基礎電學-最大功率轉移和戴維寧定理應用 - 考試
By Lucy
at 2014-03-13T19:21
at 2014-03-13T19:21
Table of Contents
問題http://ppt.cc/L8Cr
解答http://ppt.cc/NJCj
我的解法跟習題不同,但我又看不太懂習題的解法,
有人可以跟我解釋一下嗎,或者可以用讓我懂得方法解釋一下。
因為不懂為啥裡面的9歐姆都用不到,
而且6跟3歐姆應該是串聯,解答怎麼用並聯。實在搞不懂
想法
RTh = Rab=9 // 3=2.25Ω,
求ETh:ETh為3Ω兩端的電壓,即
3
ETH=24x-------=4.8V
9+3+6
E^2 4.8^2
最大功率 公式=-----=--------=2.56W
4R 4x2.25
出處:網路上找的教材內的習題
--
解答http://ppt.cc/NJCj
我的解法跟習題不同,但我又看不太懂習題的解法,
有人可以跟我解釋一下嗎,或者可以用讓我懂得方法解釋一下。
因為不懂為啥裡面的9歐姆都用不到,
而且6跟3歐姆應該是串聯,解答怎麼用並聯。實在搞不懂
想法
RTh = Rab=9 // 3=2.25Ω,
求ETh:ETh為3Ω兩端的電壓,即
3
ETH=24x-------=4.8V
9+3+6
E^2 4.8^2
最大功率 公式=-----=--------=2.56W
4R 4x2.25
出處:網路上找的教材內的習題
--
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