An Ideal Transformer Example: Difference between revisions

Revision as of 18:21, 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 $120{\sqrt {2}}\angle {0}$ ° 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:

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

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

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

Now, substituting:

$\ {Z_{th}}=3^{2}(5+j3)$

$\ =(45+j27)\Omega$

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

${i_{1}}={\frac {120{\sqrt {2}}}{45+j27}}$

@@ Line 6: / Line 6: @@
 <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>
-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>
@@ Line 20: / Line 20: @@
 Now, substituting:
-<math>{Z_{th}} = 3^2(5+j3)</math>
+<math>\ {Z_{th}} = 3^2(5+j3)</math>
-<math>=(45+j27)\Omega</math>
+<math>\ =(45+j27)\Omega</math>
 Since <math>{i_{1}}=\frac{e_{1}}{R_{th}}</math>,