# Practical Transformer on No Load

- Link: https://studyelectrical.com/2018/08/practical-transformer-on-no-load.html
- Published: 2018-08-16T07:23:00+00:00
- Author: 

Please read [Ideal Transformer and Practical Transformer](https://studyelectrical.com/2018/06/01/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.

[⌊Practical%2Btransformer%2Bon%2Bno%2Bload⌉⌊Practical%2Btransformer%2Bon%2Bno%2Bload⌉[

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.

[⌊Phasor%2Bdiagram%2B Practical%2Btransformer%2Bon%2Bno%2Bload⌉⌊Phasor%2Bdiagram%
2B Practical%2Btransformer%2Bon%2Bno%2Bload⌉[

 1. I_(w) 
 2. 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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