基本電學 - 考試
By Caroline
at 2013-05-23T21:16
at 2013-05-23T21:16
Table of Contents
http://wwwc.moex.gov.tw/ExamQuesFiles/Question/092/014425300.pdf
請問第44題我的方法哪裡有問題??
將ω=2帶入 主線圈1.5H=j3Ω 副線圈6.67mF=-j75Ω
將副線圈負載換算至主線圈 Z1/Z2=(N1/N2)^2
故Z1=(1/5)^2*(100-j75)=4-j3
主線圈總阻抗=(2+j3)+(4-j3)=6Ω
電壓有效值12/√2∠0 ÷6 = √2∠0 (電流有效值)
題目答案為 2∠0是指最大值嗎??
是的話該怎麼判斷要求有效值或最大值??
謝謝
--
請問第44題我的方法哪裡有問題??
將ω=2帶入 主線圈1.5H=j3Ω 副線圈6.67mF=-j75Ω
將副線圈負載換算至主線圈 Z1/Z2=(N1/N2)^2
故Z1=(1/5)^2*(100-j75)=4-j3
主線圈總阻抗=(2+j3)+(4-j3)=6Ω
電壓有效值12/√2∠0 ÷6 = √2∠0 (電流有效值)
題目答案為 2∠0是指最大值嗎??
是的話該怎麼判斷要求有效值或最大值??
謝謝
--
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考試
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