Understanding Norton Method: A Question on Negative I in KCL Analysis

  • Thread starter Thread starter transgalactic
  • Start date Start date
  • Tags Tags
    Method
AI Thread Summary
The discussion centers on understanding the direction of current in a node using Kirchhoff's Current Law (KCL). Participants clarify that the sign of current "I" can be negative depending on the chosen reference direction for current flow. It is emphasized that currents entering and exiting a node must balance according to KCL, and the direction of current can be defined as long as the same convention is consistently applied throughout the analysis. The confusion arises from the interpretation of voltage differentials (e-v vs. v-e) based on current direction. A suggestion is made to draw simple circuits to better visualize and understand current flow and voltage relationships.
transgalactic
Messages
1,386
Reaction score
0
why "I" is negative??

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

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??
 
Physics news on Phys.org
current in = current out
I (current into node) = (e-v)/R1 + e/R2 (current out of node)
 
i don't 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
 
you can make current go any way you want as long as you stick to the same convention throughout the circuit (this is just something to keep in mind)

In this case, you have 3 lines in your circuit with respect to that node. Current is not pointing up. That is just a symbol for current. It says current is flowing in that line going that direction. It is going 'into' the node. What goes in must come out equally. If it goes in 1A, it has to come out 1A.

Now, if you want to make current at R1 going into the node, you can do it too but then the convention changes.
ie:
going in = going out
I + (v-e)/R1 = e/R2
e/R2 - I -(v-e)/R1 = 0
e/r2 -I +(e-v)/R1 = 0

Observe the sign symbol; -(v-e) = e-v
see how it comes back the same? When current flows one way, then your I is the voltage differential (from) - (to). If you made R1 current 'going into', it will be (v-e)/r1. If you made R1 current 'going out' from node, it will be (e-v)/R1.

Hope this helps
 
what conventions?

i can't see how the currents flow using these conventions

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

It would be best if you draw a simple circuit, with voltage and 1 resistor. What is the current? What happens if you add 2 resistor in series now. What is the current and voltage between both resistor. Keep adding, another resistor in parallel with 2 resistor in series and find current/voltage ...etc

Try this site
Read thru it
http://www.allaboutcircuits.com/vol_1/chpt_6/4.html. Take a look at ohms law, simple current and voltage rules.
 
Last edited:
i know ohm laws

but here i can't construct them
because of these new symbols
they differ the normal symbols
 
transgalactic said:
i know ohm laws

but here i can't construct them
because of these new symbols
they differ the normal symbols

I'm not sure I understand your confusion. The symbol on the right is a current source. The arrow on the symbol shows which way the current is flowing through the current source. The "up" arrow means that the current is being pumped up from ground and into the top node.

The equation the way you have written it is summing all of the currents flowing *out* of that node, so the current from the current source leg of the circuit is marked as negative.
 
Back
Top