You don't get more current. Just because the output is reduced to 360V,
this doesn't mean that the output current can double. Copper is expensive,
and there's just enough for the rated current.
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I do quite a bit of transformer modification and have even built a few of my own. There are formulas for figuring the turns per volt based on the core size. The transformer you have has twice the windings you would need to achieve proper core saturation and would not work well in that instance.
If you were to take apart all the I E core laminations and unwind the bobbin, then rewind with the appropriate number of turns with the right size magnet wire, you could then have a 400VA transformer with whatever voltage output you want.
If you were to take apart all the I E core laminations and unwind the bobbin, then rewind with the appropriate number of turns with the right size magnet wire, you could then have a 400VA transformer with whatever voltage output you want.
What @rayma said. The limitation of a transformer is how hot it gets and that's a function of the current in the windings (called 'copper losses' in the transformer world).
Try a thought experiment - if your trafo did indeed manage 400VA with the primary running at half (115V) the voltage, it would need double the input current because you've lost 50% of the volts. The output current also would need to be double to run the same VA for the same reason. De-rate it to 200VA though and the input and output currents are the same hence the heating is the same.
I agree with the conclusions of rayma and abraxalito being 50% of the initial power. Also that heating sets the limit.
I have two more academic considerations that work in each direction:
The magnetization current with half the input voltage is less, probably around half as well. Thus, less primary heating from magnetization currents. Also the core losses will be less as the magnetic cycling is less. This core heating may not be fully linear. These effects could indicate that the transformer can be pushed a bit further than 50% before reaching the same temperature limit as for nominal operation.
On the other hand, the Ohmic resistance in the windings remains the same. Without loading, the secondary voltage is proportional to the input voltage and the windings ratio. When current is drawn at the secondary, the output voltage is lowered with the current in the primary and secondary windings times the total equivalent resistance in the primary and secondary windings. Thus, these current losses are dependent on the current but not voltage and remain the same at full or half input voltage. This is the same loss in voltage when we use the transformer at nominal input voltage or if we use it at half the nominal input voltage. The nominal VA rating takes that loss (voltage drop) into account. With half the input voltage, this loss becomes double when considering only half the output voltage. This relatively higher voltage drop alone indicates a bit less than 50% power from the transformer.
In conclusion, we have one effect that indicates we can take the transformer a bit further than 50% and one effect indication that we get a little less than 50% at the output. In total, the transformer "VA" at half the input voltage remains in the order of 50%.
I have two more academic considerations that work in each direction:
The magnetization current with half the input voltage is less, probably around half as well. Thus, less primary heating from magnetization currents. Also the core losses will be less as the magnetic cycling is less. This core heating may not be fully linear. These effects could indicate that the transformer can be pushed a bit further than 50% before reaching the same temperature limit as for nominal operation.
On the other hand, the Ohmic resistance in the windings remains the same. Without loading, the secondary voltage is proportional to the input voltage and the windings ratio. When current is drawn at the secondary, the output voltage is lowered with the current in the primary and secondary windings times the total equivalent resistance in the primary and secondary windings. Thus, these current losses are dependent on the current but not voltage and remain the same at full or half input voltage. This is the same loss in voltage when we use the transformer at nominal input voltage or if we use it at half the nominal input voltage. The nominal VA rating takes that loss (voltage drop) into account. With half the input voltage, this loss becomes double when considering only half the output voltage. This relatively higher voltage drop alone indicates a bit less than 50% power from the transformer.
In conclusion, we have one effect that indicates we can take the transformer a bit further than 50% and one effect indication that we get a little less than 50% at the output. In total, the transformer "VA" at half the input voltage remains in the order of 50%.
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Surely in any efficient transformer the magnetization current is a small fraction of full load current and can be ignored for copper losses (which go as the square of current anyway). 10% magnetization current means it contributes 1% of copper losses, for instance, which is in the noise as far as dissipation limits goes.
Iron losses are a non-linear function of magnetization current due to the non-linear nature of ferromagnetic materials. I'd expect iron losses to reduce to less than 50% if the primary voltage is halved. Probably easiest to measure this than try to calculate. They will also be temperature sensitive.
Iron losses are a non-linear function of magnetization current due to the non-linear nature of ferromagnetic materials. I'd expect iron losses to reduce to less than 50% if the primary voltage is halved. Probably easiest to measure this than try to calculate. They will also be temperature sensitive.
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