An Ideal Transformer Example: Difference between revisions

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* Winding 1 has a sinusoidal voltage of <math>120\sqrt{2}\angle{0}</math>° applied to it at a frequency of 60Hz.
* Winding 1 has a sinusoidal voltage of <math>120\sqrt{2}\angle{0}</math>° applied to it at a frequency of 60Hz.
* <math>\frac{N_{1}}{N_{2}}=3</math>
* <math>\frac{N_{1}}{N_{2}}=3</math>
* The combined load on winding 2 is <math>{Z_{L}}=(5+j3)\Omega</math>
* The combined load on winding 2 is <math>\ {Z_{L}}=(5+j3)\Omega</math>
===Solution===
===Solution===
<math>{e_{1}}(t)={V_{1}}\cos(\omega t)</math>
<math>\ {e_{1}}(t)={V_{1}}\cos(\omega t)</math>


<math>\omega=2\pi f</math>, so <math>\omega=120\pi</math>
<math>\ \omega=2\pi f</math>, so <math>\omega=120\pi</math>


Therefore, <math>{e_{1}}(t)={V_{1}}\cos(120\pi t)</math>
Therefore, <math>\ {e_{1}}(t)={V_{1}}\cos(120\pi t)</math>


Now the Thevenin equivalent impedance, <math>{Z_{th}}</math>, is found through the following steps:
Now the Thevenin equivalent impedance, <math>{Z_{th}}</math>, is found through the following steps:

Revision as of 19:21, 17 January 2010

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

  • Winding 1 has a sinusoidal voltage of ° applied to it at a frequency of 60Hz.
  • The combined load on winding 2 is

Solution

, so

Therefore,

Now the Thevenin equivalent impedance, , is found through the following steps:

Now, substituting:

Since ,