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Ok, so for subtractive polarity in transformers(dots at the same side of the windings) E1/E2=N1/N2

But if it's subtractive polarity E1 and E2 are out of phase so shouldn't E1/E2=-N1/N2 enter image description here

Edit: Ok here is what I tried out. I just need to understand the first step. I solved the same question for a transformer with additive polarity

enter image description here!

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  • \$\begingroup\$ You defined E1 as positive when current flows out of the left dot. Current flows in one dot and out the other. Therefore when E1 is positivve (current out of left dot) then current flows into right dor which makes E2 negative (i.e. oppposite of your defined polarity.) Some unnecessary negatives but it works. I recommend you define things so such negatives don't happen for straightforwardness. \$\endgroup\$ – DKNguyen Nov 20 '19 at 1:53
  • \$\begingroup\$ BTW, your "additive" and "subtracted" polarity have no meaning...the windings are isolated. Nothing is adding or subtracting from anything. Stop thinking about it that way. The reference of one winding is completely independent of the reference of the other. If you pick the top right terminal to be the reference for the secondary, it doesn't suddenly change the polarity of the physical transformer's windings relative to each other. \$\endgroup\$ – DKNguyen Nov 20 '19 at 1:57
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The transformer primary and secondary are isolated from each other. What have you defined as the reference potential on each side? That will determine the polarity of math. You can assign both sides in any combination you want, as long as you keep the math consistent with your definitions.

Remember the transformer is just DRAWN with dots on the opposite side, but the primary and secondary are isolated in reality. That means you could flip the entire right side of the transformer in the drawing upside down so the dots are on the same side if you wanted to...you would just also have to flip the entire right side of the circuit upside down too so things stay consistent (although it doesn't matter so much with an impedance which aren't polarized). Or you could do the same thing with the left side. It doesn't matter.

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    \$\begingroup\$ DKNguyen is correct. In this particular problem you don't care about the phase relationship between E1 and E2. \$\endgroup\$ – relayman357 Nov 19 '19 at 23:49

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