Inrush current in a transformer chain

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SUMMARY

The discussion centers on the inrush current behavior in a transformer chain, specifically regarding XFMR1 and XFMR2 with secondary voltages of 110V. It is established that the inrush current of XFMR1, denoted as xA, may decrease due to increased leakage inductance when the secondary of XFMR2 is open. Factors influencing inrush current include load circuitry, transformer core saturation, and winding capacitance, all of which complicate predictions. A detailed model is necessary to accurately assess inrush current, but under specific conditions, such as a demagnetized core, no inrush current is expected in the secondary.

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Guineafowl
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TL;DR
How does connecting two transformers in a chain affect the inrush current (at switch-on), as opposed to one transformer on its own?
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This is the proposal, except the two middle voltages are 110V, not 12V.

Say the max switch-on inrush current of XFMR1 is xA with the secondary open, would the value change significantly in the above arrangement, with the secondary of XFMR2 open?
 
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You have twice the leakage inductance, so the inrush may be a little lower.
 
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berkeman said:
You have twice the leakage inductance, so the inrush may be a little lower.
Thanks.
 
Hard to answer without knowing what causes the inrush current.

- If it's from the load circuitry, the I'm with @berkeman, more series impedance buffer means less inrush magnitude, but longer duration (probably).

- If it's from the transformer core saturation because of residual magnetization, then it's just not predictable, at least to me. But, the same or maybe more.

- If it's from winding capacitance then you get an addition of some sort. Complicated by turns ratio and leakage inductance.

The short answer is that you need a much more detailed model, which is pretty difficult. As you drew it, I'd say there's no inrush surge in any case.
 
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If the magnetic core was in demagnetized position when it was de-energized

then up to 100 A/m magnetic field the flux density is about 0 so E1=0 and E2=0 and I2=0 since the second loop E=0.

So, no inrush current is expected in the secondary.

0=U-(E+Z*Iinrush)

U is the supply voltage, E it is the electromotive force in the primary or secondary windings and Z is the short-circuit impedance.

E=Φ*ω Φ=Bfe*A [A= magnetic core cross section area].The current will rise in time from 0 to Iinrush .

If the magnetic core was in demagnetized position when it was de-energized then up to 100 A/m magnetic field the flux density is about 0 then Φ=0 so E1=0 and so E2. So, no inrush current is expected in the secondary.

Iinrush=(U-E)/Z [ if E1=0 then Iinrush=U/Z] ; Z=R+jXe Xe=leakage magnetic flux reactance Xe=Le*ω it is considered independent of main flux level.
 

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