Answer / 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.

Answer / kanchan mukherjee

p=1.732*v*i*.8
p=2hp=1.5kw
v=415
1.5*1000=1.732*415*i*.8
i=1500/1.732*415*.8
i=2.7 amp.

Answer / rajagopal

2Hp=1492Watts

I=1.732x415x0.8 =575.024

575.024/1492=0.38540482
0.385x1000 =3.85

Answer / rajasekar.a

3.85

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