The transformer is said to be at no load when the primary of a transformer is connected to the source of ac supply and the secondary is open.

We consider an ideal transformer whose secondary side is open and the primary winding is connected to a sinusoidal alternating voltage V1. Due to the alternating voltage applied to the primary winding, the alternating current will flow in the primary winding. since the primary coil is purely inductive and the secondary is open, the primary draw the magnetizing current Im only. This current magnetizes the core of the transformer. The magnitude of Im will be zero if the transformer is truly ideal. The magnetizing current Im is small in magnitude and lags behind supply voltage V1 by 90º. The magnetizing current Im produces an alternating flux Φ which is proportional to the current and in phase with it.

Le the instantaneous linking flux ,

The instantaneous value of induced emfs in the primary and secondary windings will be,

We assumed that zero ohmic resistance in the primary winding. Thus the applied voltage to the primary winding opposes the induced emf in the primary winding. So, instantaneous applied voltage to primary is,

**No-load phasor Diagram for an ideal Transformer**

From the phasor diagram it is concluded that:

- Induces emfs in primary and secondary windings, E1 and E2 lag behind the main flux Φ by π/2, so these emfs E1 and E2 are in phase with each other.
- Applied voltage to the primary winding leads the main flux by π/2 and is in phase opposition to the induced emf in the primary winding.
- Secondary voltage V2=E2 as there is no voltage drop in secondary.

There will be power loss (iron or core loss) when a varying flux is set up in magnetic material. So the input current to the primary under no load condition has to supply the Hysteresis losses and eddy current losses (iron losses) occurring in the core in addition to a small amount of copper loss occurring in the primary winding. Hence the no-load primary current Io does not lag behind applied voltage V1 by 90° but lags behind V1 by angle Φo<90°.

**Input power on no load, Po=V1IocosΦo**

where cosΦo is the primary power factor under no load condition.

**Io**=input current to the primary i.e **exciting current**

**Ie**=**active or energy component**(used to meet the iron loss in addition to small amount of copper loss occurring in the primary winding). **Range of Ie is from 2 to 5 % of full load primary current.**

**Im**=quadrature component or wattless component i.e** magnetizing component**(used to create the alternating flux in the core)

**Equivalent circuit of Transformer on no load**

R0=Resistance

X0=inductive reactance

Energy component of no-load current,Ie=V1/R0

Magnetizing component of no-load current,Im=V1/X0

Soo useful