Excercises Electric Machinery Fundamentals

Chapter 2

Problem 2-18

In [1]:
%pylab inline

Populating the interactive namespace from numpy and matplotlib


A 10-kVA 480/120-V conventional transformer is to be used to supply power from a 600-V source to a 480-V load. Consider the transformer to be ideal, and assume that all insulation can handle 600 V.

In [2]:
Sw = 10e3  # [VA]
Vp = 600   # [V]

Vh = 480   # [V] which is also the load voltage
Vl = 120   # [V]


  • Sketch the transformer connection that will do the required job.


  • Find the kilovoltampere rating of the transformer in the configuration.


  • Find the maximum primary and secondary currents under these conditions.


The transformer in Problem 2-18 is identical to the transformer in Problem 2-17, but there is a significant difference in the apparent power capability of the transformer in the two situations.

  • Why?
  • What does that say about the best circumstances in which to use an autotransformer?



For this configuration, the common winding must be the larger of the two windings. The transformer connection is shown below:


The kVA rating of the autotransformer can be found from the equation:

$$S_{IO} = \frac{N_{SE} + N_C}{N_{SE}}S_W = \frac{N_{SE} + 4N_{SE}}{N_{SE}}S_W$$

In [3]:
n = Vh/Vl  # = Nc/Nse


In [4]:
Sio = (1 + n)/1 * Sw
Sio = {:.1f} kVA

Sio = 50.0 kVA


The maximum primary current for this configuration will be:

$$I_P = \frac{S}{V_P}$$

In [5]:
Ip = Sio/Vp
Ip = {:.1f} A

Ip = 83.3 A

and the maximum secondary current is:

$$I_S = \frac{S}{V_S}$$

In [6]:
Is = Sio/Vh
s = {:.0f} A

s = 104 A


Note that the apparent power handling capability of the autotransformer is much higher when there is only a small difference between primary and secondary voltages. Autotransformers are normally only used when there is a small difference between the two voltage levels.