questions regarding this norton article..

Thread Starter

transgalactic

Joined Apr 29, 2008
74
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?
 
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Dave

Joined Nov 17, 2003
6,969
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?
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.

Look at this picture: http://sub.allaboutcircuits.com/images/00240.png

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
 

Attachments

Caveman

Joined Apr 15, 2008
471
You have to follow all of the steps:
1. Start with the schematic at the top of http://www.allaboutcircuits.com/vol_1/chpt_10/9.html
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.
 

Dave

Joined Nov 17, 2003
6,969
what happens to the load resistor (the 2 ohm resistor component)

???
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:

Norton's Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single current source and parallel resistance connected to a load.
So when you come to deduce the Norton resistance you need to remove the load resistance from the calculations.

Dave
 

Caveman

Joined Apr 15, 2008
471
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.
 
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Thread Starter

transgalactic

Joined Apr 29, 2008
74
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??
 

Dave

Joined Nov 17, 2003
6,969
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??
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
 
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