99地方四等 基本電學 第五題 - 考試
By Yedda
at 2014-05-06T14:38
at 2014-05-06T14:38
Table of Contents
99四等基本電學題目:http://ppt.cc/k5Wx
問第五題的P解
答案是 3*I^2*R(三角) = 3*0.727^2*75 = 118.92
有辦法用3*Vp*Ip*cos 表示嗎
3*Vp*0.727*cos
3*110√3*0.727*cos(?)
3*110√3*0.727*cos66.3 = 166.4
3*I^2*R(total) = 3*0.727^2*(75+30) = 166.4
感覺怪怪的 不知如何表示 = =...
還是答案有錯其實是166.4才對 ?
難道只能用3*I^2*R(三角) 解嗎 ?
求解 謝謝
--
問第五題的P解
答案是 3*I^2*R(三角) = 3*0.727^2*75 = 118.92
有辦法用3*Vp*Ip*cos 表示嗎
3*Vp*0.727*cos
3*110√3*0.727*cos(?)
3*110√3*0.727*cos66.3 = 166.4
3*I^2*R(total) = 3*0.727^2*(75+30) = 166.4
感覺怪怪的 不知如何表示 = =...
還是答案有錯其實是166.4才對 ?
難道只能用3*I^2*R(三角) 解嗎 ?
求解 謝謝
--
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考試
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