問題詳情

22 一電容器連接於一 10 V 之直流電源上,以 1 mA 之電流向電容器充電 2 秒,則該電容器儲存之電能為多少焦耳?
(A)0.01
(B)0.001
(C)0.0001
(D)0.00001

參考答案

答案:A
難度:非常簡單0.925926
統計:A(175),B(8),C(5),D(1),E(0)

用户評論

awer89】評論

Q=0.001*2(I*t)=0.002W=0.5QV=0.5*0.002*10=0.01 J

NotAgain】評論

Q = C*VQ = I*tC*V = I*t =    C*10 = 1m*2s(秒)C = 0.2mW = 0.5*0.2m*(10*10) = 0.01

項弦予】評論

Q = I * t = 1m * 2 = 2mCW = 0.5CV2= 0.5QV = 0.5*2m*10 = 10mJ  →W = 10*0.001 = 0.01J