基本電學 - 考試
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By Yedda
at 2013-05-16T23:37
at 2013-05-16T23:37
Table of Contents
1.先用三角形轉Y然後兩者並聯得到電阻為6.4
2.總電流=100/(6.4+0.6+3)=10
3.Vab=(100-3.6*10)*(6/9.6-18/19.2)=-20
-------
|0.6
100V 1.2 | |3.6
18 | |6
------ |3
提供拙見,給大家參考
※ 引述《sjacky (陳小文)》之銘言:
: 題目:http://ppt.cc/hkfX
: 我的解法:http://ppt.cc/Ysu3
: 我是算 100 * (20//12)/(20//12)+3 上部並聯的V
: V* 18/20 - V* 6/12 = Vab
: 答案是 -20
: 請問一下,我有哪邊觀念錯了嗎
: 謝謝
--
2.總電流=100/(6.4+0.6+3)=10
3.Vab=(100-3.6*10)*(6/9.6-18/19.2)=-20
-------
|0.6
100V 1.2 | |3.6
18 | |6
------ |3
提供拙見,給大家參考
※ 引述《sjacky (陳小文)》之銘言:
: 題目:http://ppt.cc/hkfX
: 我的解法:http://ppt.cc/Ysu3
: 我是算 100 * (20//12)/(20//12)+3 上部並聯的V
: V* 18/20 - V* 6/12 = Vab
: 答案是 -20
: 請問一下,我有哪邊觀念錯了嗎
: 謝謝
--
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考試
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By Queena
at 2013-05-20T06:57
at 2013-05-20T06:57
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at 2013-05-21T18:39
at 2013-05-21T18:39
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at 2013-05-22T00:45
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By Bennie
at 2013-05-23T08:20
at 2013-05-23T08:20
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