary leakage reactance, X_{1} = *2πf* L_{1}

In the same way secondary leakage inductance, L_{2} = N_{2} Φ_{2} / I_{2}

Secondary leakage reactance, X_{2} = *2πf* L_{2}

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.

## No load practical transformer on

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 I_{0} 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 I_{0} is not 90˚ behind the applied voltage V_{1} but lags it by angle Φ_{0} < 90˚.

No load input power, W_{0} = V_{1} I_{0} cos Φ_{0}

In primary side for I_{0} we get two components I_{w} and I_{m}.

- The component I
_{m}is known as magnetizing component. This component produces mutual flux Φ in the core. I_{m}lagging behind V_{1}by 90˚.

I_{m} = I_{0} sin Φ_{0}

- The component I
_{w}is known as iron loss or active or working component. It is in phase with the applied voltage V_{1}. It supplies a very small primary copper loss and iron loss.

I_{w} = I_{0} cos Φ_{0}

It is clear that I_{0} is the phasor sum of I_{m} and I_{w}.

I_{0} = √(I_{m}^{2} + I_{w}^{2})

No load power factor, cos Φ_{0} = I_{w} / I_{0}

At no load practical transformer primary copper loss *I _{0}^{2}R *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, *W _{0} *= Iron loss

As primary loss in practical transformer is quite small so it can be written at no load, V_{1} = E_{1}.

There is no load in secondary so E_{2} = V_{2}.

## Practical transformer on no load phasor diagram

Figure 4 is the phasor of practical transformer on no load. Primary small current I_{0} is phasor sum of I_{m} and I_{w}.

Primary no load current I_{0} lags by V_{1} an angle Φ_{0} < 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. **

Answer:

No load primary power, W_{0} = V_{1} I_{0} cosΦ_{0}

Iron loss component, I_{w} = I_{0} cosΦ_{0} = W_{0} / V_{1} = 128/400 = 0.32 A

No load current, I_{0} = √(I_{m}^{2} + I_{w}^{2})

Magnetizing current, I_{m} = √(I_{0}^{2} – I_{w}^{2}) = √{(1.2)^{2 }– (0.32)^{2}} = 1.16 A