Talk:Current divider

Untitled
Can someone change the current divider law picture. The law is Rx_current_unknown = Itotal_x(R_total/Rx) the picture has an extra R total at the bottom of the denominator —Preceding unsigned comment added by 99.249.120.189 (talk) 20:12, 9 January 2011 (UTC)

I am not shure what needs to be cleaned up in it. i agree that some pictures would help but since i cannot host any..

as for merging it with the resistors i would have to disagree. It does not mention how to find current through a resistor.


 * A merge to Ohm's Law would seem more appropriate. Note that merge means summarise in the other article, not just redirect. - Just zis Guy, you know? [T]/[C] AfD? 22:12, 11 December 2005 (UTC)

Ohm's law talkes about how voltage equals current divided times resistance


 * Gee, really? And how do you think the current divider (and potential divider come to that) equations are derived?  Please note: I do have some minor understanding of this, my (B.Eng) degree is in electrical engineering and I learned Ohm's law on my father's knee, since he was an electrical engineering teacher at a college :-) I'd say that at the very least Voltage divider and Current divider should be merged and actually I'd put both in Ohm's law since the three articles are inextricably linked. - Just zis Guy, you know? [T]/[C] AfD? 09:58, 12 December 2005 (UTC)

Somebody add a proof please? i had to figure it out on my own — Preceding unsigned comment added by 94.209.183.26 (talk) 16:31, 18 September 2014 (UTC)

The rules
It is true that the CDR and VDR and Ohms law are the very basics of electrical enginering. however i disagree with you comments that they should be merged into the ohms law.

Current Divider and Voltage Divider do use ohms law and are a pratical application of it. however they are seperate elements and most textbooks i read have them divided up into different sections as well. 08:19, 17 December 2005 (UTC)

Alternative Method?
There seems to be another way of using the current divider rule. The formula is slightly different. I suggest adding this to the article as well if this method is correct. Here's the link to the site: http://www.wisc-online.com/objects/index_tj.asp?objID=DCE3502 --Pavithran 08:50, 4 October 2007 (UTC)


 * Nope, it looks the same. Only difference might be that they define RT = R1 + R2. -Roger (talk) 18:05, 6 January 2008 (UTC)


 * While the answer is the same, I think that the algebraic idea of using RT = (the total resistance of the "parallel network") pedagogically makes more sense for people learning how to solve circuits. If we use a property of parallel circuits, that the voltage across each component is the same: VT = V1, and then apply Ohm's law for finding voltage from resistance and current, then IT*RT = I1*R1. From this stage, solving for the current is significantly more simple, I1 = RT/R1*IT . This solution was inspired by my Circuits I professor's lecture. ~(Can't sign for security reasons, the IP Address is at Rochester Institute of Technology). — Preceding unsigned comment added by 129.21.69.191 (talk) 17:53, 12 October 2022 (UTC)

Article should be renamed for consistency
Since current division is the dual of voltage division, I propose we rename this article for consistency. Plus I don't think I've ever heard the term "current divider rule". People usually just say "current divider". -Roger (talk) 17:01, 6 January 2008 (UTC)
 * I agree that the added word "rule" for both current and voltage cases departs from standard usage. It should be dropped in both cases. Brews ohare (talk) 17:39, 6 January 2008 (UTC)
 * In addition, the voltage divider article is more complete and should be made a model for this page as well.Brews ohare (talk) 17:43, 6 January 2008 (UTC)
 * I have rewritten the intro, made some corrections, and redirected this page to avoid the term rule. Brews ohare (talk) 18:56, 18 January 2008 (UTC)

Is there an error in the resistive divider paragraph?
According to my book the right way to find $$I_x$$ is $$I_x = \frac{R_T}{R_x} I_T$$ where $$R_T = R_1 \parallel R_2 \parallel R_3 \parallel \ldots \parallel R_n$$. If we use the formula $$I_x = \frac{R_T}{R_x + R_T} I_T$$ (the one that actually is in that paragraph) $$R_T$$ shouldn't include $$R_x$$ and this is not clear from there. I also took the values from an exercise on the book and I checked with Python that both the methods work.

>>> def par(*res):

...  return 1 / sum(1/r for r in res)

>>> r1 = 10

>>> r2 = 2

>>> r3 = 20

>>> it = 4

>>> # i1 should be equal to 0.6154 ampere

>>> # method 1

>>> rt = par(r1, r2, r3)

>>> (rt/r1) * it

0.61538461538461531

>>> # method 2

>>> rt = par(r2, r3)

>>> (rt/(r1+rt)) * it

0.61538461538461531

As you can see in the method 2 there are only R2 and R3. —Preceding unsigned comment added by 130.232.126.213 (talk) 20:24, 11 October 2008 (UTC)


 * Looks like your mistake is that you included $$R_x$$ in $$R_T$$. Both your book and the article are correct, they just have different definitions of $$R_T$$.
 * I like your use of Python BTW. I believe there are a couple circuit simulator libraries written in Python, though I haven't tried them yet. -Roger (talk) 19:38, 12 October 2008 (UTC)


 * I know that both the ways are correct, I was just pointing out that from that paragraph is not clear that $$R_x$$ mustn't be included when you calculate $$R_T$$. Usually when you calculate $$R_T$$ (total resistance) you include all the values, but with that formula you have to omit $$R_x$$. I think it would be better to show both the methods and explain that they have different definition of $$R_T$$. I also find $$I_x = \frac{R_T}{R_x} I_T$$ simpler, you can find all the currents just changing the value of $$R_x$$ without having to change and recalculate $$R_T$$ (with $$I_x = \frac{R_T}{R_x + R_T} I_T$$ you have as many different $$R_T$$ as the number of resistors in the circuit), moreover you often already have the total resistance. —Preceding unsigned comment added by 130.232.126.213 (talk) 00:03, 21 October 2008 (UTC)

I think there still is an error. Take the example $$ \frac{R_{T}}{R_{T}+R{X}}I_{T} $$, for two resistors in parallel. If we do this, and note that $$ \frac{1}{R_{T}} = \frac{1}{R_{1}}+\frac{1}{R_{2}} $$ we get, by insertion, $$ I_{1} = \frac{(\frac{1}{R_{1}}+\frac{1}{R_{2}})^{-1}}{(\frac{1}{R_{1}}+\frac{1}{R_{2}})^{-1}+R1}I_{T} $$, and extending with $$ (\frac{1}{R_{1}}+\frac{1}{R_{2}})$$, we get: $$ \frac{1}{R_{1}(\frac{1}{R_{1}}+\frac{1}{R_{2}})+1}I_{T} = \frac{R_{2}}{2R_{2}+R_{1}}I_{T} $$, which is NOT equal to what the textbook gives us, which is, as noted above, $$ \frac{R_{2}}{R_{2}+R_{1}}I_{T} $$, for the case of two transistors in parallel. So we should either change the definition for $$ I_{T} $$ in the article, or write another formula for the voltage division (the equation for division and the definition for equivalent resistance are inconsistent).

RTotal -> RX
Can someone please double check I was correct in making my last edit? I changed RTotal to RX — Preceding unsigned comment added by 2601:880:C100:69D0:C414:6EEF:8F2A:646A (talk) 22:50, 7 December 2016 (UTC)


 * You are correct. I just double checked the simple derivation from Ix Rx = (IT - Ix) RT. digital_me (talk) 23:51, 5 February 2017 (UTC)

External links modified
Hello fellow Wikipedians,

I have just modified one external link on Current divider. Please take a moment to review my edit. If you have any questions, or need the bot to ignore the links, or the page altogether, please visit this simple FaQ for additional information. I made the following changes:
 * Added archive https://web.archive.org/web/20070227223015/http://utwired.engr.utexas.edu/rgd1/lesson05.cfm to http://utwired.engr.utexas.edu/rgd1/lesson05.cfm

When you have finished reviewing my changes, you may follow the instructions on the template below to fix any issues with the URLs.

Cheers.— InternetArchiveBot  (Report bug) 16:06, 15 August 2017 (UTC)

Possible error
Hi! A friend of mine has pointed out to me that maybe $$I_X = \frac{Z_T} {Z_X}I_T \ $$ should be edited in $$I_X = \frac{Z_T} {Z_X+Z_T}I_T \ ,$$ and $$I_X = \frac{R_T}{R_{X} }I_T \ $$ in $$I_X = \frac{R_T}{R_{X}+R_T}I_T \ $$, and the sentence "where ZT refers to the equivalent impedance of the entire circuit" is probably wrong. Could someone have a look? Thank you, --Epì dosis 10:32, 20 May 2019 (UTC)

In response to above: I believe both are correct. But the original form of $$I_X = \frac{R_T}{R_X+R_T} I_T $$ is (in my mind) extremely confusing. While Nilson/Riedal use this form (I went and checked my copy as I didn't believe the reference at first), every circuit division rule I've seen (besides this one) uses the form of $$I_X = \frac{R_T}{R_X} I_T$$, because in general, $$R_T$$ always refers to the entire circuit (or at least, the entire parallel portion). Defining it to be the other way is just... odd. GetaR (talk) 00:56, 19 January 2020 (UTC)