# Help on a problem



## TTE (Sep 29, 2006)

problem 106 (norton theorem) in the NCEES sample questions manual is giving me problems. I havn't done this stuff in a long time. An explanation of the derived answer would help.

Thank you,


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## benbo (Sep 29, 2006)

I'm at work right now and don't have my book. If nobody else answers, I'll dig it out when I get home and see if I can help.


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## benbo (Sep 30, 2006)

Sorry it took me so long. I couldn't find my book.

First, there is probably some way to fit this into a neat formula. But I didn't work this using these Norton equations. A lot of times it is hard to fit these problems exactly into a given equation so I used common sense and a little circuit theory.

Second, I hate Siemens. I alway convert to ohms.

First, the Norton current is equal to the short circuit current, or 3A. Even if you didn't know this, assume the network is a resistor with a current source. Connect a short across the resistor and obviously all the current from the source flows through the short, or 3A. So, you now have either answer B or D.

Draw the picture. You have a 3A source with two resistors across it. One is unknown and the other is .2 S or 1/.2 = 5 ohms.

Since the voltage across the parallel resistors is going to be 5V, that means (by ohms law) the equivalent resistance will be R = V/I = 5/3 = 1.67 ohm.

You have two choices for the unknown --&gt; 1/.4 = 2.5 ohms, or 1/.6 = 1.67 ohm. So, which resistance, in parallel with the given 5 ohms, will give 1.67 ohm. You don't even have to calculate. It can't be the 1.67 ohm, it has to be 2.5.

But just in case, try 2.5 --&gt; (2.5*5)/(2.5+5) = 12.5/7.5 = 1.67 (this is the parallel combined value I want). So I pick B = 1/2.5 ohm = .4 S.

Hopefully you can understand my explanation, or somebody will explain the plug in method, but this is about as tough of a circuits problem as you'll get in the AM. They can have little tricks, and drawing the picture and doing a little thinking pays off.


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## TTE (Oct 2, 2006)

Thanks, I appreciated you taking the time to answer the question. The problem I was having was setting up the problem. I didnt understand where to put the unkown resistor and how to handle the known resistor.

Thanks for the help.


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## Texass (Oct 16, 2008)

sorta......but

since

s.c = 3A, this means with no load, we have 3 amps

and the eq. voltage is 5 volts with a conductance of 0.2 S (conductance = 1/R), so r = 5 ohms

now, we know v = ri and we know that v = 5volts and so we are solving for R &amp; I

lets see how much current there is through the conductance so ...

v=ri =&gt; 5/R = I ..... 5/5 = 1 amp

so one amp is being used, but there are three amps when nothing is being used, therefore we need something to absorb those 2 amps

so.... v = ri (whole system) ...... 5 = r * 2 ....... therefore r = 5/2 and g = 2/5

that means if we used the current source and the load, we would have a difference of 2 amps that need to be absorbed into the system, therefore a resistance of (5/2) would do the trick.


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