Help on a problem

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,

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.

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 --> 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 --> (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.

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.

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 & I

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

v=ri => 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.