How Does Overvoltage Affect Transformer Core Saturation and Current Flow?

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Overvoltage conditions in distribution transformers can lead to core saturation, significantly increasing magnetizing current in relation to the flux field. This current increase on the primary side behaves similarly to a locked rotor motor, where winding resistance limits current flow. The discussion highlights that saturation is unlikely to occur with no load on the secondary, as it requires a high impressed H field generated by coil currents. At full load, overvoltage should ideally not induce saturation, indicating a need for sufficient design margin. Understanding the B-H curve is crucial for evaluating the magnetic material's behavior as saturation progresses.
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I'm trying to understand what happens to a distribution transformer in relation to overvoltage conditions (in terms of saturation and current flow; not insulation failure).
I understand that once the voltage has risen high enough to drive the magnetic core past the saturation point, that the magnetizing current will rise dramatically in relation to the increase in the flux field.

1. Is the current increase (on the primary) at this point analogous to a motor at locked rotor, where the resistance of the windings themselves are the limiting factor?

2. Is there a substantial difference in the amount of this current in relation to full load vs. no load on the secondary (other than the load itself)?
 
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I don't think you will get any saturation with no load on the secondary, regardless of how high the primary voltage goes. Saturation comes from too high an impressed H field, which comes from the coil currents. And even at full load, hopefully an overvoltage in the primary won't cause saturation. That's not much design margin, IMO.

When a coil saturates, then yes, the final limiting factor is the coil windings. Look at the B -- H curve for the magnetic material to see what the effective mu does as you get farther and farther into saturation.
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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