Can you calulate a transformer secondary resistance by using load regulation %

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To calculate the internal resistance of a transformer’s secondary winding using load regulation, the voltage difference between no-load and full-load can be determined by multiplying the voltage by the regulation percentage. For a 300 VA transformer at 45 Volts with a 10% regulation, this results in a voltage difference of 4.5 volts. The secondary series resistance per winding is then calculated as 4.5 volts divided by 3.3 amps, yielding approximately 1.36 ohms. It is important to consider the primary resistance, which can be reflected to the secondary side and may significantly impact regulation. In scenarios where both resistances are equal, the secondary resistance should be taken as half of the calculated value to account for both contributions to regulation.
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If I had a step down transformer of 300 VA in size at 45-0-45 Volts@3.3 Amps with a regulation of 10% could I do the following to calculate the internal resistance of the secondary winding for my power supply circuit?

Voltage * Regulation 10% to work out the voltage difference between noload and full load.

45 Volts* .1 = 4.5 volts

Secondary series resistance per winding
4.5 Volts/3.3 amps = 1.36 ohms

I know there is inductance as well but at the moment I am ignoring it.
I need to put values into circuit maker student edition to reflect the real world.
 
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You need to take the primary resistance into account. It is reflected by the square of the turns ratio to the secondary, and is a quite significant contributor to regulation.

If a transformer is well designed, typically the reflected value of the primary resistance on the secondary side will be the same as the secondary intrinsic resistance (equal copper losses in both windings being the goal). Often this is close to the truth, but not always. However, in the absence of any possibility of measuring both windings' DC resistance, the best you can do is to take them to be equal.

Since both the secondary resistance AND the reflected primary resistance will contribute equally to the regulation (if the copper losses both windings are equal), you should take the secondary resistance to be half of your calculated value.
 
Ok I forgot about the primary resistance thanks for that.
 
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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