# Norton Resistance Question

Discussion in 'Electronic Basics' started by [email protected], Nov 18, 2008.

1. ### Guest

Hi guys,

I've gotten a lot of repsonses, but no real answers. To clariy, the
text says to remove RL and short the source. To you guys that think
this is my homework.... good thing it is not. This is actually on a
computer aided instruction course at my tech school. It states the
answer as 21 ohms. I came up with 20.55 ohms. It generally wants two
decimal places, so I would be shocked if it rounded 20.55 to 21. So, I
was picking other folks brains.

Here is the original post:

Can someone tell me (step by step) how to calculate the Norton's
resistance for this circuit? (see picture below)

Thanks!
Lee

2. ### sertGuest

wrote in
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.

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.

3. ### Guest

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

4. ### Guest

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.

5. ### Guest

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.

6. ### Rich GriseGuest

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

7. ### Guest

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!!!

8. ### Guest

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*******

9. ### Jasen BettsGuest

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

10. ### Guest

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

11. ### Rich GriseGuest

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

Have Fun!
Rich

12. ### Bob PownallGuest

I probably should have posted this link to "How To Ask Questions The
Smart Way" earlier in this thread...
http://www.catb.org/~esr/faqs/smart-questions.html  