電路學-問題Kirchhoff’s laws請幫手

Using the directions of current as indicated on the diagram:

For the large loop
4 = 5.(I1) + (10+8).(I3) -------- (1)
For the right loop:
2 = 15(I2) + (10+8).(I3) ----------- (2)
And I1 + I2 = I3
i.e. I2 = I3-I1
aubstitute into (2)
2 = 15(I3-I2) + 18(I3) = -15(I1) + 33(I3) --------------- (3)

(1)x3+(3)
14 = 87(I3)
i.e. I3 = 0.161 A
From (3): I1 = 0.221 A
I2 = (0.161 - 0.221) A = - 0.06 A

The -ve sign in front of I2 indicates that the actual current flow direction is opposite to that indicated on the diagram.
This is obvious as the 2v battery is being charged by the 4v battery.

究竟計出負數要點處理,(即點決定電流方向)




負數代表你圖中assign既方向係錯，電流應向相反方向流。

圖片參考：http://i164.photobucket.com/albums/u1/irwinting/temp/Untitled-1.jpg


I3 冇理由係負數, 因爲 I3 只係負載, 你可以先將15Ω拿掉,
1. I1
= 4 / ( 5+10+8 )
=4 / 23
=0.17391304A

2. R5Ω
=0.17391304 x 5
=0.8695652V

3. 4V - 2V - 0.8695652V
= 1.1304348V
I R15Ω
= 1.1304348/15
= 0.07536232A

所以 I2 = 0.07536232A 都唔係負電流