Norton Resistance Question

snipped-for-privacy@progressivetel.com wrote in news: snipped-for-privacy@f37g2000pri.googlegr oups.com:

There are so many tutorials on the web and in books that you shouldn't have a problem with such an easy circuit.

But, to answer your question, there are several equivalent ways to reach your desired conclusion.

The easiest here is to repeatedly convert your voltage source to a current source and vice versa.

You convert a voltage source E (in series with a resistor R) to a current source I by replacing it with a current source valued I = E/R. The same resistance is then removed and placed parallel to the current source. That leaves you with two resistances in parallel which you can combine into one and repeat the procedure converting the other way round.

That means that the first voltage source with a current source I = E/R = 50/5 = 10 A and the 5 Ohm resistance will go in parallel. You then combine the two 5 Ohm resistances into one

2.5 Ohm one and replace the 10 A source with a voltage source with E = I*R = 10*2.5 = 25 V. You proceed that way until you reach the Norton equivalent.

People here are reluctant to give away solutions to what appears to be homework. Hints yes, suggestions yes, but answers without some effort on the requester's part, no.

So...it's course work but not homework... Perhaps you could explain the distinction.

You didn't mention the further details about RL in your original question, yet you expect to obtain meaningfull answers?

20.555... actually.

And you never mentioned this as the being the motivation for your post. Why not? Are you always less than forthcoming about your motives?

You will find people here can be very helpful and are willing to help you understand things, but you have to show some initiative on your part. Show what you've tried, where you're stuck or confused.

Wow.. never mind guys.... I didn't expect this much pent up hostility. Honestly.... i will move on. I simply wanted to validate if the screen was rounding, I have the computer's answer, but I was simply trying to validate it. As I stated in my 2nd post, (and per the instruction on the computer) I remove RL and short the load. After that I calculate resistance using Ohm's Law. I got 20.55. The computer says it is 21 (multiple guess). I was wondering if my Ohm's law is flawed, or if the computer is rounding. I spent 3 hours trying to get an answer on my own. SO IT IS NOT A LACK OF EFFORT. But C'est la vie.... other things need my attention. I will move on. I do appreciate the responses.

Lee

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Hostility? What's hostile about being careful not to give away homework answers to individuals who show no evidence of effort?

A clear, complete question without alterior motives and hidden clauses would also help.

That which is claimed without evidence can be dismissed without evidence.

No problem. It's nice to see that you were able to solve your problem on your own.

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If you think it is homework, then email me on January 1st with your answer. (I would suspect that my "homework" would be due before then) Let's see if you can do it. And by the way, it's ulterior, not alterior.

Oops. My bad. Thanks for pointing that out.

Tell you what, I'll tell you now that the Norton current is 20/37 Amps. The resistance is 185/9 Ohms. If you want to see how I arrived at those numbers, first show us how you solved the problem yourself.

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The two 5 ohm resistors in parallel (R1 and R4) give you 2.5 ohms. This is added to the 10 ohm (R2). This 12.5 ohms is in parallel with an additional 10 ohm (R5). This equates to 5.55555555 ohms added to (R3) 15 ohms. 15 + 5.55 is 20.55 (which is what i got). The %^&* computer ALWAYS wants two decimal places, but this time the choices were 17, 21 and 25... I think they dropped the ball. Or broke their own rounding rule. But - THAT - is how I arrived at my answer.

And that's fine. Works for me. So they dodn't ask for the complete Norton equivalent, including the current generator value?

Given that it was a multiple choice question you're free to pick the answer closest to the correct value.

I had assumed, from your initial post, that the whole Norton model was required (both current and resistance). And since I happen to have Mathcad running here, I chose to solve it by matrix methods. First I wrote down the loop equations fo rthe three loops and solved for the current in the last loop with RL set to zero (short circuit current, which is the Norton current).

Then I removed RL to leave two loops and solved for the voltage across R5 giving the open circuit voltage. The ratio of the open circuit voltage to the short circuit current is the Norton resistance.

Loop equations (easily written by inspection):

- - - - - - | R1 + R4 -R4 0 || i1 | | V | | -R4 R2 + R4 + R5 -R5 || i2 | = | 0 | | 0 -R5 R3 + R5 +RL || i3 | | 0 | - - - - - -

The text is correct that shorting the source and removing RL, then computing the resistance at RL's terminals, is the correct method (don't short a 50V power supply in real life, though, at least not if it's capable of some serious current :-).

Consider: If the resistors are all 5% accuracy, how close is your 20.55 going to be to the 'correct' answer? Even at 1%, is there any real difference between 20.55 and 21?

Yes, the programmers of the app should have been more consistent, but that's life.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

On Tue, 18 Nov 2008 06:50:37 -0800, radlee wrote: (top posting fixed)

There's a reason for the hostility, although I don't think you rate it.

There's a word to describe people who skate through technical courses by getting others to do their homework.

That word is "incompetent".

Since technical people usually depend on the work of others, incompetent colleagues make everyone look bad. Capable incompetents do this while looking good themselves, and get promoted to management, where they _really_ make life hell for the competent folks.

So we do all we can to encourage such folks to quit tech school and go be incompetent in some entirely different field.

If you have a question about homework, say "this is homework for such and such a class", then ask for details that will help you come to the answer yourself. That way we won't think we're being asked to supply answers that you can just copy in (and believe me, plenty of folks ask for just that, particularly around end-of-term time).

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

Did they say two _decimal places_, or two _significant figures_, which you're interpreting as decimal places? (the two are not the same.)?

Thanks, Rich

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Wow... I must say that technical arrogance is not just limited to my little corner of the world. Here are the factst. I GRADUATED technical school in 1983. I am taking some core classes at the local tech school (to get my associates). A friend of mine is an adjunct instructor there and we sat down to go through some of these courses (It's been 25 years for God's sake) to see what we could remember. We were rocking along pretty well until we got here. It piqued my curiosity and I wanted to make sure I wasn't totally brain dead.

I didn't know that a simple question would breed more accusations than an episode of the Jerry Springer Show. MY BAD!!! For crying out loud, put down the scope probe, turn off the function generator, find a girl and do what comes naturally. I know the economy is bad and people are a bit uptight.. but JEEEZZZ!!!

There is no way for anyone here to know the details of your situation and background before you state them. Further, there are many instances of individuals who have concocted all sorts of stories claiming to be anything other than a student, yet looking for effort-free solutions to what appear for all the world to be homework problems. So naturally people have become somewhat thick crusted about such requests, and nearly always ask to see what the poster has tried so far to solve the problem.

Now you may very well be the finest most upstanding gentleman on the net, someone I'd be honored to share a beer with sometime, but alas there's no way to tell that from a single post that asked for free help on what appeared to be a circuit problem typical of homework questions, asking for a step by step solution no less, and in fact not even containing enough information to state the problem in its entirety (a typical student oversight) and was in fact a multiple choice question, and besides all that you were *really* looking for a justification for the problem's choice of precision in the stated answers...

A bit if a tangled web...

Anyways,

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OK!!! OK!!! I concede. I harken back to the days when the web was a friendly place. Like Main Street on a summer's day. I forget sometimes that the world has changed around me. But alas, I suppose that it is "Change We Need"..... *****SIGH*******

It's good to see evidence of an effort being made.

the impedance left of R4 is 0 + 5 = 5 left of R2 5 || 5 = 2.5 left of R5 2.5 + 10 = 12.5 left of R3 12.5 || 10 = 5+(5/9) left of RL = 15 + 5+(5/9) = 20+(5/9) = 20.56 I get (essentially) the same answer as you. my question is do they say "two significant figures" or "two decimal places."

real resistors typically aren't precision devices a good one will be within 2% of the marked value

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Past lessons were always precise. If you omit one of the 5 ohm resistors, then that gives you 15 ohms in parallel with a 10 ohm. That becomes a whole 6 ohms added to the 15 ohm, which would give a precise 21 ohms. I went back and looked at the lesson criteria and question, and there was no verbage stating otherwise. Questions before and after included decimals on similar questions. Go figure. I just wanted to be sure that I was still trainable.

Lee

(people would be significantly less "hostile" if you'd checked that first.)

Have Fun! Rich

I probably should have posted this link to "How To Ask Questions The Smart Way" earlier in this thread...

It's about asking questions on computer software and hardware problems in a way that gets you useful answers, but really - it applies to pretty much any topic.

Bob Pownall