Why does the inverse parallel resistor formula work?

Discussion in 'Math' started by SullivanTrainingSystems, May 31, 2009.

  1. SullivanTrainingSystems

    Thread Starter New Member

    May 26, 2009
    [1] I'm curious to hear what people teach to explain *why* the inverse resistor combination formula works. I understand how parallel resistors affect total resistance and hence current, but I'm interested to learn how people understand why the math formula really works. I have a personal understanding that works for me, but what are any other views?

    [2] This is a question of semantics: "Is there any such thing as a parallel circuit, singular?"
  2. Ratch

    New Member

    Mar 20, 2007

    Then all you have to do is open up a elementary book on circuit analysis where it's mathematically proven that the formula for total resistance of paralled resistors is correct. Any student that can follow the simple math should have no trouble understanding the formula,.

    Sure, why would anyone think otherwise?

  3. SullivanTrainingSystems

    Thread Starter New Member

    May 26, 2009
    Does everyone on this forum respond with rude, demeaning and rather inane comments? People like the one who just responded to this post are the very reason why I have a job - bland, irresponsible and haughty teachers and engineers who deign to respond to students who want and need help understanding a fairly complex subject - at least for us sub-humans.

    I assume your lack of a thoughtful response is because you don't actually have a thoughtful response?

    If there's a moderator available, would it be possible to let me know why in my first two attempts to participate in these forums I've encountered such insipid reactions to my honest and genuinely sincere questions?

    I had hoped this would be a place where I could share my 25+ years of experience teaching and find colleagues interested in taking this subject to a new educational plane, since my firm belief is that anyone who has knowledge should share it - with humility.

    I have no time to humor those who have such limited vocabulary or tolerance. You may consider me "chased away". I have two businesses to run and products to sell and I'm afraid my social calendar doesn't permit me time to engage in battles of wit with cretins.

    Really - do you feel better when you attempt to belittle people?
  4. studiot

    AAC Fanatic!

    Nov 9, 2007
  5. SullivanTrainingSystems

    Thread Starter New Member

    May 26, 2009
    I asked if anyone knew *why* the formula worked - not *how* it works. I've taught parallel combinations for a quarter-century. That wasn't the point. I was looking for explanations of the math - NOT the circuitry.

    And, BTW - it's impossible to have anything be a single circuit, AND parallel at the same time. There's no such thing as a parallel circuit - singular. If it's parallel, it's plural.

    That's in an elementary text in English...
  6. studiot

    AAC Fanatic!

    Nov 9, 2007
    Sorry I was only trying to be helpful.

    How many circuits would you have if you had an amplifier with parallel connected feedback and no input signal?
  7. studiot

    AAC Fanatic!

    Nov 9, 2007
    I don't need to answer on behalf of Ratch - he is admirably able to take care of himself.

    However I would have thought that anyone teaching parallel circuits for 25 years + would know that the situation arises because resistance is the reciprocal of conductance.

    Since conductance is a measure of the 'capacity' of a current channel when you parallel them you add them together.

    This is in exact accord with the hydraulic analogy two equal pipes have twice flow capacity of one - I mean mathematically exact. But you have to take the right hydraulic analogs. Taking pressure to represent voltage will get you the wrong answer.
    The secret here is to use hydraulic head in metres or whatever.

    I am quite interested in your students, however.
    I just don't understand why they can't be expected to know basic electrical quantities. These are taught in all English secondary schools.

    This morning you were offered the opportunity to contribute from your 15 years experience to someone with considerably less but nevertheless had a genuine question.

    Could you not have used your limited time to help this guy rather than to denigrate other members?
  8. SullivanTrainingSystems

    Thread Starter New Member

    May 26, 2009
    My original post asked to have people offer *their* logic and experience of the concept of parallel resistance math, so I could see how they taught it. I didn't ask for help understanding the process. I wanted to be able to benefit from the experience of others in explaining, because there are always better ways. The first response was rude, and he last post also assumed that I needed an answer. I don't. I do appreciate the difference between an ohm and a mho. I was hoping that there would be a wealth of accumulated wisdom to share. My 25+ years of experience is only relevant to the extent that it helps young people develop a valid and real grounding (no pun intended) in electrical concepts. I don't think I was out of line being perturbed at the snobbish and off putting first response, especially given the other similar responses to the other post I referred to. It was a simple question asking for a genuine response. I didn't unfairly disparage the rude response; it wasn't a denigration, it was a retort.
  9. afaik


    Nov 2, 2008
    If everyone can put down their flame throwers for a second, I would appreciate it if someone did explain the math behind the inverse resistance formula. I understand how to use the formula, but why it works is somewhat vague to me. I do understand having two equal parallel resistances will double the current and halve the resistance, but the math behind unequal resistances gets blurry to understand afterwards, even though I know how to do it. This might be because of a weakness in mathematics, but I have not found a book to properly explain it to me yet, and I have been through many and searched the web for answers also.
    Hopefully the guns don't get pointed to me, but if anyone can answer the point of this thread, it should help plenty of people. Thanks.
  10. KL7AJ

    AAC Fanatic!

    Nov 4, 2008


    Last November, I had an article in QST called "Making the Glass Half Full" in which I discussed the little-used inverse functions:conductance, susceptance, and admittance.

    In many regards, this approach is much more intuitive...it causes you to think about how big (or how many) paths are available from a source to a load, instead of how lousy the path is. The parallel formulas for resistance are really just the reciprocals of the conductance formulas. If left in the conductance mode, it's very simple....you just add the conductances....like putting parallel pipes together. To convert the final answer back to resistance, you just take the reciprocal again...voila!

  11. studiot

    AAC Fanatic!

    Nov 9, 2007
    Hi Afaik, that was a fair question, fairly put. So here goes.

    Let us go back to Ohms law, but put it in a slightly different way, that might seem more natural.

    The current through a component is proportional to the voltage applied across its ends.

    I \propto V

    We can form an equation out of this I = sV, where s is the constant of proportionality.

    I say it is more natural because we think of the current as the result of the driving action (voltage). If we do a practical experiment to test the law we take a component and impress various voltages across it and measure the resultant current, not the other way round.

    If we take a second component then current is again proportional; to the voltage, but the constant of proportionality is different, say s2, unless the components are identical in which case s1=s2.

    Now if we connect the same voltage source across both components at the same time we will have a curent I = I1 + I2 = s1V+s2V = (s1+s2)V

    But wait, if I= sV then V = \frac{1}{s}I

    But this is the definition of resistance!

    So s = \frac{1}{R}

    And further connecting the two components as above is parallel connection!

    So I = (s1+s2)V = (\frac{1}{R1} +\frac{1}{R2})V

    s is called the conductance and is measured in Siemens.
    Conductance and resistance form reciprocal pairs.

    The corresponding pair for AC work is admittance and impedance.

    It is often easier to work in the admittance version of the formulae.

    Hope this helps.
    Last edited: Jun 5, 2009
  12. SigmaCeq


    Jul 30, 2009
    I always imagined parallel resistors as being like multiple paths. The more paths you put in, the more people can fit to walk on them. You could also use pipes with water, etc. A higher resistance means the path is narrower (less people can walk on it), so of course if you put in a lot of narrow paths right next to each other, more people can walk the same way at a time. That is how I "understand" the formula, but I have a very abstract way of thinking about things at times so I don't know if that's the sort of answer your looking for.

    As for the second semantics question, I would say that if you were referring to the entire circuit as a whole, yes, the parallel circuit could be said to be singular. Of course if you are looking at the individual components inside the circuit, it would be plural, but so would every other circuit in the world since I've never heard of a circuit with one component. Beyond that I don't really understand the question...
  13. Heavydoody

    Active Member

    Jul 31, 2009
    I am not certain precisely what it is you seek. Are you attempting to explain why one divided by the resistance is the inverse of the resistance? Or that the inverse of resistance is the conductance? Or just the overall concept of calculating the resistance of a parallel circuit by adding those conductances and then taking the overall inverse to get back to resistance?

    As a student, the first time I encountered this topic I was simply told to memorize the formula and conductance was never even mentioned. As a service technician, never using this formula, I quickly forgot it. However, further studies revealed this inverse relationship between conductance and resistance, and since then, I have always been able to determine basic parallel resistance. Just like several other respondents, the multiple paths explanation provided by conductance really opened my eyes.

    Furthermore, as a technician, while I rarely had to calculate circuit resistance, the understanding of the concept made troubleshooting significantly easier. Knowing how and why additional parallel loads contributed to or detracted from the functioning of a circuit allowed me to "think for myself" so-to-speak. That is, instead of relying solely on past experiences or what I had been told, I was able to think in terms of what-ifs.

    Later on, as a trainer, I knew I had to balance between what I expected a participant to simply "trust me" on and what I had the time to explain to the extent they actually understood what was going on. In my opinion, this explanation of conductance cannot be skipped. It is well worth the time.
  14. someonesdad

    Senior Member

    Jul 7, 2009
    The math and physics we use are models for experimental knowledge. You could approach the subject from an experimental standpoint: take two resistors and an ohmmeter and measure their separate resistances and combined resistances, both in series and parallel. Perhaps do it with multiple pairs of resistors. These are experimental data and, as such, one may want to look for a pattern in the to see if some rule can explain them.

    Now, this may seem too elementary an approach, but remember that all our models need to be based on experimental facts. And, models, to be useful, need to make predictions that can be verified by experiment.

    One could make an interesting segue into some useful Ohm's Law bashing. "Ohm's Law" was originally an experimental statement from the 1820's that the voltage across a piece of metal wire was proportional to the current through it. So V = ci where c is a constant over fairly wide ranges of current and voltage. But this "Ohm's Law" is not a law -- it fails if you apply it over wide enough ranges (if you don't believe this, keep doubling the current until you do). This brings up an interesting pedagogical discussion. What one needs to point out is that R = V/i is the definition of resistance and applies under all conditions. Unfortunately, this definition also gets labeled with "Ohm's Law" and everyone assumes that's the end of the discussion. It is, almost, in a practical sense. But I think students should understand the difference between the original "Ohm's Law" with respect to metallic conductors and the definition of resistance. The lesson for the students to learn is resistance may not be independent of current or voltage over wide enough ranges. Permanent changes or damage may occur -- consider putting too large of a current through a resistor. To be truly educated, students also need to know where their laws don't apply. And, ultimately, the world is nonlinear, as they will find out when they eventually look at lots of i vs. V characteristics.

    You could also make mention of j = σE. This is a generalization of "Ohm's Law" -- and you can mention that 1) it's a vector equation in j and E and 2) sometimes the conductivity σ is a tensor. Elementary students don't need to know what vectors, tensors, or electric fields are, of course, but you can at least mention that this equation captures the interesting case where the current density may not be in the same direction as the applied electric field. This does happen, for example in anisotropic crystals.

    Another point that can be made is that if you make sensitive voltage measurements with your voltmeter across a resistor, you'll see that the voltage fluctuates. Thus, experimentally, you have to acknowledge that the voltage isn't even constant. Nor, either, is the current. This could warrant a sentence about thermal noise and its effect on e.g. resistance measurement.

    So, there are lots of things that can be discussed in relation to the fundamental experimental observations about resistors in parallel or series.

    SullivanTrainingSystems, I think you asked your question in good faith. You got a response that wasn't to your liking and you reacted back. It's hard to know what another poster's intentions were in written conversation because we lack diction, body language, etc. I recommend that if you get a response that affronts you, just ignore it and calmly continue pursuing you original topic, explaining more if you think people aren't understanding your question. If there really was an attack on what you said, your most powerful response is to ignore it and the wind goes right out of the attack's sails. Remember, there's no chip on your shoulder unless you put it there. Or, to put it another way, there's no peeing contest unless both members want to pee. :)
  15. studiot

    AAC Fanatic!

    Nov 9, 2007
    Good Golly Gosh,

    What a truly profound statement. I take my hat off to you sir!
  16. zgozvrm

    Active Member

    Oct 24, 2009
    I think this is what you are looking for...

    With 2 parallel resistances R_1 and R_2, we know that the voltage across each is E, therefore the current through them is I_1=\frac{E}{R_1} and I_2=\frac{E}{R_2}.

    We know that the 2 currents I_1 and I_2 must add to the total current I_t. I_t = I_1 + I_2

    We also know that there is some equivalent resistance R_t that is equal to the parallel combination of R_1 and R_2. The current through this equivalent resistance must have current I_t = \frac{E}{R_t}

    Therefore, we have I_t=\frac{E}{R_t}=I_1 + I_2

    But I_1=\frac{E}{R_1} and I_2=\frac{E}{R_2}

    So, \frac{E}{R_t} = \frac{E}{R_1} + \frac{E}{R_2}

    Solving for R_t, we have:
    R_t=\frac{E}{\frac{E}{R_1} + \frac{E}{R_2}}

    Then factor the E out:
    R_t = \frac{E}{E} \times \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}

    But \frac{E}{E} = 1, so finally we have:

    R_t = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}
    Last edited: Oct 24, 2009
  17. wr8y

    Active Member

    Sep 16, 2008
    First, that is a wonderful way to state the case.

    Second, I have actually encountered degreed engineers (ok, ONE engineer) that not only didn't realize this - but was greatly offended when I accidentally pointed this out in his presence.
  18. wr8y

    Active Member

    Sep 16, 2008
    Not everyone, I find the number of such responses here to be about in line with nearly every forum I visit. The desire to appear as an authority is a strong one on the internet, I don't really know why.

    I get tired of it, too. So tired of it, that on the forum I moderate, I delete or edit such comments and issue a warning! As does all who are on the moderation team. We insist that posters conduct themselves as if speaking face-to-face. We also warn everyone that "The Spaceport.us is a community of friends". It is NOT a typical internet forum. :cool:
  19. zgozvrm

    Active Member

    Oct 24, 2009
    I'd like to see this done on all Forums. I'm in such a "discussion" right now on ElectricianTalk.com. It makes me feel like forums are only for perusing, unless you enjoy defending every word you say in a post.

    It's as if you ask someone for a Kleenex, they'll respond with, "Don't you mean 'facial tissue?' Kleenex is a brand name!" Pish-posh! In most cases, I think these people know exactly what you're talking about but, like you said, want to appear more authoritative.
  20. neonstrobe

    Well-Known Member

    May 15, 2009

    I too use the inverse to illustrate parallel resistors. I'd like to see "s" adopted as a standard inverse r too.

    The approach can be illustrated for three or more resistors:

    =>R = (R1.R2.R3)/(R2.R3 + R3.R1 + R1.R2)

    and for (n) in parallel the approach is
    top line= product of all
    bottom line = sum of the products of (n-1) terms, skipping one resistor in each term each time
    You can see above that we skipped R1 in the first term, R2 in the second ...