[問題] 電路功率
【出處】大學普通物理
【題目】two cables in series are coupled and have I=740A over a distance
of 9.0km. Both cables have R=0.80歐姆. If the ingoing voltage is 42kV.
a) What is the outgoing voltage? (A: 4.1 x 10^4 V)
b) What is the power that goes in? (A: 3.1 x 10^7 W)
c) What is the lost power? (A: 8.8 x 10^5 W)
d) What is the outgoing power? (A: 3.0 x 10^7 W)
【瓶頸】(解題瓶頸或思考脈絡，請盡量詳述以利回答者知道要從何處講解指導)
對於a):
我對題目的理解是兩條串連的cables (可是我不確定是不是交流電),
因為 42kV - 740*0.8*2(條) 大約等於 40.8kV 又約等於4.1kV
可是b): 我用 power = V^2/(2R) = 1.1 x 10^9 W. 跟答案差很多
而c) 我用I^2*R*2(條) = 8.76 x 10^5 W, 跟答案滿接近的(所以是直流電的意思?)
d) 同b), 我用類似算法, 但一樣跟答案差很多
所以如果假設我 a), c)沒算錯的話, 那b), d) 不知為啥又差很多,
還煩請各位給個提示了, 感謝Orz
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恩恩
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謝謝, 我會再好好想想~
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謝謝兩位的指教~
※ 編輯: LittleIe (123.193.58.198), 07/15/2017 23:22:21
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