Please turn on javascript in your browser to play chess.
Posers and Puzzles

Posers and Puzzles

  1. Subscriber sonhouse
    Fast and Curious
    19 May '14 10:50 / 1 edit
    So I was showing the CEO of our company a simple but redrawn schematic of a three resistor problem, which, when analyses correctly shows it to be just three resistors in parallel.

    Now I have shown that problem to potential candidates for a position here and nobody incoming has passed.

    I showed it to our CEO and I was surprised he just did the problem by making an assumption without actually going through a circuit analysis and said he remembers parallel resistors and used a formula, R1*R2*R3 divided by R1+R2+R3.

    I didn't want to let him down, he's the CEO....

    That formula poops out after 2 resistor problems, the best you can get is R1*R1 divide by R1+R2 and for those problems it works every time.

    The way we were taught to do three and up parallel resistors is the formula R1 invert +R2 inv+ R3 inv and invert THAT result, which works every time for any number of resistors in parallel.

    What I noticed about the R1*R2*R3 version is using 10 ohms apiece for the three resistors, the answer should be 3.333.... ohms, 3 and a third ohms.

    But using the triple multiplication formula you get for 10 ohms, 1000 divide by 30 which happens to be exactly 10 times the real answer.

    I thought I was on to something there where I could just go divide the whole thing by the resistor value, an added step and it works with 10 ohms but if I do the same with 7 ohms it comes out as 343 (7 cubed) divide by 21 (7X3) and that comes out as 16 and a third. So now divide THAT by 7 and you get 2 and 1/3. No help. It does appear SOMETHING is going on here, 7/2.3333 is 3. Not sure what that means but there it is.

    So the question is, what is going on that the double formula works and the triple doesn't and of course quad and quints too.

    So another question, is there a reasonably simple modification of the triple and up formula that will work in all cases as good as the inversion formula, is there some trick you can add to the triple and up formulae that would make it work to solve the simple case of parallel resistors of any number and any resistance, except zero of course. Stupid Zero
  2. Subscriber sonhouse
    Fast and Curious
    19 May '14 11:58
    Originally posted by sonhouse
    So I was showing the CEO of our company a simple but redrawn schematic of a three resistor problem, which, when analyses correctly shows it to be just three resistors in parallel.

    Now I have shown that problem to potential candidates for a position here and nobody incoming has passed.

    I showed it to our CEO and I was surprised he just did the problem ...[text shortened]... ase of parallel resistors of any number and any resistance, except zero of course. Stupid Zero
    I looked closer, did the numbers with the triple version and quad version and found my answer for 7 ohms listed was right, 7 ohms in parallel IS 2 1/3 ohms.

    So the result can be (R1*R2*R3/R1+R2+R3) divide by R, the first order.

    Then a quad, (R1*R2*R3*R4/R1+R2+R3+R4) divided by R squared.

    and so forth. Now that is for equal resistors. Can the same be done if the resistors are NOT equal? What version of this formula works for all resistors regardless of value?
  3. Standard member forkedknight
    Defend the Universe
    19 May '14 16:20 / 1 edit
    Originally posted by sonhouse
    I looked closer, did the numbers with the triple version and quad version and found my answer for 7 ohms listed was right, 7 ohms in parallel IS 2 1/3 ohms.

    So the result can be (R1*R2*R3/R1+R2+R3) divide by R, the first order.

    Then a quad, (R1*R2*R3*R4/R1+R2+R3+R4) divided by R squared.

    and so forth. Now that is for equal resistors. Can the same b ...[text shortened]... sistors are NOT equal? What version of this formula works for all resistors regardless of value?
    The actual formula: Rt = 1/(1/R1 + 1/R2 + 1/R3)

    Multiply by R1/R1:
    Rt = R1 / (1 + R1/R2 + R2/R3)

    Multiply by R2/R2:
    Rt = R1*R2 / (R2 + R1 + R1*R2 / R3)

    Multiply by R3/R3:
    Rt = R1*R2*R3 / (R1*R2 + R1*R3 + R2*R3)

    Basically, you get all of the R values multiplied together in the numerator, and the sum of each combination of R values multiplied in the denominator.

    Now, if R1 = R2 = R3, you can factor R out of the denominator:
    Rt = R*R*R / R * (R + R + R)
    Rt = R / (1 + 1 + 1)

    This can be generalized for N equal resistors:
    Rt = R / N

    There are lots of formulas that can be greatly simplified with assumptions that certain values are always equal.
  4. Standard member forkedknight
    Defend the Universe
    19 May '14 16:27 / 1 edit
    I always think of it this way:

    For resistors in series, add up component resistance to get total resistance.
    Rt = R1 + R2 + ... + Rn

    For resistors in parallel, add up component conductance to get total conductance.
    1/Rt = 1/R1 + 1/R2 + ... + 1/Rn
  5. Subscriber sonhouse
    Fast and Curious
    19 May '14 17:18
    Originally posted by forkedknight
    I always think of it this way:

    For resistors in series, add up component resistance to get total resistance.
    Rt = R1 + R2 + ... + Rn

    For resistors in parallel, add up component conductance to get total conductance.
    1/Rt = 1/R1 + 1/R2 + ... + 1/Rn
    Yes, and it is pretty easy to do on a casio.

    I was just wondering what is going on with the multiplication method and why it only works as a 2 resistor thing, then goes off on a tangent when using it as a three or four resistor calculation.

    I found with equal resistors the multiplication theme gives answers that start off with R^2 for three resistors and R^3 for 4 resistors but that is a special case of equal resistances.

    It REALLY gets bollixed up when you do larger numericals for resistances, even with only 4 of uneven resistors.

    I found them off by maybe 2 to 1 at low resistances but when I did the numbers for 5 digit resistances, just made up a bunch of 5 digit resistances and did both methods, the inversion method which we know works for any value and the multiplication version. at 5 digits, the invert version gave reasonable and correct 4 and 5 digit answers.

    The same multiplication formula for 5 digits made it look like it was a trillion ohms! Just a BIT off