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equivalent circuit of tranformer referred to secondary?

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tarikelec

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Hello Everyone,
I would like to understand why when we do the equivalent circuit of transfer referred to scondary we multiply the imedance of the primary by N°2. the voltage mltiplied by N , current divided b N and vice versa when referred to the primary?
best regards
 
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to make the analysis/calculation simpler.
 

Hello Everyone,
I would like to understand why when we do the equivalent circuit of transfer referred to scondary we multiply the imedance of the primary by N°2. the voltage mltiplied by N , current divided b N and vice versa when referred to the primary?
best regards

Because that's how you relate the parameters (viz. voltage, current and impedance) of primary to the secondary.
 

Hi,

What else do you expect?
What exactly is the problem?

Klaus
 

I want to know how did the equations and the formulas are derived from the transformer to its equivalent circuit step by step because when I look at it in teh litereture I don#t understand how did they come to the result
 

I want to know how did the equations and the formulas are derived from the transformer to its equivalent circuit step by step because when I look at it in teh litereture I don#t understand how did they come to the result

The formulas are derived from laws governing physical phenomena (Faraday, in this case) and also from premises (the magnetic flow passing through the secondary arm is ideally equal to the one of the primary). The rest is purely mathematical development, but this is well documented on the Web in greater or lesser depth of details. Anyway, as stated above, it is a matter of simplification; if we are analyzing a particular part of a circuit, we have to reduce the remainder that is not the focus of attention to a simpler model. The classic example is when we simulate the load at the output of this transformer only by a resistor, although it is effectively a complex circuit.
 

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