An Ideal Transformer Example: Difference between revisions

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* 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>
Given:
<math>\ {e_{1}}(t)={V_{1}}\cos(\omega t)</math> and <math>\ \omega=2\pi f</math>


<math>\ \omega=2\pi f</math>, so <math>\ \omega=120\pi</math>
Substituting <math>\ f = 60Hz</math>, <math>\ \omega=120\pi</math>


Therefore, <math>\ {e_{1}}(t)={V_{1}}\cos(120\pi t)</math>
Therefore, <math>\ {e_{1}}(t)=120\sqrt{2}\cos(120\pi t)V</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:
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<math>{Z_{th}} = \frac{e_{1}}{i_{1}}</math>
<math>{Z_{th}} = \frac{e_{1}}{i_{1}}</math>


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


<math>=(\frac{N_{1}}{N_{2}})^2{R_{L}}</math>
So we can substitute, <math>{Z_{th}}=\frac{\frac{N_{1}}{N_{2}}{e_{2}}}{\frac{N_{2}}{N_{1}}{i_{2}}}</math>


Now, substituting:
<math>=(\frac{N_{1}}{N_{2}})^2{Z_{L}}</math>
 
Now, plugging in the given values:


<math>\ {Z_{th}} = 3^2(5+j3)</math>
<math>\ {Z_{th}} = 3^2(5+j3)</math>
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<math>\ =(45+j27)\Omega</math>
<math>\ =(45+j27)\Omega</math>


Since <math>{i_{1}}=\frac{e_{1}}{R_{th}}</math>,
Since this is an ideal transformer, it can be modeled by this simple circuit:
[[Image: Ideal_Circuit.jpg]]
 
Therefore, <math>{i_{1}}=\frac{e_{1}}{Z_{th}}</math>,
 
<math>{i_{1}}=\frac{120\sqrt{2}}{45+j27} A</math>


<math>{i_{1}}=\frac{120\sqrt{2}}{45+j27}</math>
===Contributors===
===Contributors===


[[Lau, Chris|Christopher Garrison Lau I]]
[[Lau, Chris|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

Latest revision as of 17:34, 24 January 2010

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

  • Winding 1 has a sinusoidal voltage of 1202∠0° applied to it at a frequency of 60Hz.
  • N1N2=3
  • The combined load on winding 2 is ZL=(5+j3)Ω

Solution

Given: e1(t)=V1cos⁡(ωt) and ω=2πf

Substituting f=60Hz, ω=120π

Therefore, e1(t)=1202cos⁡(120πt)V

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

Zth=e1i1

Since this is an ideal transformer e1=N1N2e2 and i1=N2N1i2

So we can substitute, Zth=N1N2e2N2N1i2

=(N1N2)2ZL

Now, plugging in the given values:

Zth=32(5+j3)

=(45+j27)Ω

Since this is an ideal transformer, it can be modeled by this simple circuit:

Therefore, i1=e1Zth,

i1=120245+j27A

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

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