✔ 最佳答案
Use P = V^2 / R to calculate the internal resistances of 2 light bulbs.
The one with higher resistance should be brighter as they are connected in series, i.e. they draw the same current.
2007-01-14 17:08:39 補充:
kaiyan0226, please also consider the current that two bulbs draw. The resistances of them are different.
2007-01-15 17:27:31 補充:
謝謝Chan Chi Wang的回應, 但我要補充一點, 我沒有說過答案不是B.用公式 P = V^2 / R,可得出A和B燈泡的電阻分別為3.6和7.2歐姆,因此B燈泡的電阻確是大於A燈泡的,而燈泡的電壓可用 P' = I^2 * R計出, (注意: 先前的P值跟這個P'值有不同意思)由於串聯緣故,通過兩個燈泡的電流一樣,電壓跟電阻成正比,又B燈泡的電阻大於A燈泡的,所以B燈泡輸出的電壓較A燈泡的為大.
2007-01-16 23:08:19 補充:
不好意思, 用錯了名詞, 以上的電壓應該改為功率(Power).
2007-01-16 23:10:21 補充:
方便起見, 我重答一次吧.用公式 P = V^2 / R,可得出A和B燈泡的電阻分別為3.6和7.2歐姆,因此B燈泡的電阻確是大於A燈泡的,而燈泡的功率可用 P' = I^2 * R計出, (注意: 先前的P值跟這個P'值有不同意思)由於串聯緣故,通過兩個燈泡的電流一樣,功率跟電阻成正比,又B燈泡的電阻大於A燈泡的,所以B燈泡輸出的功率較A燈泡的為大.