Why Is I Negative? A Closer Look at the Norton Method

[transgalactic]

why "I" is negative??

http://img391.imageshack.us/my.php?image=img8855ew6.jpg

 

[Phys988]

I is negative because you consider it from - to + (as you show in your paper).

 

[transgalactic]

i know that KCL says that the sum of the currents that goes in
equals the sum of the currents that goes out

here i don't know what current goes into the node
and what current goes out??

 

[huckmank]

here i don't know what current goes into the node
and what current goes out??

And you don't need to. The sum of currents entering a node is equal to zero, the sum of currents leaving the node is also zero. Pick one and stay consistent.

If you choose to sum the currents leaving the node, then by the passive sign convention you have (e-v)/R1 + e/R2 - I = 0. If you choose to sum the currents entering the node you have (v-e)/R1 + (0-e)/R2 + I = 0, which is exactly the same as the previous equation once you sort out the signs.

So, to answer your question, you are summing the currents leaving the node. Since I is entering the node it follows that -I is leaving it.