標題:
current electricity
發問:
最佳解答:
Let I1 be the current in the branch AB flowing from B to ALet I2 be the current in the brach CD flowing from D to CThen, current in the branch EF is (I1 + I2), flowing from E to F For the loop connectin the 3v and 2v cells 3 - 2 = (I1).(1+3) - (I2).(1+2) i.e. 1 = 4(I1) - 3(I2) ------------------------ (1) For the loop connecting the 3v and 1v cells 3 - 1 = (I1).(4) + (I1+I2).(1+1)i.e. 1 = 3(I1) + (I2) -------------------- (2) From equations (1) and (2), solve for I1 and I2 gives I1 = 0.308 A I2 = 0.0769 A hence, (I1+I2) = 0.3849 A Therefore, p.d. between A and B = [3 - (1+3) x 0.308] v = 1.769 v p.d. between C and D = [2 - (1+2) x 0.0769] v = 1.769 vp.d. between E and F = [1+ (!+1) x 0.3849] v = 1.769 v
其他解答:7638E7481407D16B
current electricity
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圖片參考：http://imgcld.yimg.com/8/n/HA00081791/o/701103250116213873395000.jpg Find the potential difference between point A and B, C and D ,E and F represently And please show step最佳解答:
Let I1 be the current in the branch AB flowing from B to ALet I2 be the current in the brach CD flowing from D to CThen, current in the branch EF is (I1 + I2), flowing from E to F For the loop connectin the 3v and 2v cells 3 - 2 = (I1).(1+3) - (I2).(1+2) i.e. 1 = 4(I1) - 3(I2) ------------------------ (1) For the loop connecting the 3v and 1v cells 3 - 1 = (I1).(4) + (I1+I2).(1+1)i.e. 1 = 3(I1) + (I2) -------------------- (2) From equations (1) and (2), solve for I1 and I2 gives I1 = 0.308 A I2 = 0.0769 A hence, (I1+I2) = 0.3849 A Therefore, p.d. between A and B = [3 - (1+3) x 0.308] v = 1.769 v p.d. between C and D = [2 - (1+2) x 0.0769] v = 1.769 vp.d. between E and F = [1+ (!+1) x 0.3849] v = 1.769 v
其他解答:7638E7481407D16B
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