# Example: Ideal Transformer Exercise

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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. The equations used are those derived in class by Professor Frohne.

$e_2=\frac{N_2}{N_1}e_1=\frac{2000}{500}(120)=480\text{ V}$

$i_2=\frac{N_1}{N_2}i_1=\frac{500}{2000}(2+3j)=\left(\frac{1}{2}+\frac{3}{4}j\right)\text{ A}$

$Z_L=\frac{e_2}{i_2}=\frac{480}{\frac{1}{2}+\frac{3}{4}j}=\mathbf{(295.4-443.1j)\ \Omega\ =(532.5\angle{-56.3^\circ})\ \Omega}$

$Z_{th}=\left(\frac{N_1}{N_2}\right)^2Z_L=\left(\frac{500}{2000}\right)^2(295.4-443.1j)=\mathbf{(18.5-27.7j)\ \Omega\ =(33.3\angle{-56.3^\circ})\ \Omega}$

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
• Jimmy Apablaza-Lorca

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.