1 Answers

Given,

φ= 3

P = 12Kw P. f = 0.6 lead

VL=208V

IL =?, Zp =?

cos θ = 0.6

θ=-35.13° (Lead)

Soln :  

P=3 VLILcos θ

12=3 ×0.208 ×IL×0.6

At star connection IL=Ip=55.51Amp

VP=VL3=2083=120V

Zp=VpIp=12055.51 =2.16 Amp

Zp=2.16 -35.13° (Ans)

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