Note: many snips follow, not annotated as such, in the interest of
attempted brevity.
n is the number of resistors that you plan on using.... should be clear from
the context of my original post.
It certainly is not clear what you are trying to say. Hence my followup.
Nope... look you can have 2 possible configurations: All n of the identical
resistors in parallel or all in series.
Okey dokey, we will work from that premise.
Obviously your eating the wrong pie because your forgetting that you can
have constant current.
Attempts at obfuscation will not help you. The proof is in the pudding,
which in electronics, is the math. Read on!
I didn't know I was going to have to demonstrate it in such simple terms. I
thought most people here would be educated enough to see the what is going
on and I wouldn't have to take baby steps.
You are trying to explain away your inability to answer questions by
insulting the audience for asking them. I asked you to show your work in
order to see if you are capable of basic algebra. You apparently are not.
Sheesh... I think you are being a little dense.
I think you are being a little evasive. You've made what I consider to be
a vague and misleading statement about power transmission. Please answer
the questions.
I will, for the sake of your enlightenment be more precise and clear.
Well, you've failed to do that. Your post is only more vague and
misleading. If you could confine yourself to terse statements that answer
particular and direct questions, we might get somewhere.
I
figured most here would have basic knowledge of elementary circuit analysis
but i guess I was wrong ;/
Your condescending attitude is not helping your case. The point is, your
explanations make no sense. *You* might know what you are talking about,
but I'll wager few others do. I pointed out the error here:
I showed this to be clearly wrong, and you have responded with insults as
to my intelligence. Why? Perhaps because you are hoping that the error
will go away if you respond with enough new, obtusely presented, argument.
Well, it hasn't, and you've made many more in your insult ridden response.
Suppose we have n1 resistors each with resistance Ra and wattage Wa in
parallel. I'm not going to try and draw it because its obvious and if you
can't picture it in your brain then maybe you don't need to be in
electronics.
More insults, no rectification of previous errors.
the current going through each resistor is I/n1. This is basic kirchoff's
current law.
Pedantically, it is not; you've modified it and not shown how you arrived
at your derivation. Basic Kirchoff's current law would be:
0 = I1 + I2 + I3 + ... + In
This can be seen from another much simpler method.
What follows is *simpler?*!!
Its quite simple and any 3rd grader should have no problem with the math and
electronics to this point.
More insults. Well, better suck up your pride, because you are about to
become rather embarrassed.
For some real world examples to make all the egg heads understand? Ok...
More insults. You must be very insecure.
******* CONTENT ALERT! BEGINNING TO GET TO THE POINT! ********
Given 10 resistors of 100 ohms in parallel how can we get an equivilent
circuit(same resistance and wattage dissipation) in series?
Simple. Add the inverses of the individual resistances, and invert. This
is a basic formula, and works even if the resistances might happen to be
not all identical.
Rs = 1 / (1/Rp1 + 1/Rp2 + ... + 1/Rpn)
If the denominators happen to all be identical, as is the case in our
mutually accepted premise, then you may simply use 'n' as the numerator and
reduce:
Rs = 1 / (n/Rpi) = Rpi/n
That's extremely simple, isn't it? Now you know the equivalent series
resistance. For dissipation, add the individual legs.
Pd = PdR1 + PdR2 + ... + PdRn
Simple, and it works, again, even if the resistances are not identical. In
the case of identical values,
Pd = n*PdRi
******* CONTENT ALERT! CRUX OF THE MATTER REACHED! ********
Aha! I've found your hypothesis! You buried it well!
Rb represents the resistance in series and Ra in parallel.
But, I thought we were solving for series resistance? So I guess you mean
to say:
Rb = Ra / n^2
So as a simple example, if Ra = 1000 and n = 10, then you say:
Rb = 1000 / 10^2 = 1000 / 100 = 10
making Rb = 10, but by the "product over sum" method (which is taught at
the basic level, and should be understood by most readers of the group):
Rs = Rp^n / Rp*n = 1000^2 / 1000*10 = 100
or from the reduced inverse-sum form:
Rs = Rp / n = 1000 / 10 = 100
and by either equation the series resistance is 100. I've shown the math
behind my work; you've posted an equation that appears to be nonsense, and
supported it only with more vague wording and error-filled calculations.
But on the other hand if you have 100 100 ohm resistors in series and want
to convert that to a parallel(for whatever reason) then you need 100 100000
ohm resistors for parallel. (sinec 100*100 = 1000 = 100000/100).
Fascinating. Rather than bury these glaring math errors within convoluted
meandering of faulty logic, I'd simply calculate:
Rp = Rs * n = 100 * 100 = 10,000
Whereas you would:
Ra = Rb * n^2 = 100 * 10,000 = 1,000,000
So ineffect going from parallel to series will decrease the resistance you
need while going from series to parallel will increase it(obviously).
What is obvious is that you have no idea what you are talking about.
Thats the mistake you seem to be making. Its easy to show a counter example:
Hmm? Did I make a mistake somewhere? Please point it out. I am open to
constructive criticism, albeit possibly curmudgeonly.
The first is 2 the power dissipated in either of the others.
More gibberish. I will now tell you what I expect from you. I expect
apologies for your numerous insults, and a retraction of your nonsensical
parallel-to-series conversion equation:
Ra = n^2*Rb
I don't think that's too much to ask.
