# NCEES Problem 106



## Volts006 (Jan 20, 2007)

Can anyone explain the solution to to this problem?

I think I am following the solution, but don't understand the reasoning.

It appears that they are just using ohm's law. They take 3A from the Isc subtract current when the load is applied and divide by the voltage at the load terminals.


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## jdd18vm (Jan 20, 2007)

gms

Scroll down to an Oct 2nd post by TTE. That was on the same quetion. I am going to relook myself. Im still not sure on that one either...hell the only was i HAVE been sure on are the NEC ones.

GL


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## Volts006 (Jan 20, 2007)

Ok, thanks jdd. I'll take a look at that previous post.


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## Volts006 (Jan 20, 2007)

Ok, thanks jdd. I'll take a look at that previous post.


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## Volts006 (Jan 21, 2007)

Thanks, jdd. I wonder if there is a way to rename that topic to "NCEES Problem 106". So, if others need it they can find it easier. For now here's the link and copy of his explanation

But, I do understand benbo's explanation. It makes a lot of since.

http://engineerboards.invisionzone.com/ind...t&amp;p=5022053

Maybe one of the moderators can explain, but the copy will do for now.

*From benbo, "*

_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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