An Ideal Transformer Example

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Consider a simple, transformer with two windings. Find the current provided by the voltage source.

  • Winding 1 has a sinusoidal voltage of 120\sqrt{2}\angle{0}° applied to it at a frequency of 60Hz.
  • \frac{N_{1}}{N_{2}}=3
  • The combined load on winding 2 is \ {Z_{L}}=(5+j3)\Omega

Solution

Given: \ {e_{1}}(t)={V_{1}}\cos(\omega t) and \ \omega=2\pi f

Substituting \ f = 60Hz, \ \omega=120\pi

Therefore, \ {e_{1}}(t)=120\sqrt{2}\cos(120\pi t)V

Now the Thevenin equivalent impedance, \ {Z_{th}}, is found through the following steps:

{Z_{th}} = \frac{e_{1}}{i_{1}}

Since this is an ideal transformer {e_{1}}=\frac{N_{1}}{N_{2}}{e_{2}} and {i_{1}}=\frac{N_{2}}{N_{1}}{i_{2}}

So we can substitute, {Z_{th}}=\frac{\frac{N_{1}}{N_{2}}{e_{2}}}{\frac{N_{2}}{N_{1}}{i_{2}}}

=(\frac{N_{1}}{N_{2}})^2{Z_{L}}

Now, plugging in the given values:

\ {Z_{th}} = 3^2(5+j3)

\ =(45+j27)\Omega

Since this is an ideal transformer, it can be modeled by this simple circuit: Ideal Circuit.jpg

Therefore, {i_{1}}=\frac{e_{1}}{Z_{th}},

{i_{1}}=\frac{120\sqrt{2}}{45+j27} A

Contributors

Christopher Garrison Lau I

Reviwed By

Andrew Sell - Chris, everything looks fine, though I would do some extra formatting if possible to help make the problem flow a little smoother as you read it, and locate the picture a little higher to help bring the solution together.

Tyler Anderson - Looks good.

Read By

John Hawkins