基本電學 - 考試
By Lydia
at 2013-04-03T14:27
at 2013-04-03T14:27
Table of Contents
第一題:
此題欲求R有大功率輸出時之電流。
把R拿掉,求兩端的Vth。利用節電壓法,在電流源6A上頭令節點V,則:
根據KCL:
6 = V-80 / 5 + V-50 / 20
-->V = 178
則:80V上的電流I= 178-80 / 5 = 19.6(A)
則:Vth = 5*19.6 - 50 = 48 (V)
Rth = 5//20 = 4 歐姆
電路圖請自己畫:
欲使R有最大功率輸出,則R = Rth = 4歐姆
則R的電流為 : 48 / 4+4 = 4(A)
第二題:
答案是否有誤?
Vth = 7.2 (V)
Rth = 4歐姆
欲求t = 3ms 的IL,我算的答案是 0.593 (A)
※ 引述《funya (天意)》之銘言:
: http://ppt.cc/X05z
: 求R最大功率時電流為多少~答案是4A~
: 我自己算RTH是4 Vth=不知道怎麼算
: 最大功率RTH=R 電阻是8 這樣推答案VTH是32嗎?
: http://ppt.cc/pS7W
: 求t=3ms時IL=?
: 我算出時間常數=7.5m,
: 一般來說大概都是整數吧?
: 是我算錯了嗎?
: 答案是0.379A
: 懇請高手幫忙
--
此題欲求R有大功率輸出時之電流。
把R拿掉,求兩端的Vth。利用節電壓法,在電流源6A上頭令節點V,則:
根據KCL:
6 = V-80 / 5 + V-50 / 20
-->V = 178
則:80V上的電流I= 178-80 / 5 = 19.6(A)
則:Vth = 5*19.6 - 50 = 48 (V)
Rth = 5//20 = 4 歐姆
電路圖請自己畫:
欲使R有最大功率輸出,則R = Rth = 4歐姆
則R的電流為 : 48 / 4+4 = 4(A)
第二題:
答案是否有誤?
Vth = 7.2 (V)
Rth = 4歐姆
欲求t = 3ms 的IL,我算的答案是 0.593 (A)
※ 引述《funya (天意)》之銘言:
: http://ppt.cc/X05z
: 求R最大功率時電流為多少~答案是4A~
: 我自己算RTH是4 Vth=不知道怎麼算
: 最大功率RTH=R 電阻是8 這樣推答案VTH是32嗎?
: http://ppt.cc/pS7W
: 求t=3ms時IL=?
: 我算出時間常數=7.5m,
: 一般來說大概都是整數吧?
: 是我算錯了嗎?
: 答案是0.379A
: 懇請高手幫忙
--
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考試
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