[考題] 基電

看板Examination作者 (無敵浩克)時間12年前 (2013/04/12 22:17), 編輯推噓4(4推 0噓 18→)
留言22則, 9人參與, 最新討論串1/1
http://wwwc.moex.gov.tw/ExamQuesFiles/Question/101/101080_81040.pdf 第5題第1小項我算的結果 a.ZL=(4+j3)//-j3=3.75歐姆 Pmax=Emax^2/4*ZL=9.6w b.ZL=5//-j3=2.57歐姆 Pmax=Emax^2/4*ZL=14w 兩種方式來驗證，怎算出結果不同,是我的觀念錯了嗎? -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 111.253.208.236
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04/12 22:47, , 1F
你b. 5是phasor的絕對值 但是-j3可不是phasor喔
04/12 22:47, 1F
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04/12 22:53, , 2F
請問答案是多少??
04/12 22:53, 2F
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04/12 23:06, , 3F
我算出PL(max)=18
04/12 23:06, 3F
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04/12 23:40, , 4F
(9/4)*4=9W
04/12 23:40, 4F
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04/12 23:41, , 5F
Vth=-9j Rth=2.25-3J PL(Max)=9W
04/12 23:41, 5F
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04/12 23:42, , 6F
請教一下第三題有電阻被短路嗎?
04/12 23:42, 6F
推
04/12 23:45, , 7F
我覺得是求ZL看進去的Rth和Eth,欲使ZL得到最大功率
04/12 23:45, 7F
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04/12 23:47, , 9F
id=RF-20120615172849-4BE
04/12 23:47, 9F
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04/12 23:48, , 10F
ZL=Rth*=2.25-j3 Eth=7.2ㄥ-126.8度 Pmax=5.76W,對嗎?
04/12 23:48, 10F
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04/12 23:50, , 11F
最大功率是指Rth的共軛負數2.25-3j Zl=2.25+3j R=4.5歐姆
04/12 23:50, 11F
推
04/12 23:58, , 12F
第三題 五個元件為並聯
04/12 23:58, 12F
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04/13 00:10, , 13F
被題目唬住了了解
04/13 00:10, 13F
推
04/13 00:15, , 14F
打錯了,ZL=2.25+j3
04/13 00:15, 14F
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04/13 00:19, , 15F
Zth=j3//(4-j3)
04/13 00:19, 15F
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04/13 00:19, , 16F
所以我算9.6w這一題會拿滿分嗎
04/13 00:19, 16F
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04/13 00:23, , 17F
會0分= =
04/13 00:23, 17F
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04/13 00:23, , 18F
答案是9W
04/13 00:23, 18F
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04/13 00:24, , 19F
ZL=j3//(4-j3)=2.25+j3,PL=(9/4)*4=9W
04/13 00:24, 19F
推
04/13 00:27, , 20F
又把Eth算錯了,其實這題考的是戴維寧,考試真的要很專心
04/13 00:27, 20F
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04/13 07:35, , 21F
算錯了 P=I^2*R (2)^2*2.25=9W才對
04/13 07:35, 21F
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10/11 22:15, , 22F
我算出PL(max)= https://daxiv.com
10/11 22:15, 22F
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