# questions regarding this norton article..

Discussion in 'Homework Help' started by transgalactic, May 16, 2008.

1. ### transgalactic Thread Starter Active Member

Apr 29, 2008
74
0

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????

2. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
8
To put many resistors between the lines, simply reduce the many resistors to their equivalent resistance.

3. ### transgalactic Thread Starter Active Member

Apr 29, 2008
74
0
i cant figure out how did they get 0.8 ohm???

4. ### Dave Retired Moderator

Nov 17, 2003
6,960
164
The 4$\Omega$ and 1$\Omega$ resistors are in parallel where the lumped resistance is given by:

$R = \frac{R_{1} x R_{2}}{R_{1} + R_{2}}$

And:

$R_{1} = 4\Omega$

$R_{2} = 1\Omega$

Dave

5. ### transgalactic Thread Starter Active Member

Apr 29, 2008
74
0
our two resistors connected one after the other
we suppose to add their value

what you did is presenting them as parallel
why did you did that?

Last edited: May 16, 2008
6. ### Dave Retired Moderator

Nov 17, 2003
6,960
164
You do add the resistances if the resistors in series; however these resistors are in parallel. Why?

If the two resistors were in series, then the current through the two device would have to flow through both resistors, whereas in parallel the current can flow through either of the two resistors independently because they have common nodes.

There are two current paths from the two nodes over which the resistance is measured - the first path is through the 4 $\Omega$ resistor and the second independent path is through the 1 $\Omega$. Two independent current paths indicates the resistors are in parallel.

Would it help if I drew the resistor arrangement as in the attachment?

Dave

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7. ### transgalactic Thread Starter Active Member

Apr 29, 2008
74
0
what happens to the load resistor (the 2 ohm resistor component)

???

8. ### Caveman Senior Member

Apr 15, 2008
471
1
You have to follow all of the steps:
2. B1 and R1 can be converted to their norton equivalent. This replaces B1 and R1 with a current source, I1 and parallel resistor, R4. R4 is 4 ohms.
3. B2 and R3 can be converted to their norton equivalent. This replaces B2 and R3 with a current source, I2 and parallel resistor, R5. R5 is 1 ohm.
4. Since I1 and I2 are in parallel, you can just add their currents together.
5. Since R4 and R5 are in parallel, you can replace them with their parallel equivalent, which I will call Req. Dave's posting is just showing how to do that.
6. The circuit you have now is a single current source in parallel with Req and in parallel with the original load resistor, R2 = 2 Ohms. This is the last schematic on http://www.allaboutcircuits.com/vol_1/chpt_10/9.html.

Now let's go back to exactly what you are doing. You are trying to come up with an equivalent circuit to everything that is not the load resistor. The reason you would want to do that is that you can define easily what would happen if you changed the load resistor. This is what the table at the bottom of that page shows with Rloads of 2, 4, and 8 ohms.

9. ### Dave Retired Moderator

Nov 17, 2003
6,960
164
Further to Caveman's post above: Remember the point of the Norton theorem, you are trying to deduce the Norton Equivalent Circuit (a current source in parallel with an equivalent resistance) connected to a load - this load is independent of the equivalent resistance.

To quote the e-book:

So when you come to deduce the Norton resistance you need to remove the load resistance from the calculations.

Dave

10. ### transgalactic Thread Starter Active Member

Apr 29, 2008
74
0
ok i understand the part of the resistors

i cant understand how to determine the size of the current source??

11. ### Caveman Senior Member

Apr 15, 2008
471
1
The equivalent current source is the amount of current that would flow if the two nodes were short circuited. So if you are trying to replace B1 and R1, then you shorted R1 to ground, you would get a current of B1/R1 = I1.

I've seen that you are having trouble with a lot of concepts here and in other threads. And I don't mind helping out, but I would like to suggest that you get a good book that has answers to problems in the back. If you insist on learning from online sources only, you should go to various sources to get different explanations of the concepts. That way you have a good chance of understanding at least one of the explanations. Sometimes it makes sense when explained one way vs. another.

For example why don't you read the wikipedia article: http://en.wikipedia.org/wiki/Norton's_theorem.
It goes into the details a bit and it is a slightly different than the AAC explanation.

Last edited: May 16, 2008
12. ### transgalactic Thread Starter Active Member

Apr 29, 2008
74
0
thanks
correct me if i am wrong
we short out our load resistor
and count how much voltage will go threw this line
i did this using kirshgof equations

and i got the same result
am i correct??

13. ### Dave Retired Moderator

Nov 17, 2003
6,960
164
How much current through the line - it is a short circuit so there will be no voltage across anything.

You are right, you use KCL to work out the current in each branch.

Dave

14. ### Dave Retired Moderator

Nov 17, 2003
6,960
164
If you want to practice your skills, try the Socratic worksheet: http://www.ibiblio.org/kuphaldt/socratic/output/thev.pdf

Note it covers Thevenin and Maximum Power Transfer theorems as well as Norton theorems, but it should be clear which questions are Norton's theorems.

Dave