Please read Ideal Transformer and Practical Transformer before continuing.

Consider a practical transformer on no load i.e., secondary on open-circuit as shown in figure below.

Consider a practical transformer on no load i.e., secondary on open-circuit as shown in figure below.

The primary will draw a small current I_{0} to supply

(i) the iron losses and

(ii) a very small amount of copper loss in the primary.

Hence the primary no load current I_{0} is not 90° behind the applied voltage V_{1} but lags it by an angle Φ_{0} < 90° as shown in the phasor diagram.

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

As seen from the phasor diagram, the no-load primary current I_{0 }can be resolved into two rectangular components viz.

- I
_{w} - I
_{m}

(i) The component I

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

_{w}= I_{0}cosΦ_{0}(ii) The component I

_{m}lagging behind V_{1}by 90° and is known as magnetizing component. It is this component which produces the mutual flux Φ in the core.I

_{m}= I_{0}sinΦ_{0}Clearly, I

_{0}is phasor sum of I_{m}and I_{w}.I

_{0}= √ (I_{m}^{2}+I_{w}^{2})No load power factor, cosΦ

_{0}= I_{w}∕I_{0}It is emphasized here that no load primary copper loss (i.e. I

_{0}^{2}R_{1}) is very small and may be neglected. Therefore, the no load primary input power is practically equal to the iron loss in the transformer i.e.,No load input power, W

_{0}= Iron loss*.*

__Note__At no load, there is no current in the secondary so that V

_{2}= E_{2}.On the primary side, the drops in R

_{1}and X_{1}, due to I_{0}are also very small because of the smallness of I_{0}.Hence, we can say that at no load, V

_{1}= E_{1.}
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