WAZIPOINT Engineering Science & Technology: Transformer on No-load

## Thursday, November 10, 2022

The answer to the question lies in one of the assumptions made for an ideal transformer. The core of an ideal transformer is supposed to have infinite permeability. So, no MMF is needed to produce working flux. But, in a practical transformer, the permeability is not infinite so we need a certain current for a certain amount of MMF which will produce the required working flux.

No load current has two components (a)Iron loss component and (b) Magnetizing Component.

Iron loss component: It is the component of no load current responsible for resistive loss in the core…

Magnetizing Component: It is the component of no load current responsible for hysteresis loss in the core.

Hench, the primary current I0 is vector summation of Iµ & Iw, So, we can write that I0 = ( Iµ 2 +  Iw2) and is not a 90⁰ behind V1, but lags it by an angle φ < 90⁰ Which is shown in the figure. And no-load input power, W0 = V1 I0Cose φ0.

The magnitude of the no-load primary current is very small as compared to the full-load primary current.

It is 1% of the full-load current. As I0 is very small, the no-load primary Cu loss is negligible which means that no-load primary input is practically equal to the iron loss in the transformer.

The no-load loss of a transformer arises at its core, a part that experiences lower heating than the transformer windings depending on the quality of lamination and the thickness and resistance of the core.

In the case of no load, the second terminal of the transformer is open means the circuit is not complete on the secondary side. This situation clearly indicates that no path is available for the current to flow on the secondary side. And if there is no current flowing in the secondary side, there is no de-magnetizing flux generated which means there is no need to draw more current from the source. So primary current would contain only the exciting current (i.e. no-load current)

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