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Transformers Turn ratio versus Impedence ratio

  1. Oct 29, 2013 #1
    Hello,

    This is my first post, so I hope this is in the right section and format.

    I've looked all over the web, but I can't find anything that addresses my question clearly.

    I know that for an ideal transformer to not lose any power, the turn ratio has to be the square root of the impedence ratio. My question is what happens when this is not the case? I would think the voltage and current ratios wouldn't be affected because of the nature of a transformer, but this means that one of the impedences would have to magically change to match to I and V on its respective side.

    Basically, if (NP/NS)^2 = RP/RS for max power efficiency, what happens when this isn't true?

    Thanks in advance,
    Hollumber
     
  2. jcsd
  3. Oct 29, 2013 #2
    After researching this so much, I finally realized where I was going wrong.

    (NP/NS)^2 = RP/RS is used to calculate the turn ratio of the transformer to get max power to the load resistor (RS).

    If this ratio doesn't hold true, then Req = RS(NP/NS)^2, where Req is the equivalent resistance that the primary circuit (which includes the voltage source) will experience, resulting in lower power in the secondary circuit (containing the load resistor).
     
  4. Oct 29, 2013 #3

    marcusl

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    Gold Member

    You can think of it in terms of an equivalent circuit. Losses can be modeled by an equivalent resistance in series with (or in parallel across) the secondary of an ideal transformer. The transformer impedance ratio is still the square of the turns ratio. The power dissipation can be found by analyzing the behavior of the entire circuit, including the source and its internal resistance, the transformer (ideal transformer plus loss resistance), and the load circuit. I and V across the ideal transformer element still have the ideal ratio, but some of the power is dissipated in the loss resistance and doesn't make it into the load.
     
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