An Ideal Transformer Example: Difference between revisions

From Class Wiki
Jump to navigation Jump to search
Cdxskier (talk | contribs)
No edit summary
Cdxskier (talk | contribs)
No edit summary
Line 17: Line 17:
Since this is an ideal transformer <math>{e_{1}}=\frac{N_{2}}{N_{1}}{e_{2}}
Since this is an ideal transformer <math>{e_{1}}=\frac{N_{2}}{N_{1}}{e_{2}}


So we can substitute, </math><math>=\frac{\frac{N_{1}}{N_{2}}{e_{2}}}{\frac{N_{2}}{N_{1}}{i_{2}}}</math>
So we can substitute, <math>=\frac{\frac{N_{1}}{N_{2}}{e_{2}}}{\frac{N_{2}}{N_{1}}{i_{2}}}</math>


<math>=(\frac{N_{1}}{N_{2}})^2{R_{L}}</math>
<math>=(\frac{N_{1}}{N_{2}})^2{R_{L}}</math>

Revision as of 10:50, 21 January 2010

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

  • Winding 1 has a sinusoidal voltage of 12020° 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=N2N1e2Sowecansubstitute,<math>=N1N2e2N2N1i2

=(N1N2)2RL

Now, substituting:

Zth=32(5+j3)

=(45+j27)Ω

Since i1=e1Rth,

i1=120245+j27A

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

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.

Read By

John Hawkins