Norton Resistance Question

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)

http://www.progressivetel.com/~radney4/NT.jpg

Thanks!
Lee

 

[Greg Neill]

radlee@progressivetel.com wrote:

Hi guys,

I've gotten a lot of repsonses, but no real answers.

People here are reluctant to give away solutions to what appears
to be homework. Hints yes, suggestions yes, but answers without
some effort on the requester's part, no.

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.

So...it's course work but not homework...
Perhaps you could explain the distinction.

You didn't mention the further details about RL in your original
question, yet you expect to obtain meaningfull answers?

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.

20.555... actually.

And you never mentioned this as the being the motivation for your
post. Why not? Are you always less than forthcoming about your
motives?

You will find people here can be very helpful and are willing to
help you understand things, but you have to show some initiative
on your part. Show what you've tried, where you're stuck or
confused.

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)

http://www.progressivetel.com/~radney4/NT.jpg

Thanks!
Lee

 

[sert]

radlee@progressivetel.com wrote in
news:5c09eaf0-185a-41b9-b144-9a55fecc6ab8@f37g2000pri.googlegr
oups.com:

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

There are so many tutorials on the web and in books that you
shouldn't have a problem with such an easy circuit.

But, to answer your question, there are several equivalent
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.

 

[Greg Neill]

radlee@progressivetel.com wrote:

Wow.. never mind guys.... I didn't expect this much pent up
hostility.

Hostility? What's hostile about being careful not to give away
homework answers to individuals who show no evidence of effort?

A clear, complete question without alterior motives and hidden
clauses would also help.

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.

That which is claimed without evidence can be dismissed without
evidence.

But C'est
la vie.... other things need my attention. I will move on. I do
appreciate the responses.

No problem. It's nice to see that you were able to solve your
problem on your own.

 

[Greg Neill]

radlee@progressivetel.com wrote:

On Nov 18, 10:07 am, "Greg Neill" <gneil...@MOVEsympatico.ca> wrote:
rad...@progressivetel.com wrote:
Wow.. never mind guys.... I didn't expect this much pent up
hostility.

Hostility? What's hostile about being careful not to give away
homework answers to individuals who show no evidence of effort?

A clear, complete question without alterior motives and hidden
clauses would also help.

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.

That which is claimed without evidence can be dismissed without
evidence.

But C'est
la vie.... other things need my attention. I will move on. I do
appreciate the responses.

No problem. It's nice to see that you were able to solve your
problem on your own.

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.

Oops. My bad. Thanks for pointing that out.

Tell you what, I'll tell you now that the Norton current is 20/37 Amps.
The resistance is 185/9 Ohms. If you want to see how I arrived at those
numbers, first show us how you solved the problem yourself.

 

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



On Nov 18, 9:35 am, "Greg Neill" <gneil...@MOVEsympatico.ca> wrote:

rad...@progressivetel.com wrote:
Hi guys,

I've gotten a lot of repsonses, but no real answers.

People here are reluctant to give away solutions to what appears
to be homework.  Hints yes, suggestions yes, but answers without
some effort on the requester's part, no.

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.

So...it's course work but not homework...
Perhaps you could explain the distinction.

You didn't mention the further details about RL in your original
question, yet you expect to obtain meaningfull answers?

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.

20.555... actually.

And you never mentioned this as the being the motivation for your
post.  Why not?  Are you always less than forthcoming about your
motives?

You will find people here can be very helpful and are willing to
help you understand things, but you have to show some initiative
on your part.  Show what you've tried, where you're stuck or
confused.



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)

http://www.progressivetel.com/~radney4/NT.jpg

Thanks!
Lee

 

On Nov 18, 10:07 am, "Greg Neill" <gneil...@MOVEsympatico.ca> wrote:

rad...@progressivetel.com wrote:
Wow.. never mind guys.... I didn't expect this much pent up
hostility.

Hostility?  What's hostile about being careful not to give away
homework answers to individuals who show no evidence of effort?

A clear, complete question without alterior motives and hidden
clauses would also help.

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.

That which is claimed without evidence can be dismissed without
evidence.

But C'est
la vie.... other things need my attention.  I will move on.  I do
appreciate the responses.

No problem.  It's nice to see that you were able to solve your
problem on your own.

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.

 

[Greg Neill]

radlee@progressivetel.com wrote:

On Nov 18, 10:40 am, "Greg Neill" <gneil...@MOVEsympatico.ca> wrote:
rad...@progressivetel.com wrote:
On Nov 18, 10:07 am, "Greg Neill" <gneil...@MOVEsympatico.ca> wrote:
rad...@progressivetel.com wrote:
Wow.. never mind guys.... I didn't expect this much pent up
hostility.

Hostility? What's hostile about being careful not to give away
homework answers to individuals who show no evidence of effort?

A clear, complete question without alterior motives and hidden
clauses would also help.

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.

That which is claimed without evidence can be dismissed without
evidence.

But C'est
la vie.... other things need my attention. I will move on. I do
appreciate the responses.

No problem. It's nice to see that you were able to solve your
problem on your own.

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.

Oops. My bad. Thanks for pointing that out.

Tell you what, I'll tell you now that the Norton current is 20/37 Amps.
The resistance is 185/9 Ohms. If you want to see how I arrived at those
numbers, first show us how you solved the problem yourself.

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.

And that's fine. Works for me. So they dodn't ask for the complete
Norton equivalent, including the current generator value?

Given that it was a multiple choice question you're free to pick
the answer closest to the correct value.

I had assumed, from your initial post, that the whole Norton model was
required (both current and resistance). And since I happen to have
Mathcad running here, I chose to solve it by matrix methods. First
I wrote down the loop equations fo rthe three loops and solved for
the current in the last loop with RL set to zero (short circuit
current, which is the Norton current).

Then I removed RL to leave two loops and solved for the voltage across
R5 giving the open circuit voltage. The ratio of the open circuit
voltage to the short circuit current is the Norton resistance.

Loop equations (easily written by inspection):

- - - - - -
| R1 + R4 -R4 0 || i1 | | V |
| -R4 R2 + R4 + R5 -R5 || i2 | = | 0 |
| 0 -R5 R3 + R5 +RL || i3 | | 0 |
- - - - - -

 

On Nov 18, 10:40 am, "Greg Neill" <gneil...@MOVEsympatico.ca> wrote:

rad...@progressivetel.com wrote:
On Nov 18, 10:07 am, "Greg Neill" <gneil...@MOVEsympatico.ca> wrote:
rad...@progressivetel.com wrote:
Wow.. never mind guys.... I didn't expect this much pent up
hostility.

Hostility? What's hostile about being careful not to give away
homework answers to individuals who show no evidence of effort?

A clear, complete question without alterior motives and hidden
clauses would also help.

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.

That which is claimed without evidence can be dismissed without
evidence.

But C'est
la vie.... other things need my attention. I will move on. I do
appreciate the responses.

No problem. It's nice to see that you were able to solve your
problem on your own.

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.

Oops.  My bad.  Thanks for pointing that out.

Tell you what,  I'll tell you now that the Norton current is 20/37 Amps..
The resistance is 185/9 Ohms.  If you want to see how I arrived at those
numbers, first show us how you solved the problem yourself.

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.

 

[Tim Wescott]

On Tue, 18 Nov 2008 03:33:33 -0800, radlee wrote:

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)

http://www.progressivetel.com/~radney4/NT.jpg

Thanks!
Lee

The text is correct that shorting the source and removing RL, then
computing the resistance at RL's terminals, is the correct method (don't
short a 50V power supply in real life, though, at least not if it's
capable of some serious current .

Consider: If the resistors are all 5% accuracy, how close is your 20.55
going to be to the 'correct' answer? Even at 1%, is there any real
difference between 20.55 and 21?

Yes, the programmers of the app should have been more consistent, but
that's life.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

 

[Tim Wescott]

On Tue, 18 Nov 2008 06:50:37 -0800, radlee wrote:
(top posting fixed)

On Nov 18, 9:35 am, "Greg Neill" <gneil...@MOVEsympatico.ca> wrote:
rad...@progressivetel.com wrote:
Hi guys,

I've gotten a lot of repsonses, but no real answers.

People here are reluctant to give away solutions to what appears to be
homework.  Hints yes, suggestions yes, but answers without some effort
on the requester's part, no.

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.

So...it's course work but not homework... Perhaps you could explain the
distinction.

You didn't mention the further details about RL in your original
question, yet you expect to obtain meaningfull answers?

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.

20.555... actually.

And you never mentioned this as the being the motivation for your post.
 Why not?  Are you always less than forthcoming about your motives?

You will find people here can be very helpful and are willing to help
you understand things, but you have to show some initiative on your
part.  Show what you've tried, where you're stuck or confused.



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)

http://www.progressivetel.com/~radney4/NT.jpg

Thanks!
Lee
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

There's a reason for the hostility, although I don't think you rate it.

There's a word to describe people who skate through technical courses by
getting others to do their homework.

That word is "incompetent".

Since technical people usually depend on the work of others, incompetent
colleagues make everyone look bad. Capable incompetents do this while
looking good themselves, and get promoted to management, where they
_really_ make life hell for the competent folks.

So we do all we can to encourage such folks to quit tech school and go be
incompetent in some entirely different field.

If you have a question about homework, say "this is homework for such and
such a class", then ask for details that will help you come to the answer
yourself. That way we won't think we're being asked to supply answers
that you can just copy in (and believe me, plenty of folks ask for just
that, particularly around end-of-term time).

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

 

[Rich Grise]

On Tue, 18 Nov 2008 07:59:33 -0800, radlee wrote:

On Nov 18, 10:40 am, "Greg Neill" <gneil...@MOVEsympatico.ca> wrote:
rad...@progressivetel.com wrote:
On Nov 18, 10:07 am, "Greg Neill" <gneil...@MOVEsympatico.ca> wrote:
rad...@progressivetel.com wrote:
Wow.. never mind guys.... I didn't expect this much pent up
hostility.

Hostility? What's hostile about being careful not to give away
homework answers to individuals who show no evidence of effort?

A clear, complete question without alterior motives and hidden
clauses would also help.

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.

That which is claimed without evidence can be dismissed without
evidence.

But C'est
la vie.... other things need my attention. I will move on. I do
appreciate the responses.

No problem. It's nice to see that you were able to solve your problem
on your own.

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.

Oops.  My bad.  Thanks for pointing that out.

Tell you what,  I'll tell you now that the Norton current is 20/37
Amps. The resistance is 185/9 Ohms.  If you want to see how I arrived
at those numbers, first show us how you solved the problem yourself.

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.

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

 

On Nov 18, 1:06 pm, Tim Wescott <t...@justseemywebsite.com> wrote:

On Tue, 18 Nov 2008 06:50:37 -0800, radlee wrote:

(top posting fixed)





On Nov 18, 9:35 am, "Greg Neill" <gneil...@MOVEsympatico.ca> wrote:
rad...@progressivetel.com wrote:
Hi guys,

I've gotten a lot of repsonses, but no real answers.

People here are reluctant to give away solutions to what appears to be
homework.  Hints yes, suggestions yes, but answers without some effort
on the requester's part, no.

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.

So...it's course work but not homework... Perhaps you could explain the
distinction.

You didn't mention the further details about RL in your original
question, yet you expect to obtain meaningfull answers?

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.

20.555... actually.

And you never mentioned this as the being the motivation for your post..
 Why not?  Are you always less than forthcoming about your motives?

You will find people here can be very helpful and are willing to help
you understand things, but you have to show some initiative on your
part.  Show what you've tried, where you're stuck or confused.

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)

http://www.progressivetel.com/~radney4/NT.jpg

Thanks!
Lee
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

There's a reason for the hostility, although I don't think you rate it.

There's a word to describe people who skate through technical courses by
getting others to do their homework.

That word is "incompetent".

Since technical people usually depend on the work of others, incompetent
colleagues make everyone look bad.  Capable incompetents do this while
looking good themselves, and get promoted to management, where they
_really_ make life hell for the competent folks.

So we do all we can to encourage such folks to quit tech school and go be
incompetent in some entirely different field.

If you have a question about homework, say "this is homework for such and
such a class", then ask for details that will help you come to the answer
yourself.  That way we won't think we're being asked to supply answers
that you can just copy in (and believe me, plenty of folks ask for just
that, particularly around end-of-term time).

--

Tim Wescott
Wescott Design Serviceshttp://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says..
See details athttp://www.wescottdesign.com/actfes/actfes.html

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!!!

 

[Greg Neill]

radlee@progressivetel.com wrote:

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!!!

There is no way for anyone here to know the details of your
situation and background before you state them. Further,
there are many instances of individuals who have concocted
all sorts of stories claiming to be anything other than a
student, yet looking for effort-free solutions to what appear
for all the world to be homework problems. So naturally
people have become somewhat thick crusted about such requests,
and nearly always ask to see what the poster has tried so far
to solve the problem.

Now you may very well be the finest most upstanding gentleman
on the net, someone I'd be honored to share a beer with
sometime, but alas there's no way to tell that from a single
post that asked for free help on what appeared to be a circuit
problem typical of homework questions, asking for a step by
step solution no less, and in fact not even containing enough
information to state the problem in its entirety (a typical
student oversight) and was in fact a multiple choice question,
and besides all that you were *really* looking for a justification
for the problem's choice of precision in the stated answers...

A bit if a tangled web...

Anyways,

 

On Nov 18, 3:13 pm, "Greg Neill" <gneil...@MOVEsympatico.ca> wrote:

rad...@progressivetel.com wrote:
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!!!

There is no way for anyone here to know the details of your
situation and background before you state them.  Further,
there are many instances of individuals who have concocted
all sorts of stories claiming to be anything other than a
student, yet looking for effort-free solutions to what appear
for all the world to be homework problems.  So naturally
people have become somewhat thick crusted about such requests,
and nearly always ask to see what the poster has tried so far
to solve the problem.

Now you may very well be the finest most upstanding gentleman
on the net, someone I'd be honored to share a beer with
sometime, but alas there's no way to tell that from a single
post that asked for free help on what appeared to be a circuit
problem typical of homework questions, asking for a step by
step solution no less, and in fact not even containing enough
information to state the problem in its entirety (a typical
student oversight) and was in fact a multiple choice question,
and besides all that you were *really* looking for a justification
for the problem's choice of precision in the stated answers...

A bit if a tangled web...

Anyways,

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*******

 

[Jasen Betts]

On 2008-11-18, radlee@progressivetel.com <radlee@progressivetel.com> wrote:

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.

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

 

On Nov 19, 3:21 am, Jasen Betts <ja...@xnet.co.nz> wrote:

On 2008-11-18, rad...@progressivetel.com <rad...@progressivetel.com> wrote:

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.

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

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

 

[Rich Grise]

On Tue, 18 Nov 2008 12:22:43 -0800, radlee wrote:

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*******

http://www.google.com/search?hl=en&q=%22Norton+Resistance%22&btnG=Search
(people would be significantly less "hostile" if you'd checked that
first.)

Have Fun!
Rich

 

[Bob Pownall]

radlee@progressivetel.com wrote:
<snip>

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*******

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

It's about asking questions on computer software and hardware problems
in a way that gets you useful answers, but really - it applies to pretty
much any topic.

Bob Pownall