# Transformer Example Problem

Problem:

A step down transformer has a winding of $N_1 = 10 \text{ turns and } N_2 = 2$ turns. (a) If the input voltage is 1200V, what is the resulting output voltage? (b) If the input and output currents are $5A \mbox{ and } 20A$, respectively, what is the current loss due to the leakage inductance, $i_{m}(t)$.

 Figure 1: Model for an ideal transformer. Figure 2: Magnetic circuit of an ideal transformer.

Solution:

a) By modifying equation 5-39 (Mohan 5-22) we can obtain an equation for the output voltage. That is,

$e_2 = \left( \frac{N_2}{N_1} \right) e_1$.

With the information provided we can now determine the output voltage:

$e_2 = \left( \frac{N_2}{N_1} \right) e_1 = \left( \frac{2}{10} \right) 1200V = 240V$

b) From the figure above, we need to find the true value of current (i.e. the current that is not lost by leakage inductance). To accomplish this we will use equation 5-40 (Mohan 5-23) to obtain the value of $i_{1}^{'}$. That is,

$i_{1}^{'} = \left( \frac{N_2}{N_1} \right) i_{2} = \left( \frac{2}{10} \right) 20A = 4A$.

Now from equation 5-42 (Mohan 5-23) we can obtain the current loss due to the leakage inductance, which is

$i_{m} = i_{1} - i_{1}^{'} = 5A - 4A = 1A$.

Therefore, the current loss in our transformer is $1A$, which means this specific transformer is very "leaky."