# Transformers & Total Mutual Flux

Discussion in 'General Electronics Chat' started by shespuzzling, Mar 22, 2010.

1. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
88
1
My question is about an ideal transformer under load. I understand transformer operation at no-load. The flux is what it is due to the magnetizing current flowng through the primary, and that flux creates an induced emf across the secondary. When a load is connected and current now flows in the secondary is where I get messed up. Most books tell you that the mmf due to the current flowing in the secondary causes and equal and opposite mmf in the primary and that these two buck each other (I'm assuming that means they totally cancel each other out so that the total flux through the core is still what it was when the secondary was an open circuit). However, using the equation:

mmf=(Phi)*Reluctance

You get Phi=mmf/Reluctance and this equals NIA(mu)/length

In order for phi due to current 2 to be equal in magnitude with phi due to current 1, then, since mu is the same, NIA/length would have to be equal for both the secondary and the primary, not just NI. Unless of course the area and length are the same for both primary and secondary, but I don't think that is always the case.