基本電學 功率因素 - 考試
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By Valerie
at 2017-07-22T17:36
at 2017-07-22T17:36
Table of Contents
※ 引述《isso (小烏龜)》之銘言:
: 請問各位大大這題
: https://imgur.com/a/auEr8
: 我寫到這邊發現跟答案不一樣
: 不知道哪邊寫錯了 謝謝各位大大
: https://imgur.com/a/wHGiy
藉這題請教諸位大德,想了很久一直卡關...囧
這一題我的理解是:直接求出阻抗的相角即可,
因為所謂落後跟超前,完全是由負載決定,所以
根本也不用管電源長怎樣。
第一種解法直接求導納,算出來答案是 C=1/32 F (正確)
主要是第二種算法,第一項 1/(2+2j) 沒有先化簡成 1/4(1-j),
直接硬算後...
最後用 tan-1(4C/(1-4C))-tan-1(2/2) =tan-1(-3/4)
得到答案是 C=1/20 F(錯誤)
http://imgur.com/a/u2PFd
我知道複數要先化簡再計算...
但是第二種算法到底錯在哪裡啊?
驗算好多次還是想不透...囧
懇請高人指點啊...
--
: 請問各位大大這題
: https://imgur.com/a/auEr8
: 我寫到這邊發現跟答案不一樣
: 不知道哪邊寫錯了 謝謝各位大大
: https://imgur.com/a/wHGiy
藉這題請教諸位大德,想了很久一直卡關...囧
這一題我的理解是:直接求出阻抗的相角即可,
因為所謂落後跟超前,完全是由負載決定,所以
根本也不用管電源長怎樣。
第一種解法直接求導納,算出來答案是 C=1/32 F (正確)
主要是第二種算法,第一項 1/(2+2j) 沒有先化簡成 1/4(1-j),
直接硬算後...
最後用 tan-1(4C/(1-4C))-tan-1(2/2) =tan-1(-3/4)
得到答案是 C=1/20 F(錯誤)
http://imgur.com/a/u2PFd
我知道複數要先化簡再計算...
但是第二種算法到底錯在哪裡啊?
驗算好多次還是想不透...囧
懇請高人指點啊...
--
Tags:
考試
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By Heather
at 2017-07-27T11:27
at 2017-07-27T11:27
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By Sandy
at 2017-07-30T05:00
at 2017-07-30T05:00
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By Valerie
at 2017-07-31T15:14
at 2017-07-31T15:14
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By Aaliyah
at 2017-08-01T22:29
at 2017-08-01T22:29
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By Lily
at 2017-08-03T11:13
at 2017-08-03T11:13
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By Margaret
at 2017-08-04T14:21
at 2017-08-04T14:21
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By Quanna
at 2017-08-06T12:30
at 2017-08-06T12:30
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