基本電學 - 考試
By Damian
at 2013-06-19T00:21
at 2013-06-19T00:21
Table of Contents
※ 引述《demonazx (肥貓)》之銘言:
: 出處:102年佐級基本電學
: http://wwwc.moex.gov.tw/ExamQuesFiles/Question%5C102/102080_4906.pdf
: 請問一下
: 第14.24.26.40怎做
: 謝謝了
第14題我的解法是利用節點電壓法
(V2-60)/6 + V2/6 + (V2+3V2)/6=0
解出V2=5,V1=3V2+V2=20V
第24題是用戴維寧等效取由8Ω看進去的等效電阻,
因為要最大功率所以要取等效電阻為8Ω
5+6//RL=8,得到RL=6Ω
第34題i(t)=sin(1000t+90)mA,ω=1000所以XC=-j500Ω
v=i*XC=1m∠90*500∠-90=0.5∠0=0.5sin(1000t)
第40題先將4+j2//4-j2=2.5Ω,在利用分壓
[2.5+j2/(2+2.5+j2)]*12∠0=7.8∠14.7
這是我的算法啦...跟大家現醜了Orz
--
: 出處:102年佐級基本電學
: http://wwwc.moex.gov.tw/ExamQuesFiles/Question%5C102/102080_4906.pdf
: 請問一下
: 第14.24.26.40怎做
: 謝謝了
第14題我的解法是利用節點電壓法
(V2-60)/6 + V2/6 + (V2+3V2)/6=0
解出V2=5,V1=3V2+V2=20V
第24題是用戴維寧等效取由8Ω看進去的等效電阻,
因為要最大功率所以要取等效電阻為8Ω
5+6//RL=8,得到RL=6Ω
第34題i(t)=sin(1000t+90)mA,ω=1000所以XC=-j500Ω
v=i*XC=1m∠90*500∠-90=0.5∠0=0.5sin(1000t)
第40題先將4+j2//4-j2=2.5Ω,在利用分壓
[2.5+j2/(2+2.5+j2)]*12∠0=7.8∠14.7
這是我的算法啦...跟大家現醜了Orz
--
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考試
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