102鐵特高員三級電路學 - 考試
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By Michael
at 2013-09-28T19:05
at 2013-09-28T19:05
Table of Contents
這題是102年鐵特高員三級電路學的第四題:http://ppt.cc/V8t~
我得出的方程組是
100 = (4+j3)*I1-j8*I2
0 = (-j8)*I1 + (10+j18)*I2
而公職王的解答是:http://ppt.cc/uera
100 = (4+j3)*I1 - j8*I2
0 = (-j4)*I1 + (10-j2)*I2
第一個方程式都一樣,問題在第二個,
按照I2的方向,
I1在阻抗j6及互感j2所產生的電壓,
應是與I2產生的電壓極性相反才對
不曉得是我觀念有問題還是解答錯了,
請各位高手幫幫忙~3Q
--
我得出的方程組是
100 = (4+j3)*I1-j8*I2
0 = (-j8)*I1 + (10+j18)*I2
而公職王的解答是:http://ppt.cc/uera
100 = (4+j3)*I1 - j8*I2
0 = (-j4)*I1 + (10-j2)*I2
第一個方程式都一樣,問題在第二個,
按照I2的方向,
I1在阻抗j6及互感j2所產生的電壓,
應是與I2產生的電壓極性相反才對
不曉得是我觀念有問題還是解答錯了,
請各位高手幫幫忙~3Q
--
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考試
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By Daph Bay
at 2013-09-30T21:39
at 2013-09-30T21:39
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By Rae
at 2013-10-01T04:12
at 2013-10-01T04:12
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