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| jclause 2005-08-01, 7:21 pm |
|
Hi
A question for the Gurus, please:
The voltage divider rule for two resistors in series is:
V1 = R1 / (R1 + R2) x Vs
V1 = voltage across R1
Vs = applied voltage
Is there a simple power divider rule for two resistors in series?
Thanks.
| |
| jabara 2005-08-01, 7:21 pm |
|
"jclause" <jc@the.web> wrote in message
news:11et5ccjsmhe329@news.supernews.com...
>
>
> Hi
>
> A question for the Gurus, please:
>
> The voltage divider rule for two resistors in series is:
>
> V1 = R1 / (R1 + R2) x Vs
>
> V1 = voltage across R1
> Vs = applied voltage
>
> Is there a simple power divider rule for two resistors in series?
>
Yes, simple algebra.
power = v^2 / R or I * V or I^2 * R
Solve for the current
You do it.
| |
| Palindr☻me 2005-08-01, 8:21 pm |
| jabara wrote:
> "jclause" <jc@the.web> wrote in message
> news:11et5ccjsmhe329@news.supernews.com...
>
>
>
> Yes, simple algebra.
> power = v^2 / R or I * V or I^2 * R
> Solve for the current
> You do it.
>
>
If you need another hint:
1) Write the equation for total power in terms of total
resistance and current.
2) write the equation for R1 power in terms of R1 and current.
3) Use the two equations above to get the ratio you want.
--
Sue
| |
| Don Kelly 2005-08-01, 9:21 pm |
| ----------------------------
"jclause" <jc@the.web> wrote in message
news:11et5ccjsmhe329@news.supernews.com...
>
>
> Hi
>
> A question for the Gurus, please:
>
> The voltage divider rule for two resistors in series is:
>
> V1 = R1 / (R1 + R2) x Vs
>
> V1 = voltage across R1
> Vs = applied voltage
>
> Is there a simple power divider rule for two resistors in series?
>
> Thanks.
>
>------------------------------
Ptotal =I^2(R1 +R2) =I^2R1 +I^2R2 =P1 +P2 so P1/Ptotal =R1/R1+R2
Without knowing I (=V/R1+R2) at least you get a ratio
Is this what you want?
--
Don Kelly @shawcross.ca
remove the X to answer
| |
| jclause 2005-08-02, 2:21 pm |
| In article <EbyHe.86268$s54.17640@pd7tw2no>, dhky@shaw.ca says...
>
>----------------------------
>"jclause" <jc@the.web> wrote in message
>news:11et5ccjsmhe329@news.supernews.com...
>Ptotal =I^2(R1 +R2) =I^2R1 +I^2R2 =P1 +P2 so P1/Ptotal =R1/R1+R2
> Without knowing I (=V/R1+R2) at least you get a ratio
>Is this what you want?
Thank you. I was looking for a simple equation using voltage
and resistance terms only. What do you think of the following?
R1=5 ohms R2=10 ohms
|--------\/\/\/-------\/\/\/----
| |
E = 30 volts |
|-------------------------------
In case ASKI gets screwed up, two resistors (5 and 10 ohm) and
30 volts all in series.
I=E/R1+R2 = 2 amps
P total = I^2 R1+R2 = 60 watts
PI = I^2 R1 = 20 watts
P2 = I^2 R2 = 40 watts
Check:
Total power being dissipated = P1+P2=20+40=60 watts
Now the equation I came up with using E and R only:
P1 = (E/R1+R2)^2 * R1 = 20 watts
P2 = (E/R1+R2)^2 * R2 = 40 watts
Note the first term is just the current squared. No monkeying around
"hunting" current and working it in. How does this look to you? TIA.
JC the elder
| |
| jclause 2005-08-02, 2:21 pm |
| In article <42ee9aa6$0$1XXXX$892e7fe2@authen.white.readfreenews.net>,
nospam@invalid.com says...
>
>Yes, simple algebra.
>power = v^2 / R or I * V or I^2 * R
>Solve for the current
>You do it.
Thanks.
| |
| jclause 2005-08-02, 2:21 pm |
| In article <11et8os485l7e96@corp.supernews.com>, sb382638@hotmail.com.invalid
says...
>
>If you need another hint:
>
>1) Write the equation for total power in terms of total
>resistance and current.
>2) write the equation for R1 power in terms of R1 and current.
>3) Use the two equations above to get the ratio you want.
>
>--
>Sue
Thanks. All hints appreciated.
| |
| Don Kelly 2005-08-05, 1:21 am |
| "jclause" <jc@the.web> wrote in message
news:11ev7h8r2do4v64@news.supernews.com...
> In article <EbyHe.86268$s54.17640@pd7tw2no>, dhky@shaw.ca says...
>
>
>
>
> Thank you. I was looking for a simple equation using voltage
> and resistance terms only. What do you think of the following?
>
> R1=5 ohms R2=10 ohms
> |--------\/\/\/-------\/\/\/----
> | |
> E = 30 volts |
> |-------------------------------
>
>
> In case ASKI gets screwed up, two resistors (5 and 10 ohm) and
> 30 volts all in series.
>
> I=E/R1+R2 = 2 amps
> P total = I^2 R1+R2 = 60 watts
> PI = I^2 R1 = 20 watts
> P2 = I^2 R2 = 40 watts
> Check:
> Total power being dissipated = P1+P2=20+40=60 watts
>
> Now the equation I came up with using E and R only:
>
> P1 = (E/R1+R2)^2 * R1 = 20 watts
> P2 = (E/R1+R2)^2 * R2 = 40 watts
>
> Note the first term is just the current squared. No monkeying around
> "hunting" current and working it in. How does this look to you? TIA.
>
> JC the elder
> --------------------------------------
In fact, you have "hunted" for and found the current as E/(R1+R2)=2 Amps =I
no matter what you call it so you have done exactly the same here as done
above. What are you gaining? By explicitly using I you will have a better
conceptual understanding rather than simply plug in and turn the crank
--
Don Kelly @shawcross.ca
remove the X to answer
----------------------------
| |
| jclause 2005-08-05, 1:21 pm |
| In article <pfBIe.119498$s54.67647@pd7tw2no>, dhky@shaw.ca says...
>
>"jclause" <jc@the.web> wrote in message
>news:11ev7h8r2do4v64@news.supernews.com...
>
>In fact, you have "hunted" for and found the current as E/(R1+R2)=2 Amps =I
>no matter what you call it so you have done exactly the same here as done
>above. What are you gaining? By explicitly using I you will have a better
>conceptual understanding rather than simply plug in and turn the crank
-------------------------------
But I ask for a simple equation and your answer was convoluted. I would
have thought that you would have picked up on the following:
You said:
P1/Ptotal=R1/R1+R2
then
P1=Ptotal*(R1/R1+R2)
or
P1=(E/R1+R2)^2*(R1+R2)*(R1/R1+R2) (messy route)
Cancelling (R1+R2)
P1=(E/R1+R2)^2 R1 (note you did not mention this)
Einstein said things should be simple as possible, but no simpler. :-)
Anyway.. If one doesn't know that I=E/Rtotal, he probably shouldn't be
trying to apportion power anyway. Might get shocked.
JC the elder
| |
| Don Kelly 2005-08-06, 10:21 pm |
| "jclause" <jc@the.web> wrote in message
news:11f73jcc9erm1ec@news.supernews.com...
> In article <pfBIe.119498$s54.67647@pd7tw2no>, dhky@shaw.ca says...
>
> -------------------------------
>
>
> But I ask for a simple equation and your answer was convoluted. I would
> have thought that you would have picked up on the following:
>
> You said:
> P1/Ptotal=R1/R1+R2
> then
> P1=Ptotal*(R1/R1+R2)
> or
> P1=(E/R1+R2)^2*(R1+R2)*(R1/R1+R2) (messy route)
> Cancelling (R1+R2)
> P1=(E/R1+R2)^2 R1 (note you did not mention this)
>
> Einstein said things should be simple as possible, but no simpler. :-)
>
> Anyway.. If one doesn't know that I=E/Rtotal, he probably shouldn't be
> trying to apportion power anyway. Might get shocked.
>
> JC the elder
>
>
Least convoluted:
I=V/(Rtotal) (1)
Pn =(I^2)*Rn for resistor n (2)
Can it be simpler?
--
Don Kelly @shawcross.ca
remove the X to answer
----------------------------
| |
| jclause 2005-08-08, 3:21 pm |
| In article <htcJe.128454$5V4.114381@pd7tw3no>, dhky@shaw.ca says...
>
>"jclause" <jc@the.web> wrote in message
>news:11f73jcc9erm1ec@news.supernews.com...
------
[color=darkred]
>Least convoluted:
>I=V/(Rtotal) (1)
>Pn =(I^2)*Rn for resistor n (2)
>
>Can it be simpler?
Least convoluted? Your original response was
"Ptotal =I^2(R1 +R2) =I^2R1 +I^2R2 =P1 +P2 so P1/Ptotal =R1/R1+R2"
And still no provision for current. Arghhhh.......
Also yout last response above is misleading. You left out the
step to sum R1 and R2. When this step is added, your method is:
P1=V/(R1+R2)^2 R1
My method was:
P1=(E/R1+R2)^2 R1
You tried a bit of artfulness by leaving out the summing step, and I
had to lead you to your arrangement of terms, therefore I win. :-)
JC the elder
| |
| Don Kelly 2005-08-08, 11:21 pm |
| "jclause" <jc@the.web> wrote in message
news:11ff706gv8sl96c@news.supernews.com...
> In article <htcJe.128454$5V4.114381@pd7tw3no>, dhky@shaw.ca says...
>
> ------
>
>
>
>
>
>
> Least convoluted? Your original response was
>
> "Ptotal =I^2(R1 +R2) =I^2R1 +I^2R2 =P1 +P2 so P1/Ptotal =R1/R1+R2"
---------
>
> And still no provision for current. Arghhhh.......
---
The ratio is independent of current (as I originally mentioned)
--------
>
> Also yout last response above is misleading. You left out the
> step to sum R1 and R2. When this step is added, your method is:
>
> P1=V/(R1+R2)^2 R1
>
> My method was:
>
> P1=(E/R1+R2)^2 R1
>
>
> You tried a bit of artfulness by leaving out the summing step, and I
> had to lead you to your arrangement of terms, therefore I win. :-)
>
> JC the elder
>
>
Admittedly I did leave out the summation and it has to be done, as you point
out- but if you know enough to do this, then you know enough not to rely on
a bookful of formulae for trivial cases.
However, as to winning or losing,
The computational work is the same in both cases but I don't have to
remember a formula for each specific trivial case.
My main point is that it is better to know what you are doing than plugging
into canned formulae. Too often such formulae are misused, or forgotten- so
reduce the baggage to what's needed. Even "Ohm's Law" is often misused.
This is such a trivial situation that a specific formula is a waste of time
and energy.
--
Don Kelly @shawcross.ca
remove the X to answer
----------------------------
| |
| jclause 2005-08-09, 1:21 pm |
| In article <qNTJe.149467$%K2.76097@pd7tw1no>, dhky@shaw.ca says...
>
>"jclause" <jc@the.web> wrote in message
>news:11ff706gv8sl96c@news.supernews.com...
>---------
>
>---
>The ratio is independent of current (as I originally mentioned)
>--------
>
>Admittedly I did leave out the summation and it has to be done, as you point
>out- but if you know enough to do this, then you know enough not to rely on
>a bookful of formulae for trivial cases.
I believe you wanted to make a very brief equation, at least that's
what you did, but you "cheated" leaving out provision for current.
>However, as to winning or losing,
>The computational work is the same in both cases but I don't have to
>remember a formula for each specific trivial case.
Are you kidding? You would have had to remember the equation for current
plus derive it in your original equation:
"Ptotal =I^2(R1 +R2) =I^2R1 +I^2R2 =P1 +P2 so P1/Ptotal =R1/R1+R2"
Sorry but this was strung out, messy, and no provision for current,
whereas my equation gave everything needed:
"P1 = (E/R1+R2)^2 * R1 = 20 watts. Note the first term is just
the current squared."
>My main point is that it is better to know what you are doing than plugging
>into canned formulae.
Please explain how your original equation would show a newcomer to
"know what you are doing" better than mine, which *does* give current
and even briefly noted how by adding " Note the first term is just
the current squared."
>Too often such formulae are misused, or forgotten- so reduce the
>baggage to what's needed. Even "Ohm's Law" is often misused.
Now you parrot *exactly* what I did, I reduced the baggage with
nothing pertinent left out. You should have taken your own advice.
>This is such a trivial situation that a specific formula is a waste
>of time and energy.
Sorry, but "Ohms law" type of equations are not trivial. In fact, they
are so basic there is no need to confuse a situation with long and
strung out expressions that can be made simple as Einstein suggested.
Therefore I still suggest that I won. :-)
JC the elder.
| |
| Don Kelly 2005-08-09, 11:21 pm |
|
"jclause" <jc@the.web> wrote in message
news:11fhjji31v0jk6a@news.supernews.com...
> In article <qNTJe.149467$%K2.76097@pd7tw1no>, dhky@shaw.ca says...
>
>
> I believe you wanted to make a very brief equation, at least that's
> what you did, but you "cheated" leaving out provision for current.
>
>
>
>
>
> Are you kidding? You would have had to remember the equation for current
> plus derive it in your original equation:
>
> "Ptotal =I^2(R1 +R2) =I^2R1 +I^2R2 =P1 +P2 so P1/Ptotal =R1/R1+R2"
>
> Sorry but this was strung out, messy, and no provision for current,
> whereas my equation gave everything needed:
>
>
> "P1 = (E/R1+R2)^2 * R1 = 20 watts. Note the first term is just
> the current squared."
>
>
>
>
> Please explain how your original equation would show a newcomer to
> "know what you are doing" better than mine, which *does* give current
> and even briefly noted how by adding " Note the first term is just
> the current squared."
>
>
>
>
> Now you parrot *exactly* what I did, I reduced the baggage with
> nothing pertinent left out. You should have taken your own advice.
>
>
>
>
> Sorry, but "Ohms law" type of equations are not trivial. In fact, they
> are so basic there is no need to confuse a situation with long and
> strung out expressions that can be made simple as Einstein suggested.
> Therefore I still suggest that I won. :-)
>
> JC the elder.
>
>
The relationship for resistors in series is not Ohm's law . It depends on
Kirchoff's Laws and the concept of linearity- these are not trivial but not
exactly difficult. Would it not be better to concentrate on Kirchoff's laws
and also the rather intellectually trivial "E=IR" (which is not ohms law
unless R is constant)?. Learning these would eliminate the necessity for a
long list of formulae- replacing rote memory by some thinking/
understanding.
Please also note that my original reply was to your original question and
was not intended to produce a neat canned formula but simply to point out a
relationship which you apparently could not figure out at the time. . I
could have given your result directly but chose not to, on the assumption
that you knew how to find the current I.
I may have misread your query- if so- I am sorry - but my point still
stands. If you "have won" it is because you appear to have learned
something. Fair enough.
Bye, have fun
--
Don Kelly @shawcross.ca
remove the X to answer
----------------------------
| |
| jclause 2005-08-11, 4:21 pm |
| In article <SpdKe.165774$s54.141408@pd7tw2no>, dhky@shaw.ca says...
>
>
>"jclause" <jc@the.web> wrote in message
>news:11fhjji31v0jk6a@news.supernews.com...
>
>The relationship for resistors in series is not Ohm's law .
And where did I say it was?
>It depends on Kirchoff's Laws and the concept of linearity- these are
>not trivial but not exactly difficult. Would it not be better to
>concentrate on Kirchoff's laws and also the rather intellectually
>trivial "E=IR" (which is not ohms law unless R is constant)?
??? Ohms law states: R=V/I. R is the variable, certainly not constant.
>Learning these would eliminate the necessity for a
>long list of formulae- replacing rote memory by some thinking/
>understanding.
My solution was not a "long list of formulae" is was simply:
P1=(E/R1+R2)^2*R1
>Please also note that my original reply was to your original question and
>was not intended to produce a neat canned formula
Sorry but I did ask "Is there a simple power divider rule for two
resistors in series?" Note the word simple. You gave a rather
convoluted reply.
>but simply to point out a
>relationship which you apparently could not figure out at the time. . I
>could have given your result directly but chose not to, on the assumption
>that you knew how to find the current I.
>I may have misread your query- if so- I am sorry - but my point still
>stands. If you "have won" it is because you appear to have learned
>something. Fair enough.
Did you not note the smiley, i.e. :-) ? In any event, please state
specifically what you taught me. Should be interesting. :-)
JC the elder.
| |
| Don Kelly 2005-08-12, 1:21 am |
| Last response to this thread below:
"jclause" <jc@the.web> wrote in message
news:11fn8h0eablnia8@news.supernews.com...
> In article <SpdKe.165774$s54.141408@pd7tw2no>, dhky@shaw.ca says...
>
>
> And where did I say it was?
------
You didn't and I may have read this into what you said when you went on
about the complexity of Ohms law.>
>
>
>
> ??? Ohms law states: R=V/I. R is the variable, certainly not constant.
_------------
Strictly speaking "Ohm's Law" as originally given and still valid holds for
R constant. It described a relationship, found by experiment (at the time
and for the materials considered) which was a linear relationship between E
and I --that is R independent of either E or I. Nonlinear materials do exist
but for such materials and then R is dependent on current or voltage.
In such a case, the standard circuit analysis theorems are ratshit except
for Kirchoff's Laws as they are based on linearity. In the simple series
circuit it is not a big problem.
Example: take R1= 2 ohms and R2 =2*(I*0.5) and what is R1 + R2 -- You can
say that E =2*(1+I*0.5)*I - so Rtotal =2(1+I*0.5). Fine- Now, given E=10V,
what is I? , what Rtotal do you use to find I?
In fact the concept of "R" is rather useless,except for visualisation, in
this case
Now consider the case where a "real" circuit with a nonlinear element
exists- it gets messy.
>
>
>
> My solution was not a "long list of formulae" is was simply:
> P1=(E/R1+R2)^2*R1
---
And if you do the equivalent for other trivial situations such as parallel
resistors, etc. the list gets longer.
---
>
>
>
>
>
> Sorry but I did ask "Is there a simple power divider rule for two
> resistors in series?" Note the word simple. You gave a rather
> convoluted reply.
-----
I had hoped that what I gave would help you work it out for yourself. You
did so- good on you.
----------.
>
>
>
>
> Did you not note the smiley, i.e. :-) ? In any event, please state
> specifically what you taught me. Should be interesting. :-)
>
> JC the elder.
>
-----------
Lets see now, you wanted a canned "plug in and turn the crank" answer. I
didn't provide it but did point out a way. Somehow, whether it is because of
what I said or otherwise, you have figured out your formula. Did you learn
anything from me? I don't know but you did go ahead and figure it out
yourself so you obviously have learned something- which is good. It also
shows that you can think- which is also good. I am happy with that.
My points about canned formulae for simple situations still holds.
(P.S. The current in the example is approximately 2.06A).
Don Kelly @shawcross.ca
remove the X to answer
----------------------------
| |
| jclause 2005-08-12, 3:21 pm |
| In article <7lVKe.196462$s54.12514@pd7tw2no>, dhky@shaw.ca says...
In article <7lVKe.196462$s54.12514@pd7tw2no>, dhky@shaw.ca says...
>
>Last response to this thread below:
>
>"jclause" <jc@the.web> wrote in message
>news:11fn8h0eablnia8@news.supernews.com...
>------
>You didn't and I may have read this into what you said when you went on
>about the complexity of Ohms law.>
No. I "went on" about the complex way you responded.
>_------------
>Strictly speaking "Ohm's Law" as originally given and still valid holds for
>R constant. It described a relationship, found by experiment (at the time
>and for the materials considered) which was a linear relationship between E
>and I --that is R independent of either E or I. Nonlinear materials do exist
>but for such materials and then R is dependent on current or voltage.
>In such a case, the standard circuit analysis theorems are ratshit except
>for Kirchoff's Laws as they are based on linearity. In the simple series
>circuit it is not a big problem.
>
>Example: take R1= 2 ohms and R2 =2*(I*0.5) and what is R1 + R2 -- You can
>say that E =2*(1+I*0.5)*I - so Rtotal =2(1+I*0.5). Fine- Now, given E=10V,
>what is I? , what Rtotal do you use to find I?
> In fact the concept of "R" is rather useless,except for visualisation, in
>this case
>Now consider the case where a "real" circuit with a nonlinear element
>exists- it gets messy.
Halliday (6th ed.) says Ohms law states that current is
proportional to the potential applied. The example I gave
(which you snipped out) clearly follows this rule.
Sorry, but there is no need for all the "complexity" you are
into in ***the example I gave***.
>
>---
>And if you do the equivalent for other trivial situations such as parallel
>resistors, etc. the list gets longer.
>---
>-----
>I had hoped that what I gave would help you work it out for yourself. You
>did so- good on you.
Thanks, but I had it worked out 50+ years ago. Had intended the post
as a ***lead-in*** for more complicated stuff - just building up to such.
>-----------
>Lets see now, you wanted a canned "plug in and turn the crank" answer. I
>didn't provide it but did point out a way. Somehow, whether it is because of
>what I said or otherwise, you have figured out your formula. Did you learn
>anything from me? I don't know but you did go ahead and figure it out
>yourself so you obviously have learned something- which is good. It also
>shows that you can think- which is also good. I am happy with that.
>
>My points about canned formulae for simple situations still holds.
>(P.S. The current in the example is approximately 2.06A).
Mr. Kelly, if I may........ I have respect for your abilities.
Seriously, you are damned sharp. However, you tend to be a bit
condescending at times, with assumption the other person is unlearned
at best... Plus.. (my opinion follows)... your store of knowledge gets
in the way of Mr. Einstein's credo that things should be simple as possible.
In any event, thank you for your time and response.
JC the elder
| |
| Don Kelly 2005-08-13, 2:21 am |
| "jclause" <jc@the.web> wrote in message
news:11fpmok5qc6ma31@news.supernews.com...
> In article <7lVKe.196462$s54.12514@pd7tw2no>, dhky@shaw.ca says...
>
>
> In article <7lVKe.196462$s54.12514@pd7tw2no>, dhky@shaw.ca says...
>
>
> No. I "went on" about the complex way you responded.
>
>
>
>
> Halliday (6th ed.) says Ohms law states that current is
> proportional to the potential applied. The example I gave
> (which you snipped out) clearly follows this rule.
> Sorry, but there is no need for all the "complexity" you are
> into in ***the example I gave***.
>
>
>
>
> Thanks, but I had it worked out 50+ years ago. Had intended the post
> as a ***lead-in*** for more complicated stuff - just building up to such.
>
>
>
>
>
> Mr. Kelly, if I may........ I have respect for your abilities.
> Seriously, you are damned sharp. However, you tend to be a bit
> condescending at times, with assumption the other person is unlearned
> at best... Plus.. (my opinion follows)... your store of knowledge gets
> in the way of Mr. Einstein's credo that things should be simple as
> possible.
> In any event, thank you for your time and response.
>
> JC the elder
>
In spite of what I said before, I have to respond to this. I have no
problem with Halliday but simply pointed out that Ohm actually came up with
a "Law" which was that E is directly proportional to current- Halliday
simply left out the "directly' as many others have done. I also attempted,
unsuccessfully that generalising R to a non-linear function of current or
voltage leads to the elimination of the basis of most of circuit analysis-
linearity.
If I have pontificated, I am sorry. I tend to do so.
If I disagree with you regarding what is simple - that's life.
Nothing that I have said, in this situation, is complex, but I simply don't
believe in canned formulae unless I know just how they are obtained (and if
I know that- why bother?).
If you want to discuss deeper aspects- fine by me- but put in a stronger
lead in ( I also have had over 50 years experience-on both the giving and
recieving end- so that makes us both old farts .
Possibly under a different thread title.
Cheers,
--
Don Kelly @shawcross.ca
remove the X to answer
----------------------------
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| jclause 2005-08-15, 2:21 pm |
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Xref: number1.nntp.dca.giganews.com alt.engineering.electrical:163709
In article <YlfLe.202322$s54.183232@pd7tw2no>, dhky@shaw.ca says...
>
>
>
>Nothing that I have said, in this situation, is complex
Agreed, but you apparently miss my point. What you wrote was
more complex then needed. Sorry, but I will not indulge in this
roundabout game of math. My goal is to obtain the result with
the path to it made short and clear, and Ohms law is more
instrumental in the process than any light I can shine on my
math capabilities.
>If you want to discuss deeper aspects- fine by me- but put in a stronger
>lead in
Exactly how strong would you like it to be? :-)
>I also have had over 50 years experience-on both the giving and
>recieving end- so that makes us both old farts.
I try to hold farting to a minimum, note. It disturbs the cat.
JC the elder
| |
| Don Kelly 2005-08-15, 9:21 pm |
| As to complexity - yes - i did give a long answer and pointed out a way for
you to go assuming your original query was genuine. In fact, the approach
that I gave later, which you found complex required exactly the same
numerical steps and effort as your canned formula and was just as explicit,
with, in fact, for most, less memorisation. However, there is likely no
point going after that any more.
As to Ohm's Law- please recognise that you were, in fact , using something
a bit more fundamental than Ohm's Law which is simply the expression of a
<linear>* relationship between the voltage across an energy dissipating
element and the current through it - i.e valid for what is called an "ohmic"
resistance. It is, in practical terms, no more than a useful mathematical
idealisation (as is all circuit theory). Math is involved, whether or not
you like it - particularly when you go beyond elementary DC analysis.
As for deeper aspects- Whatever strength you want.
* references on hand- can be quoted if you wish.
--
Don Kelly @shawcross.ca
remove the X to answer
----------------------------
"jclause" <jc@the.web> wrote in message
news:11g1gkkomi9382a@news.supernews.com...
> In article <YlfLe.202322$s54.183232@pd7tw2no>, dhky@shaw.ca says...
>
>
> Agreed, but you apparently miss my point. What you wrote was
> more complex then needed. Sorry, but I will not indulge in this
> roundabout game of math. My goal is to obtain the result with
> the path to it made short and clear, and Ohms law is more
> instrumental in the process than any light I can shine on my
> math capabilities.
>
>
>
>
> Exactly how strong would you like it to be? :-)
>
>
>
>
> I try to hold farting to a minimum, note. It disturbs the cat.
>
> JC the elder
>
>
| |
| jclause 2005-08-16, 8:21 pm |
| In article <pD9Me.26773$vj.20638@pd7tw1no>, dhky@shaw.ca says...
>
>As to complexity - yes - i did give a long answer and pointed out a way for
>you to go assuming your original query was genuine. In fact, the approach
>that I gave later, which you found complex
Sorry but you are apparently confused. It was your first response
I found more complex than need be. Go back and check that which
you clipped.
>required exactly the same numerical steps and effort as your canned
>formula and was just as explicit, with, in fact, for most, less
>memorisation.
Less memorization than (E/Rtotal)^2*R ? Show me how, please.
Here's what I gave:
P1 = (E/R1+R2)^2 * R1 = 20 watts
P2 = (E/R1+R2)^2 * R2 = 40 watts
>Math is involved, whether or not you like it -
Now where did I say I didn't like math? My point was that I prefer
to not use more math than necessary, at the expense of glory... :-)
Why be provocative?
>particularly when you go beyond elementary DC analysis.
??? Where were we beyond elementary analysis?
JC the elder
| |
| Don Kelly 2005-08-17, 1:21 am |
| ----------------------------
"jclause" <jc@the.web> wrote in message
news:11g4sed1td9rb60@news.supernews.com...
> In article <pD9Me.26773$vj.20638@pd7tw1no>, dhky@shaw.ca says...
>
>
> Sorry but you are apparently confused. It was your first response
> I found more complex than need be. Go back and check that which
> you clipped.
>
>
>
>
> Less memorization than (E/Rtotal)^2*R ? Show me how, please.
> Here's what I gave:
>
> P1 = (E/R1+R2)^2 * R1 = 20 watts
> P2 = (E/R1+R2)^2 * R2 = 40 watts
>
>
>
>
>
> Now where did I say I didn't like math? My point was that I prefer
> to not use more math than necessary, at the expense of glory... :-)
>
> Why be provocative?
>
>
>
>
> ??? Where were we beyond elementary analysis?
>
> JC the elder
>
>
Lets see: first add R1 and R2 (Rtotal)
secondly find E/(R1+R2) (I)
thirdly square the above (I^2)
fourthly multiply the result by R1 to get P1
fiftly , do it again using R2 to get P2
To find P1 using your formula requires 1 addition, 2 multiplications(
including the squaring), and 1 division.
To find P2- you are explicitly repeating the whole process to double the
effort.
- <unless>,
(as I would expect you to do) you keep the intermediate
results -particularly I^2 (whether or not you name them, or not, you put
them away where you can get them when needed) All that I have done is to
give names to the intermediate results. This requires no mathematical,
numerical, or mental, effort (but helps to indicate what they really are).
..
As for memorization- when it takes less effort and time to work it out from
scratch (using either fundamental principles or more broadly useful
expressions) than to look up or memorise a formula- why bother with the
latter? That is the case here where a specific formula saves no effort
computationally at the cost of loss of understanding and an additional rote
item to memorise. Your best tool is your mind, why waste it on rote
memorisation of trivia?
You seem to have found my first response confused. To quote:
"Ptotal =I^2(R1 +R2) =I^2R1 +I^2R2 =P1 +P2 so P1/Ptotal =R1/R1+R2
Without knowing I (=V/R1+R2) at least you get a ratio
Is this what you want?"
It was not intended as a canned formula but simply to point out an approach-
It is now obvious that you wanted a canned formula which avoids any need to
understand the bases for the formula- your loss.
I could have supplied this but chose not to do so as it was a useless
exercise. Note that others had the sense not to bother. Good on them.
Note that the first line says I^2R1 + I^2R2 =P1 + P2 and the second defines
I- how is this confusing when it is so obvious? Since it appears to be
so -my loss as you apparently had problems with this.
As to math - use what is necessary and no more as you say but in order to
get your simple canned formula, someone has to use the math and understand
what is going on. I have used no more math (and actually a bit less algebra,
because of less algebraic massaging) than you have. I have simply named the
intermediate steps which are an integral part of the solution process even
if not explicitly set out as such.
Unfortunately, circuit analysis IS applied mathematical modelling and
mathematics is the language of science -that is a fact of life. One uses the
mathematical model which most closely meets the required balance between
complexity and accuracy in the particular situation that is being
considered.
You might wonder (or not) why I am suspicious of "canned" formulae. It is
because I have seen too many cases where people plug numbers into such
formulae and believe the results even when they have used an inapplicable
formula (Not to mention the tendency to slavishly believe the output of
their calculators without remembering GIGO).
I did not say that we had gone beyond elementary DC analysis - we haven't
even got off square one in that respect.
I'm not intending to be provocative but, if I am, then be provoked into
really thinking and questioning the familiar rote formulae--disagree with
me- that's how we both learn and evade Alzeimer's. It is also more fun.
The above answer is, as I expect, confusing -try thinking.
If you want to get into the meat of circuit analysis, fine. otherwise say
so. If not, then MYST 4 awaits me-more challenge.
Don Kelly @shawcross.ca
remove the X to answer
| |
| jclause 2005-08-17, 4:21 pm |
| In article <qmyMe.31636$vj.644@pd7tw1no>, dhky@shaw.ca says...
>
>----------------------------
>"jclause" <jc@the.web> wrote in message
>news:11g4sed1td9rb60@news.supernews.com...
>Lets see: first add R1 and R2 (Rtotal)
>secondly find E/(R1+R2) (I)
>thirdly square the above (I^2)
>fourthly multiply the result by R1 to get P1
>fiftly , do it again using R2 to get P2
>To find P1 using your formula requires 1 addition, 2 multiplications(
>including the squaring), and 1 division.
>
>To find P2- you are explicitly repeating the whole process to double the
>effort.
>
>- <unless>,
>(as I would expect you to do) you keep the intermediate
>results -particularly I^2 (whether or not you name them, or not, you put
>them away where you can get them when needed) All that I have done is to
>give names to the intermediate results. This requires no mathematical,
>numerical, or mental, effort (but helps to indicate what they really are).
>.
>
>As for memorization- when it takes less effort and time to work it out from
>scratch (using either fundamental principles or more broadly useful
>expressions) than to look up or memorise a formula- why bother with the
>latter? That is the case here where a specific formula saves no effort
>computationally at the cost of loss of understanding and an additional rote
>item to memorise. Your best tool is your mind, why waste it on rote
>memorisation of trivia?
>
>
>You seem to have found my first response confused. To quote:
>"Ptotal =I^2(R1 +R2) =I^2R1 +I^2R2 =P1 +P2 so P1/Ptotal =R1/R1+R2
> Without knowing I (=V/R1+R2) at least you get a ratio
>Is this what you want?"
>
>It was not intended as a canned formula but simply to point out an approach-
>It is now obvious that you wanted a canned formula which avoids any need to
>understand the bases for the formula- your loss.
> I could have supplied this but chose not to do so as it was a useless
>exercise. Note that others had the sense not to bother. Good on them.
>Note that the first line says I^2R1 + I^2R2 =P1 + P2 and the second defines
>I- how is this confusing when it is so obvious? Since it appears to be
>so -my loss as you apparently had problems with this.
A cover up by implying I had problems. You clip and rewrite history.
>As to math - use what is necessary and no more as you say but in order to
>get your simple canned formula, someone has to use the math and understand
>what is going on. I have used no more math (and actually a bit less algebra,
>because of less algebraic massaging) than you have. I have simply named the
>intermediate steps which are an integral part of the solution process even
>if not explicitly set out as such.
>Unfortunately, circuit analysis IS applied mathematical modelling and
>mathematics is the language of science -that is a fact of life. One uses the
>mathematical model which most closely meets the required balance between
>complexity and accuracy in the particular situation that is being
>considered.
>
> You might wonder (or not) why I am suspicious of "canned" formulae. It is
>because I have seen too many cases where people plug numbers into such
>formulae and believe the results even when they have used an inapplicable
>formula (Not to mention the tendency to slavishly believe the output of
>their calculators without remembering GIGO).
>
>I did not say that we had gone beyond elementary DC analysis - we haven't
>even got off square one in that respect.
No but you imply so, just as you imply above that I had problems with
elementary theory.
>I'm not intending to be provocative but, if I am, then be provoked into
>really thinking and questioning the familiar rote formulae--disagree with
>me- that's how we both learn and evade Alzeimer's. It is also more fun.
>
>The above answer is, as I expect, confusing -try thinking.
See what I mean? Try thinking yourself - about being less provocative.
But I suspect that's what you're here for - to show your store of knowledge
(which is good) and to imply others can't think, (with the hidden
implication that you can), which is not good.
>If you want to get into the meat of circuit analysis, fine. otherwise say
>so. If not, then MYST 4 awaits me-more challenge.
Your challenge is to be less provocative.
JC the elder.
| |
| jclause 2005-08-17, 4:21 pm |
| In article <u9JMe.2834$FA3.1401@news-server.bigpond.net.au>,
danbenfred@bigpond.com says...
>
>Don...why bother even to get involved in this..
>
>A guy asks a question...you answer in good faith...and then get criticised
>for it!
>
>JClause, get off Dons back. After quoting Prof Einstein (not Mr) time and
>time again out of context I'd think you also would have better things to do
>than argue over trivia. Don's got more in his little finger methinks than
>you have in your entire body...
>enuf said
Feel better now?
JC the elder
| |
| jclause 2005-08-17, 5:21 pm |
| In article <qmyMe.31636$vj.644@pd7tw1no>, dhky@shaw.ca says...
>
In article <qmyMe.31636$vj.644@pd7tw1no>, dhky@shaw.ca says...
PS - You said:
"You might wonder (or not) why I am suspicious of "canned" formulae."
But you used two "canned formulae":
P1/Ptotal =R1/R1+R2
I (=V/R1+R2)
So then, is it OK to use them in pairs? Or does it depend on
who's using them?
JC the elder
| |
| Don Kelly 2005-08-18, 2:21 am |
| "Daniel Indyk" <danbenfred@bigpond.com> wrote in message
news:u9JMe.2834$FA3.1401@news-server.bigpond.net.au...
> Don...why bother even to get involved in this..
>
> A guy asks a question...you answer in good faith...and then get criticised
> for it!
---------
Thanks for the advice.
--
Don Kelly @shawcross.ca
remove the X to answer
----------------------------
| |
| jclause 2005-08-18, 3:21 pm |
| In article <PpUMe.36548$vj.34413@pd7tw1no>, dhky@shaw.ca says...
>
>"Daniel Indyk" <danbenfred@bigpond.com> wrote in message
>news:u9JMe.2834$FA3.1401@news-server.bigpond.net.au...
>---------
>
>Thanks for the advice.
>
>Don Kelly @shawcross.ca
>remove the X to answer
Note Daniel Indyk misspells *criticized* above same as you.
Is this coincidence, since you have a problem with the words
criticize and criticized.
From google:
"You have a concept- I have criticised it"
- Feb 2 2004, 7:01 pm by Don Kelly
"I have given you specific targets to criticise"
- Feb 1, 8:15 am by Don Kelly
"If you find fault with the work I have shown, then criticise it"
- Feb 4 2004, 3:10 am by Don Kelly
"It is easy to criticise on the basis of extended knowledge"
- Sep 11 2001, 1:42 am by Don Kelly
-----------------------------------
In any event, it is your provocation that I criticize. How about being
accountable and comment on the following?
You said:
"You might wonder (or not) why I am suspicious of "canned" formulae."
However you used two "canned formulae":
P1/Ptotal =R1/R1+R2
I (=V/R1+R2)
So then, is it OK to use them in pairs?
Or does it depend on who's using them?
-----------
You said:
"you went on about the complexity of Ohms law"
No, this is untrue.
I said
"Sorry but this was strung out, messy, and no provision for current"
I did not say ohms law was complex.
You clip and rewrite history.
---------------
Also you said:
"Note that the first line says I^2R1 + I^2R2 =P1 + P2 and the second defines
I- how is this confusing when it is so obvious? Since it appears to be
so -my loss as you apparently had problems with this."
This is a cover up for your messy notation.
Again, you clip and rewrite history.
JC the elder
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Xref: number1.nntp.dca.giganews.com alt.engineering.electrical:163834
In article <11g9ifb6tbo3g27@news.supernews.com>, jc@the.web says...
> In article <PpUMe.36548$vj.34413@pd7tw1no>, dhky@shaw.ca says...
>
>
>
> Note Daniel Indyk misspells *criticized* above same as you.
> Is this coincidence, since you have a problem with the words
> criticize and criticized.
Simply UKUS. "Criticise" = UK. "Criticize"= US. It's not unusual for
right-pondians to use an 's' where left-pondians use a 'z'.
http://www.m-w.com/cgi-bin/dictiona...a=criticised&x=
19&y=14
<snip>
--
Keith
| |
|
| On 8/17/05 9:47 PM, in article PpUMe.36548$vj.34413@pd7tw1no, "Don Kelly"
<dhky@shaw.ca> wrote:
> "Daniel Indyk" <danbenfred@bigpond.com> wrote in message
> news:u9JMe.2834$FA3.1401@news-server.bigpond.net.au...
> ---------
>
> Thanks for the advice.
I was going to respond to the question when I first sq it. Then I saw that
it was answered well. I thought to myself that that should put an end to the
thread. Now, it seems that not only is the horse being beat after it is
dead, but the horse's flesh is rotting while enduring continued beatings.
Bill
| |
| jclause 2005-08-19, 1:21 pm |
| In article <MPG.1d6eaa67a5515b0798976b@news3.prserv.net>, krw@att.biz says...
>
>In article <11g9ifb6tbo3g27@news.supernews.com>, jc@the.web says...
criticised[color=darkred]
>
>Simply UKUS. "Criticise" = UK. "Criticize"= US. It's not unusual for
>right-pondians to use an 's' where left-pondians use a 'z'.
>
>http://www.m-w.com/cgi-bin/dictiona...a=criticised&x=
>19&y=14
>
>
><snip>
>
>--
> Keith
Thank you.
JC the elder
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