An Ideal Transformer Example

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 $\ {Z_{L}}=(5+j3)\Omega$

Solution

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

Substituting $\ f=60Hz$, $\ \omega =120\pi$

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

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

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

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

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

Now, plugging in the given values:

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

$\ =(45+j27)\Omega$

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

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

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.

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John Hawkins