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

## Contents

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

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

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

### 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.