✔ 最佳答案
圖片參考：
http://xb9.xanga.com/ae7c7622d4036158088995/z118658398.bmp
Q :
R = 1Ω, VR = 0 ~ 4Ω, Bulb = 4Ω & B = 3V
A :
if the VR set to 4Ω and parallel with the bulb 4Ω = Ra = 1/(1/4 + 1/4) = 1/0.5 = 2Ω, then series with R1Ω = 3Ω(Rt), It = V/Rt = 3/3 = 1A, the Va will be = IxRa = 1x2=2V, so the current through the bulb is Va x Rb = 2/4=0.5A.
if the VR reduced to 2Ω and parallel with the bulb 4Ω = Ra = 1/(1/2 + 1/4) = 1/0.75 = 1.33Ω, then series with R1Ω = 2.33Ω(Rt), It = V/Rt = 3/2.33 = 1.29A, the Va will be = IxRa = 1.29x1.33=1.71V, so the current through the bulb is Va x Rb = 1.71/4=0.4275A.
0.5 - 0.4275 = 0.0725
電流係佐0.0725A，所以個 bulb 暗佐啦。
2007-11-19 00:15:26 補充：
The power in the bulb : Va x Ib = 1.2 x 0.5 = 1Watt2.1.71 x 0.4275 = 0.731Watt
2007-11-19 18:16:19 補充：
電流細佐0.0725A，所以個 bulb 暗佐啦。
2007-11-19 18:18:19 補充：
最好你將有關資料備份，不然有關問題被人檢舉，移除，你就喊都無眼淚。
