# Norton's Theorem Question

Discussion in 'General Electronics' started by Mrs. Kerchief, Nov 10, 2003.

1. ### Mrs. KerchiefGuest

Why do you not use Rn to compute the resistance when you're trying to
arrive at In? (Or do you?) Am I wrong thinking you arrive at In
using the first resistor closest to the output terminals (as with
Thevenin's Theorem)?

My DC instructor spends one class per theorem, obviously not too
little for other students; and obviously I'm never going to be an
electrical engineer. I have a B average and would like to get my
associate's in computer technology. (Plus I would dearly love to
understand material I have shelled out big bucks to assimilate.)

Anyway, if Rn=Thn, could someone tell me how you arrive at -48.2mA for
In in the following problem? The Rn here was given at 56.9. (The 33
resistor is depicted as being in series with the 3 Volt source; the
other two are depicted in parallel.)

|-------------------------------------------------------A
| | |
| | |
| | |
330 Ohms 100 Ohms 200 Ohms
| | |
| | |
| | |
+ 3 V | |
| | |
| | |
|-------------------------------------------------------B

Also, as long as we're on Norton's, and if anyone is so inclined,
could you help with In for the following?
R3
|-----15K----|
R1 | |
--------22K-----------|Is|----------------------------A
| | 10mA |
| | |
| R2 | |R4
|12V 8.2K Ohms 10K Ohms
| | |
| | |
| | |
|-----------------------------------------------------B

2. ### Ken TaylorGuest

You need to be more careful with your notation - assuming the answer is
right, the 200 Ohm resistor is actually 220 Ohm (check the problem) - I only
got that because it's a standard value and 200 isn't (no prizes to anyone
who points out precision resistor series! You also stated that the
series resistor was 33 Ohms when it's shown as 330 - you aren't being
careful enough with the problem. Take a bit more time getting the facts down
correctly.

Have another crack at it now and get back to us.

Ken

3. ### Mrs. KerchiefGuest

Sorry for the top-post. Yes, you're correct--my notation was wrong in
the ways you stated. As for the first schematic, I honestly don't
know what happened when the post appeared; everything's thrown out of
whack.

Anyway, if you or anyone is so inclined, emendations below.

*************************************************
(The 330 Ohms resistor is in series with the battery; the others are
in parallel. I consistently get the right Norton equivalent
resistance but only once, apparently by fluke, was able to compute
Norton equivalent current correctly. The answers to the below were
-48.2mA (In) and 56.9 (Rn).
_____________________________________________________A
| | |
| | |
| | |
| | |
330 Ohms | |
| | 220 Ohms
| 100 Ohms |
| | |
+ 3 V | |
= | |
| | |
| | |
| | |
____________________|___________________|_________________B

Thanks again.

4. ### FrazeGuest

I'm going to have to assume that something is wrong with the values of
your circuit. Since the RN is correct I would say that the values and
configuration of you resistors are right. For an IN of -48.2mA the
supply would have to be -15.9V. Could you recheck the original
problem in the book and let us know? Anyway, here is Norton in a
nuttshell. To find the norton current source you would short
terminals A and B and calculate the current that would flow through
terminals A and B. RN is the same as RTH but it is now shunted across
the current source. Lets say that the values on your first circuit
were correct. If you shorted the terminals you would basically have a
3 volt supply with 330 ohms in series. This would source 9.091mA, the
direction would be determined by the polarity of the voltage source
and whether you were using conventional or electron current flow. If
you would repost the circuit after checking it in the book I might be
able to explain it further.
As for the second circuit, you would need to tell me the polarity of
the sources and let me know what type current flow you are using in
Hope this helps.
Fraze

5. ### Mrs. KerchiefGuest

Thank you for taking the time to respond. We're using
conventional current in the course, and in the first problem the
polarity of the 3 Volts is positive on top, negative on the bottom of
the schematic. I think the In in this problem *must* be wrong as
given in the answer section in the back of the book. (I've found
another wrong question in this text, Principles of Electric Current,
by Thomas Floyd.)
The polarity for this second problem is the same (positive on top,
negative on bottom). As the answer for this question is not given in
the back of the book, I was simply asking for help because I wanted to
check my answers. The question was not assigned for homework. Again,
thank you, Fraze.