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 ${e}_1=120\mathrm{\angle }{0}^{\circ }\text{ V, RMS}$ 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.

$e_2=\frac{N_2}{N_2}{e}_1=\frac{2000}{500}(120)=480\text{ V}$

$i_2=\frac{N_1}{N_2}{i}_1=\frac{500}{2000}(2+3j)=(\frac{1}{2}+\frac{3}{4}j)\text{ A}$

$$\begin{gathered} Z_L=\frac{e_2}{i_2}=\frac{480}{\frac{1}{2}+\frac{3}{4}j}=\mathbf{(}\mathbf{2}\mathbf{9}\mathbf{5}\mathbf{.}\mathbf{4}\mathbf{-}\mathbf{4}\mathbf{4}\mathbf{3}\mathbf{.}\mathbf{1}\mathbf{j}\mathbf{)} \\ \mathrm{\Omega } \\ \mathbf{=}\mathbf{(}\mathbf{5}\mathbf{3}\mathbf{2}\mathbf{.}\mathbf{5}\mathrm{\angle }\mathbf{-}\mathbf{5}\mathbf{6}\mathbf{.}{\mathbf{3}}^{\mathbf{\circ }}\mathbf{)} \\ \mathrm{\Omega } \end{gathered}$$

$$\begin{gathered} Z_{th}={\left(\frac{N_1}{N_2}\right)}^2{Z}_L={\left(\frac{500}{2000}\right)}^2(295.4-443.1j)=\mathbf{(}\mathbf{1}\mathbf{8}\mathbf{.}\mathbf{5}\mathbf{-}\mathbf{2}\mathbf{7}\mathbf{.}\mathbf{7}\mathbf{j}\mathbf{)} \\ \mathrm{\Omega } \\ \mathbf{=}\mathbf{(}\mathbf{3}\mathbf{3}\mathbf{.}\mathbf{3}\mathrm{\angle }\mathbf{-}\mathbf{5}\mathbf{6}\mathbf{.}{\mathbf{3}}^{\mathbf{\circ }}\mathbf{)} \\ \mathrm{\Omega } \end{gathered}$$

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

$$\begin{gathered} Z_{th}=\frac{e_1}{i_1}=\frac{120}{2+3j}=\mathbf{(}\mathbf{1}\mathbf{8}\mathbf{.}\mathbf{5}\mathbf{-}\mathbf{2}\mathbf{7}\mathbf{.}\mathbf{7}\mathbf{j}\mathbf{)} \\ \mathrm{\Omega } \\ \mathbf{=}\mathbf{(}\mathbf{3}\mathbf{3}\mathbf{.}\mathbf{3}\mathrm{\angle }\mathbf{-}\mathbf{5}\mathbf{6}\mathbf{.}{\mathbf{3}}^{\mathbf{\circ }}\mathbf{)} \\ \mathrm{\Omega } \end{gathered}$$

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