A 10 kw, 400 v, 3phase induction motor with star delta starter, having full load efficiency as 0.86, the full load pf is 0.8 and short circuit current is 30 a at 100 v. Find the ratio of starting to full load current?
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Answer:
1.9 A
Explanation:
efficiency= power output / power input
0.86= 10 kw/ power input
power input= 11.62 kw
power input= √3 VL IL cos fi
11.62=√3 *400*IL*0.8
IL=20.98 A
Isc at 100 V=30 A....given
Isc at 400 V= 120 A
Ist= 120/3= 40 A
Now Ist/IfL= 40/20.98 = 1.9 A
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