a question on norton method..

Thread Starter

transgalactic

Joined Apr 29, 2008
74
http://img391.imageshack.us/my.php?image=img8855ew6.jpg

why "I" is negative??
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 dont know what current goes into the node
and what current goes out??
i dont know how you decided the direction of each current

i am new to this stuff
and i think that the "I" current is pointed up
so it goes into the node
the second current the goes threw "R1" also come into the node
because the currect goes from the plus of the battery to minus of the battery

where did i go wrong??

i cant see the conventions here

i cant see how the currents flow using these conventions

in what case we have e-v and in what v-e
???
 

Caveman

Joined Apr 15, 2008
471
http://img391.imageshack.us/my.php?image=img8855ew6.jpg

why "I" is negative??
i know that KCL says that the sum of the currents that goes in
equals the sum of the currents that goes out
You pick the direction of the currents. If you pick backwards from actual current flow, the answer will come out negative. It will be correct.

In this case, you are saying that all of the currents are going out. Since none are going in (because you said), the other side is 0.

\[\frac{e - V}{R_{1}\] is the current going out through R1.
\[\frac{e}{R_{1}\] is the current going out through R2.
And since you are talking about currents going out, you must use -I

You could just put it on the other side of the equation and it would be positive. But through simple algebra you can see that they are the same.
 

Caveman

Joined Apr 15, 2008
471
It is not called the norton method. That was my point. And it doesn't really use KVL. It primarily uses KCL at the various nodes to provide a methodical process for solving these kinds of circuits.
 

Thread Starter

transgalactic

Joined Apr 29, 2008
74
this example is for using norton method on a circuit

where does the norton stuff come out??
 
Last edited:

Dave

Joined Nov 17, 2003
6,960
this example is for using norton method on a circuit

where does the norton stuff come out??
The premise of Norton Theorem is that you can reduce (simplify) a linear circuit to a single current source and parallel resistance (the Norton Equivalent Circuit) driving a load.

See: http://www.allaboutcircuits.com/vol_1/chpt_10/9.html

Also see the related Thevenin Theorem: http://www.allaboutcircuits.com/vol_1/chpt_10/8.html

And finally the Norton-Thevenin Equivalences: http://www.allaboutcircuits.com/vol_1/chpt_10/10.html

Dave
 
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