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clear
//
//case a
f=150
p=2
//assume the diameter of the stator bore is d meter
n=120*50/2 //where n is rotor speed
printf("\n n= %0.0f rpm",n)
pi=3.14
d=(120*60)/(%pi*3000)
printf("\n D= %0.3f m",d)
//case b
k=2
l=1
o=k*d**2*n*l
printf("\n output of the alternator= %0.3f KVA",o)
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