An Ideal Transformer Example: Difference between revisions

Revision as of 12:19, 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 $120{\sqrt {2}}\angle {0}$ ° applied to it at a frequency of 60Hz.
${\frac {N_{1}}{N_{2}}}=3$
The combined load on winding 2 is ${Z_{L}}=(5+j3)\Omega$

${e_{1}}(t)={V_{1}}\cos(\omega t)$

$\omega =2\pi f$ , so $\omega =120\pi$

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

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

${Z_{th}}={\frac {e_{1}}{i_{1}}}$

@@ Line 1: / Line 1: @@
 Consider a simple, transformer with two windings. Find the current provided by the voltage source.
 * 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===
-<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 <math>{Z_{th}}</math> is the impedance seen by the voltage source supplying winding 1.
+Now the Thevenin equivalent impedance, <math>{Z_{th}}</math>, is found through the following steps:
+<math>{Z_{th}} = \frac{e_{1}}{i_{1}}</math>