Creating a higer wattage resistor

As others have said, yes, you can do this. Two other options:

1. Wire "n" resistors of R/n ohms in series (as was said elsewhere in this thread).
2. Take 4 resistors of R ohms each. Make two 2-resistor series strings. Wire those two strings in parallel. Now you have a total resistance equal to R, and it will dissipate

Regards,

Mark

p.s. Extra credit problem: figure out how to wire 9 (or 16) resistors of R ohms, to get a net resistance of R with 9 (or 16) times the power.

p.p.s. Using series / parallel combinations of resistors is also a quick and dirty way to get a more precise value from lower-precision resistors. I.e., get within 1% of a design value when all you have is a box full of 5% resistors.

Note: many snips follow, not annotated as such, in the interest of attempted brevity.

On Sun, 27 Aug 2006 00:57:58 -0500, in message , "Abstract Dissonance" scribed:

It certainly is not clear what you are trying to say. Hence my followup.

Okey dokey, we will work from that premise.

Attempts at obfuscation will not help you. The proof is in the pudding, which in electronics, is the math. Read on!

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.

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.

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.

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.

More insults, no rectification of previous errors.

Pedantically, it is not; you've modified it and not shown how you arrived at your derivation. Basic Kirchoff's current law would be:

What follows is *simpler?*!!

More insults. Well, better suck up your pride, because you are about to become rather embarrassed.

More insults. You must be very insecure.

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

Aha! I've found your hypothesis! You buried it well!

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.

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

What is obvious is that you have no idea what you are talking about.

Hmm? Did I make a mistake somewhere? Please point it out. I am open to constructive criticism, albeit possibly curmudgeonly.

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.

On Sun, 27 Aug 2006 10:37:16 -0700, in message , Alan B scribed:

I made a mistake here. Product over sum apparently only works for n = 2.

On Sun, 27 Aug 2006 10:37:16 -0700, in message , Alan B scribed:

Here's a breakdown of the brain fart. For two parallel resistances:

Rs = 1 / (1/i + 1/j)

Multiplying all terms by the product of the denominators gives:

Rs = ij / (1/ij/i + 1/ij/j) = ij / (j + i)

For three legs, using the same technique:

Rs = 1 / (1/i + 1/j + 1/k)

Rs = ijk / (jk + ik + ij)

and so on.

Example

Rs = 1000^3 / (1000^2 + 1000^2 + 1000^2) = 333.3

being identical to the simplification in my post of

Rs = Rp / n = 1000 / 3 = 333.3

Sorry for the error.

yes, as I have have said several times but you seem to pick and choose what you want to read. Electrically they might be the same but not necessarily practically. I pointed it out but you don't care to use your brain to much and only find obvious mistakes that are unimportant in the logic.

simple mathematical mistake. 100 10 ohm resistors in series. The calculation error doesn't defeat the logic as you seem to think. You have similar mentality to people who point out spelling and grammar mistakes as proof of a logical error.

Sure. So? I said this in my previous post how its a difference of the square.

No, I'm making simple calculation mistakes. I explained everything correctly in the original post. Just like you I made some oversights and mistakes because no one is perfect and I wrote this in a hurry at 1AM. Forgive me for not living up to your standards of perfection.

Yep. So stop replying to me then.

So.. the point is that you cannot use your brain to decipher the logical argument and can only see immaterial mistakes. So... So keep on point them out of it makes you better but you won't learn anything.

yes, and I have a degree in applied mathematics. Sure I still make stupid calculation errors... but the logic is solid and I prefer to work in the abstract while you think its important to work in the concrete. Maybe I did make some calculation errors but it is not important. I'm not going to keep on trying to explain something that is perfectly clear(ok, maybe not but use your brain and don't expect to be spoon fed everything).

Nope, I explained it perfectly clear. You jsut don't like the insult and hence you try to divert attention away from my proof that follows. I'm kinda getting sick and tired of explaining things so I'll stop here. Everything I said in the first post holds and if you dont' like it then tough. I have better things to do with my time than explain minutiae. Maybe I wasn't clear but so what. I'm not hear to teach you anything.

Hmm.. why so many mistakes? Ateast you wern't doing it at 1am. If you want to bitch bout triffle mathematical calculation errors then so be it but expect me to do the same. Again, the everything I said in the original post logically holds. In the second post I may have made some calculation errors but hte formulas still hold. Anyways, I got better things to do.

On Sun, 27 Aug 2006 21:43:39 -0500, in message , "Abstract Dissonance" scribed:

The logic is not sound, and you are not properly supporting your position with mathematics. And you really should stop with the insults, or you are going to get slapped some more.

On Sun, 27 Aug 2006 21:43:39 -0500, in message , "Abstract Dissonance" scribed:

'Nuff said.

No one can understand what you are saying. I'll re-word it for you:

Rp = Rs*N^2

Where N is the number of identical resistors, each of resistance Rs in series, or Rp in parallel.

Is that what you are trying to say?

Ed

--- So... You can't make a point without tripping all over yourself, your math stinks, and your language skills are... well, let's just say they're lacking...

Then, on top of all that you expect everyone to try to find a needle of sense in the haystack of your post while trying to decipher your "immaterial mistakes"?

LOL, it sounds to me like a lot of work for no reward at all!

-- John Fields Professional Circuit Designer

On Sun, 27 Aug 2006 15:01:44 -0700, in message , Alan B scribed:

Damn. That last should be:

Moral of the story, boys and girls, always check your work.

Carbon resistor manufacturers used to do it this way:

They would create a large batch of resistors of value X. The resistors in the batch would then be measured for value X, and the ones with the tightest tolerance would be pulled and labeled as tight tolerance (1% 2%). Next the 5% resistors would be pulled, then the 10% resistors, and finally the 20% resistors.

So for 5% and looser tolerance resistors, rather than see a nice gaussian distribution , you ended up with a 2-lobed distribution curve, with 'dip' at the base value. If you had a large sample of 1000 Ohm,

I have seen this in my experience.

I don't know if manufacturing processes have changed over the decades to increase target accuracy, or if this practice described is still performed. I don't know if this applies to metal film resistors, either.

Tom

John, yes, that goes without saying. But why should that be a problem? You're at least the second person in recent months to suggest to me that 1% resistance measurements are not easy to come by.

Because of your post, as well as an earlier comment some months back by Phil Allison, I've dug out my DMM's manual to look up the accuracy. It turns out to be 0.3% to 1% for resistance measurements, depending on just where the reading is. (If you want, see my "p.s." below for how I calculated this)

So it's 1% or better. At least the manufacturer claims it is, in the specs. Are most DMM's not nearly this good?

Regards, and TIA for any information,

Mark

p.s. Details of figuring the meter's accuracy:

For my meter (Extech model 380282), the stated resistance accuracy is

Last time this subject was mentioned, I measured a handful of 10K 5% resistors from my employer's stock - mostly Philips SFR25 - but both metal film and carbon film, I think (brown body and light blue body).

From a sample of 50 or so, all but two were within 1 % (that's +/-

It seems from that test that current manufacturing techniques allow the makers to produce resistors of any desired value to a very close tolerance, so the ancient practice of searching your resistor stock for a specific (non-standard) value won't work any more.

Aha! No, let me explain what I meant (which is not necessarily what I said :-)

Suppose you want 1k ohms, within 1% accuracy. First you get a 5%, 1k resistor and __measure it__. If it's not within 1% of 1k, use an additional resistor to get there. If the "1k" is really 9500, then a

Mark