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Given the following text description, write Python code to implement the functionality described below step by step
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Description:
 Excercises Electric Machinery Fundamentals
Chapter 2
Problem 2-17
Step1:  Description
A 10-kVA 480/120-V conventional transformer is to be used to supply power from a 600-V source to a 120-V load. Consider the transformer to be ideal, and assume that all insulation can handle 600 V.
Step2:  (a)
Sketch the transformer connection that will do the required job.
(b)
Find the kilovoltampere rating of the transformer in the configuration.
(c)
Find the maximum primary and secondary currents under these conditions.
SOLUTION
(a)
For this configuration, the common winding must be the smaller of the two windings.
The transformer connection is shown below
Step3:  (c)
The maximum primary current for this configuration will be
Step4:  and the maximum secondary current is
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Python Code:
%pylab notebook
Explanation: Excercises Electric Machinery Fundamentals
Chapter 2
Problem 2-17
End of explanation
Vp = 600  # [V]
Vl = 120  # [V] which is also the load voltage
Vh = 480  # [V] 
Sw = 10e3 # [VA]
Explanation: Description
A 10-kVA 480/120-V conventional transformer is to be used to supply power from a 600-V source to a 120-V load. Consider the transformer to be ideal, and assume that all insulation can handle 600 V.
End of explanation
n = Vh/Vl  # = Nse/Nc
Sio = (n + 1)/n * Sw
print('''
Sio = {:.1f} kVA
==============
'''.format(Sio/1000))
Explanation: (a)
Sketch the transformer connection that will do the required job.
(b)
Find the kilovoltampere rating of the transformer in the configuration.
(c)
Find the maximum primary and secondary currents under these conditions.
SOLUTION
(a)
For this configuration, the common winding must be the smaller of the two windings.
The transformer connection is shown below:
<img src="figs/Problem_2-17.jpg" width="40%">
(b)
The kVA rating of the autotransformer can be found from the equation:
$$S_{IO} = \frac{N_{SE} + N_C}{N_{SE}}S_W = \frac{4N_C + N_C}{4N_C}S_W$$
End of explanation
Ip = Sio/Vp
print('''
Ip = {:.2f} A
============
'''.format(Ip))
Explanation: (c)
The maximum primary current for this configuration will be:
$$I_P = \frac{S}{V_P}$$
End of explanation
Is = Sio/Vl
print('''
s = {:.0f} A
=========
'''.format(Is))
Explanation: and the maximum secondary current is:
$$I_S = \frac{S}{V_S}$$
End of explanation
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