16. The figure shows an array of identical real batteries, each with emf 0.050 V and internal resistance 0.20 ohm. There are 400 branches, each with 200 batteries. What is the current ( A ) through the external resistor, which has resistance R = 0.40 Ω?
(A) 2.5,
(B) 5.6,
(C) 20,
(D) 12,
(E) 58.
(A) 2.5,
(B) 5.6,
(C) 20,
(D) 12,
(E) 58.
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統計: A(0), B(0), C(1), D(0), E(0) #3332075
統計: A(0), B(0), C(1), D(0), E(0) #3332075
詳解 (共 2 筆)
#7399481
所以這題要先了解串聯電壓會疊加,內電阻也會疊加
並聯電壓不會增加,內電阻會平分變小
所以第一步算出一整排的電壓和電阻
每一排200個電池串聯
單排電壓:200*0.05=10V
單排內電阻:200*0.2歐姆=40歐姆
第二步算出400排並聯的總電壓和總內電阻
總電壓:並聯電壓不變,仍然是10V
並聯內電阻會變小,所以把電阻除以總排數40歐姆/400=0.1歐姆
最後一步套用歐姆定律求電流

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