Answer Posted / gaurang bhavsar- jr. asst. ele

Elect. Power(3-ph)=1.732*V*I*p.f.Now,
2h.p.=2*0.746=1.492=1.5 kw. So, If a voltage system of 3-
ph,415V, 50 Hz is taken, assuming a p.f. of 0.8 then--
1.5=1.732*0.415*I*0.8 which gives I=2.6085 Amp of normal
current. The starting current will be approx. 3 times of
normal current i.e.-7.82 Amp.

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