電學 題目 http://img845.imageshack.us/i/ffffffffffff.jpg/ 圖片在 http://img858.imageshack.us/i/dddddddddddddd.jpg/ 我一開始的算法是 用歐姆定律 算出 電容板之前所剩電壓 然後那個電壓對其電容充電後得Q 但我看了解答似乎不是 解答是 Q =εC = (40 V)(2.0 ×10^-6F) = 80 μC 其中有一個註解是說 If the switch has been closed for a long time, the capacitor is fully charged and there is no current flowing through the right branch that contains the capacitor. So, a voltage of 60 V appears across the 60 Ω resistor and a voltage of 40 V appears across the 40 Ω resistor. That is, maximum voltage across the capacitor is 40 V. 為什麼 充電充很久 那個電容就不會有電流 為什麼就可以無視那一個branch 直接取與那個branch並聯所得的40V帶入那個電容得Q ?? --

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