A. 資料不足，無法計算。
B. 2P
C. 4P
D. 8P
E. 16P
圖：http://www.flickr.com/photos/harrykmb80_89b/8430439694/in/photostream
更新1:
詳細解釋，謝謝！
物理問題一問(電和磁)
2013-01-31 8:37 am
圖(A)顯示兩個相同的電阻器以串聯連接一個內電阻可略、電動勢為V的電池。每個電阻器消秏的功率為P。若該兩電阻器現以並聯連接，如圖(B)所示，每個電阻器消秏的功率是多少？
A. 資料不足，無法計算。
B. 2P
C. 4P
D. 8P
E. 16P
圖：http://www.flickr.com/photos/harrykmb80_89b/8430439694/in/photostream
A. 資料不足，無法計算。
B. 2P
C. 4P
D. 8P
E. 16P
圖：http://www.flickr.com/photos/harrykmb80_89b/8430439694/in/photostream
回答 (2)
2013-01-31 4:10 pm
✔ 最佳答案
In diagram A, voltage across each resistor = V/2Hence, power consumed by each resistor P = (V/2)^2/R = V^2/4R
In diagram B, since the two resistor is connected in parallel, voltage across each resistor = V
New power consumed by each resistor P' = V^2/R
Therefore, P'/P = 4
i.e. P' = 4P
The answer is option C.
2013-01-31 4:21 pm
Power, P = I^2R. For R to be constant, P is directly proportional to I^2.
In (a), Current flowing through R, = V/(2R) = V/2R
In (b), Current flowing through R =V/R
Let new power be P'
so P'/P = [ (V/R)/(V/2R)]^2 = 4
so P' = 4P (C).
In (a), Current flowing through R, = V/(2R) = V/2R
In (b), Current flowing through R =V/R
Let new power be P'
so P'/P = [ (V/R)/(V/2R)]^2 = 4
so P' = 4P (C).
收錄日期: 2021-04-25 22:38:57
原文連結 [永久失效]:
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