The energization transient creates a decaying dc offset in the excitation current. When the excitation current peaks in the same direction as the dc offset (for instance all positive peaks), the transformer is far into saturation and the exciting current is ineffective... must go higher to match the applied voltage. Maybe easier to explain with equaitons:
Vapplied = N1* d/dt(Phi)
= N1 * dPhi/di * di/dt.
When core enters saturation, we are on the saturated (flat) portion of Phi vs i curve where dPhi/di is very low. di/dt must become very large to compensate.
A few interesting notes on transformer inrush based on previous post.
Because inrush is dependant on where on the waveform the transformer is switched, it will be different every time. As previously stated, highest if you switch in at voltage peak.
Because three phases aren't at the same voltage the inrush will be different in the three phases.
If the transformer is energized at the zero crossing there will be no inrush in that phase.
I haven't looked into it, but I beleive there are devices out there now that switch in the three phases seperatly at zero crossings for no inrush.
To illustrate the inrush effect, consider a single-phase transformer energized at t=0 with an ideal voltage source. Therefore, when the transformer is energized, the transient voltage generated is e(t)= d M(t)/dt.
Where: e(t) = source voltage; M(t) = magnetic flux linkage of the transformer.
For initial cond @ t = 0; M(0) = a.
Voltage Source : e(t)= Vm. Cos(wt+Q).
The solution is: M(t)= (a-Vm/w. sinQ) + Vm/w.sin(wt+Q).
The first term is the dc bias. For steady state, Q=0 the inrush current is 1% of rated current, while for Q=pi/2 it may reach a peak of 800 % of rated current or higher. The inherent system resistance forces the inrush current to decay slowly.
