功率因數改善問題 - 考試
By Sierra Rose
at 2016-09-09T06:37
at 2016-09-09T06:37
Table of Contents
http://imgur.com/hyTTYoN
97年 國營職員 電機甲 電機機械
S=5000kVA, pf=0.6(lag). ∴電流落後電壓 53.13°
∵ S=VI*
∴S=5000kVA ∠53.13°= 3000kW + j4000kVAR
加設2000kVAR
=> S'=(3000kW + j4000kVAR)+ j2000kVAR
= 3000kW + j6000kVAR = 6708 kVA ∠63.43°
=> pf'=cos(63.43°)=0.447 (lag)
跟答案不符,請問觀念有那裡不對?
謝謝
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