0
\$\begingroup\$

enter image description here

If I get the transformer equivalent circuit of the above circuit did I really get a purely resistive circuit with a resistance of n 2 times RLoad . I am asking because I realized that in the oscilloscope screen there is delay(phase shift) between input voltage and input current. If I could not get purely resistive circuit why this equivalent circuit fails?

\$\endgroup\$
1
  • \$\begingroup\$ It’s only resistive for risetime of magnetic coupling time constant \$\endgroup\$ Commented May 21, 2021 at 18:48

1 Answer 1

0
\$\begingroup\$

The equivalent circuit will only be \$(\frac{N1}{N2})^2R_{load}\$ if it was an ideal transformer.

Real trasformers have nonideal components. For example, the joule loss on the widings, or the core hysteresis. Those phenomena produce several effects, that must be accountable, in order to get the accurete equivalent circuit.

I sugest the reading Electric Machinery Fundamentals by Stephen Chapman, to better understand how to experimentally obtain your transformer's equivalent circuit.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.