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//Chapter-1, Example 1.14, Page 1.35
//=============================================================================
clc
clear
//INPUT DATA
Eb=225;//Back emf in V
IL=40;//Line current in A
Rsh=150;//Field resistance in ohm
Ish=1.67;//Field current in A
//CALCULATIONS
V=(Ish*Rsh);//Terminal applied voltage in V
Ia=(IL-Ish);//Armature current in A
Ra=(V-Eb)/Ia;//Armature resistance in ohm
Ia=(V/Ra);//Maximum armature current in A
//OUTPUT
mprintf('i)Armature resistance is %3.2f ohm \nii)Armature current will be maximum at the moment of start up and it is %3.2f A',Ra,Ia)
//=================================END OF PROGRAM==============================
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