Example: Ideal Transformer Exercise

Author

John Hawkins

Problem Statement

An ideal transformer has a primary winding with 500 turns and a secondary winding with 2000 turns. Given that $\ e_{1}=120\angle {0^{\circ }}{\text{ V, RMS}}$ and $\ i_{1}=(2+3j){\text{ A}}$, find the load impedance, $\ Z_{L}$ and the Thevenin equivalent, $\ Z_{th}$.

Solution

We could find the Thevenin impedance directly, but we will save that until the end as a checking mechanism. First, we will find the actual load impedance by finding the current and voltage in the secondary winding and finding their ratio.

$e_{2}={\frac {N_{2}}{N_{2}}}e_{1}={\frac {2000}{500}}(120)=480{\text{ V}}$

$i_{2}={\frac {N_{1}}{N_{2}}}i_{1}={\frac {500}{2000}}(2+3j)=\left({\frac {1}{2}}+{\frac {3}{4}}j\right){\text{ A}}$

$Z_{L}={\frac {e_{2}}{i_{2}}}={\frac {480}{{\frac {1}{2}}+{\frac {3}{4}}j}}=\mathbf {(295.4-443.1j)\ \Omega \ =(532.5\angle {-56.3^{\circ }})\ \Omega }$

$Z_{th}=\left({\frac {N_{1}}{N_{2}}}\right)^{2}Z_{L}=\left({\frac {500}{2000}}\right)^{2}(295.4-443.1j)=\mathbf {(18.5-27.7j)\ \Omega \ =(33.3\angle {-56.3^{\circ }})\ \Omega }$

As mentioned at the beginning, this should be the impedance found using the ratio of the primary voltage and current. Using this method, we find that

$Z_{th}={\frac {e_{1}}{i_{1}}}={\frac {120}{2+3j}}=\mathbf {(18.5-27.7j)\ \Omega \ =(33.3\angle {-56.3^{\circ }})\ \Omega }$

This is the same answer as above, which verifies the solutions.

Reviewed By

Tyler Anderson - it may be helpful to the readers if you referenced what equations you are using. For example: $e_{2}={\frac {N_{2}}{N_{2}}}e_{1}$ $EQ(5-39)$ Otherwise it looks sound to me.