a question on norton method..

**transgalactic** · 15th May 2008 05:22 PM

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
???

**Diver300** · 15th May 2008 08:16 PM

In this case, all the currents are taken as the currents leaving the node.

They all add up to zero. That means some will be negative.

The current leaving the node to the battery via R1 is (e-V)/R1
(this could be negative or positive, depending on lots of things)

The current leaving the node through R2 is e/R2
(This will be positive)

The current leaving the node through the current souce is -I
(This will be negative)

I hope this helps.

**transgalactic** · 16th May 2008 07:28 AM

Norton's Theorem : DC NETWORK ANALYSIS

in this article they present only one resistor between the lines
i can put many resistor between the lines
how do i recognize the load resistor??

after that they calculated they said that there is a total current 14A
i cant understand how did they get this number

and how did they decided that the norton equivalent must be 0.8????

**jacob.zurasky** · 17th May 2008 06:52 AM

The Norton current is calculated by replacing the load with a short circuit and then figuring out how much current flows through the short circuit. There are two loops in the figure shown with the short. The total voltage drop around the loop must be equal to zero. Each loop only has two components, a source and a resistor. So all the voltage provided by the source only has one component to drop all its voltage across. So each resistor sees the full sources voltage across it. So VR1 = 28v and VR3 = 7v. Then using I=V/R, you can calculate the current through each resistor. IR1 = 28v / 4 = 7A, and IR3 = 7v / 1 = 7A.

Then, using KCL, the sum of currents entering a node must equal the sum of the currents leaving the node. There are 14A leaving the node, so there also must be 14A entering the node.

The 0.8 ohms comes from calculating the circuit's resistance as seen by the load. Use Superposition and remove the sources, voltage sources become shorts, current sources become open circuits. The resulting circuit is only 2 resistors in parallel. (1*4)/(1+4) = 0.8 ohms.

Another good way to find the Norton Resistance is to remove the load. Calculate the voltage open circuit...Voc. Then calculate current shorted across the load connections...Isc. Rt = Rn = Voc/Isc. Voltage open circuit can be found by using voltage division. The voltage across R1 and R3 is 28-7=21v. The voltage across R1 can be found by 21v * (1/(1+4)) = 4.2v. So the total voltage across the load connections equals 4.2v + 7v = 11.2v.

Rt = Rn = (11.2v / 14A) = 0.8 ohms

-Jacob

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