An Ideal Transformer Example: Difference between revisions

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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 $120 \sqrt{2} ∠ 0$ ° applied to it at a frequency of 60Hz.
$\frac{N_{1}}{N_{2}} = 3$
The combined load on winding 2 is $Z_{L} = (5 + j 3) Ω$

Solution

Given: $e_{1} (t) = V_{1} \cos (ω t)$ and $ω = 2 π f$

Substituting $f = 60 H z$ , $ω = 120 π$

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

Now the Thevenin equivalent impedance, $Z_{t h}$ , is found through the following steps:

$Z_{t h} = \frac{e_{1}}{i_{1}}$

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, $Z_{t h} = \frac{\frac{N_{1}}{N_{2}} e_{2}}{\frac{N_{2}}{N_{1}} i_{2}}$

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

Now, plugging in the given values:

$Z_{t h} = 3^{2} (5 + j 3)$

$= (45 + j 27) Ω$

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

Therefore, $i_{1} = \frac{e_{1}}{Z_{t h}}$ ,

$i_{1} = \frac{120 \sqrt{2}}{45 + j 27} A$

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

@@ Line 2: / Line 2: @@
 * 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>
-* 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===
-<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:
 <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>
-<math>{Z_{th}} = 3^2(5+j3)</math>
+Now, plugging in the given values:
-<math>=(45+j27)\Omega</math>
+<math>\ {Z_{th}} = 3^2(5+j3)</math>
-Since <math>{i_{1}}=\frac{e_{1}}{R_{th}}</math>,
+<math>\ =(45+j27)\Omega</math>
-<math>{i_{1}}=\frac{120\sqrt{2}}{45+j27}</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>
+===Contributors===
+[[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

An Ideal Transformer Example: Difference between revisions

Latest revision as of 17:34, 24 January 2010

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An Ideal Transformer Example: Difference between revisions

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