A real transformer

[ahmeeeeeeeeeed]

HELLO

 

In a real transformer , Why Ns Is= Np Ip ???

 

Please explain simply

Edited April 12, 2011 by ahmeeeeeeeeeed

[TonyMcC]

You give a formula appropriate to an ideal (not real or practical) transformer.

In a real transformer Ns.Is will be less than Np.Ip

[ahmeeeeeeeeeed]

Ok may be not so real >> but I mean a transformer where the P(out) Is less than P (in)

 

like http://www.scribd.com/doc/31506096/Transformer

[Danijel Gorupec]

Hmm... I am not sure what exactly you are considering the real transformer (or ideal transformer), but...

 

Your equation Ns Is = Np Ip cannot be generally true. Imagine a transformer that has its primary connected to AC voltage source while secondary is open. The primary current will be non-zero, while secondary current will be zero. Your equation will be very wrong. (Note that in this example the primary current is in near 90 degree phase difference to primary voltage and therefore no real power is used by the transformer.)

 

Your equation will be very close to true if the transformer is used as current transformer: Its primary winding is excited by AC current (not AC voltage), while secondary winding is almost short-circuited (that is, AC voltage over the secondary winding is near zero).

 

(Because of secondary current, the short-circuited secondary winding will create flux on its own. This flux will exactly opposite the flux created by primary winding. As flux created by any windings is proportional to I and N, there is your equation.... Why the secondary winding wants to exactly oppose the primary flux? The secondary winding just don't like flux changes - any flux change would create EMF (voltage) and this voltage instantly increases current until the flux change is exactly opposed. Because it is short-circuited, there is nothing else to 'absorb' the EMF.)