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

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SUMMARY

The discussion centers on the application of Kirchhoff's Current Law (KCL) in analyzing circuit currents, specifically addressing the confusion surrounding the direction of current flow and the resulting negative values in calculations. Participants clarify that the direction of current can be defined arbitrarily, but must remain consistent throughout the analysis. The negative current value arises when the current source's direction is opposite to the assumed positive direction. Understanding these conventions is crucial for accurate circuit analysis.

PREREQUISITES
  • Basic understanding of Kirchhoff's Current Law (KCL)
  • Familiarity with circuit symbols and current direction conventions
  • Knowledge of Ohm's Law
  • Ability to analyze simple circuits with resistors and voltage sources
NEXT STEPS
  • Study the principles of Kirchhoff's Current Law in detail
  • Learn about circuit analysis techniques using nodal analysis
  • Explore the impact of current source direction on circuit equations
  • Practice drawing and analyzing circuits with multiple resistors and sources
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Electrical engineering students, circuit designers, and anyone involved in circuit analysis who seeks to deepen their understanding of current flow conventions and KCL applications.

transgalactic
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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??
 
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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.
 

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