An Ideal Transformer Example: Difference between revisions

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

$e_{1} (t) = V_{1} \cos (ω t)$

$ω = 2 π f$ , so $ω = 120 π$

Therefore, $e_{1} (t) = V_{1} \cos (120 π t)$

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

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

$= \frac{\frac{N_{1}}{N_{2}} e_{2}}{\frac{N_{2}}{N_{1}} i_{2}}$

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

Now, substituting:

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

$= (45 + j 27) Ω$

Since $i_{1} = \frac{e_{1}}{R_{t h}}$ ,

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

@@ 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>
+<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: