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Resistors in Parallel

Discussion in 'Electronics Homework Help' started by chopnhack, Sep 3, 2014.

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  1. chopnhack

    chopnhack

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    Apr 28, 2014
    [​IMG]

    This is an algebra issue - I have written what I thought would be the correct formula for Network C - the computer doesn't like the input method, claims that it failed to parse the equation....

    It liked my other two!!

    Mind you, this has to be in terms of R

    Parallel resistance - there are two parallel circuits.

    [(R1*R2)/(R1+R2)]*2

    but in terms of R:
    [(R^2)(2R)]*2

    This is more of a "think like a computer" issue than electrical basics knowledge.
     
  2. Gryd3

    Gryd3

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    Jun 25, 2014
    For giggles can you post the answer for A and B as well?
    Could the square brackets through it off?
     
  3. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Yeah, I imagine it doesn't like the square brackets. Just use parentheses for all nesting levels.
     
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  4. chopnhack

    chopnhack

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    I tried brackets, parenthesis, no parenthesis - a few variables, tried long form without algebraic reduction of terms...

    Giggles:

    a: R+R+R
    b: R/3

    I used brackets here for clarity only. ;-)
     
  5. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    You tried it without "^2"? Just addition, subtraction, multiplication and division symbols?
     
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  6. chopnhack

    chopnhack

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    Apr 28, 2014
    Yes, I tried this:

    (R*R)/(R+R)+(R*R)/(R+R)

    I tried this:

    ((R*R)/(R+R))*2

    No avail, I am pretty confident that a parallel resistor equation is (R1*R2)/(R1+R2)
    I think I will pass it up to the TA's for some assistance.
     
  7. Gryd3

    Gryd3

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    Resistors in Parallel:

    1/Rt = 1/Ra + 1/Rb + 1/Rc + ... + 1/Rx
    Rt = 1 / ( 1/Ra + 1/Rb + 1/Rc + ... + 1/Rx )

    (R1*R2)/(R1+R2) is the same as 1/( 1/R1 + 1/R2 ) a little algebra will allow you to change the form of the equation to something easier to work with or use.
    I'm curious why is causing the answer to fail.
     
  8. Laplace

    Laplace

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    Apr 4, 2010
    What if the computer recognized the parallel function ║; then you could write the answer as R+(R║2R)
     
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  9. chopnhack

    chopnhack

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    Just tried that and still no go.. Very strange. I am waiting on some of the TA's to recognize my post. Maybe tomorrow we will have an answer! Thanks for the thought :)
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Yes, I think it does if its the one I think it is, but it should like the others too.

    I'm not sure that its too fond of extranious spaces and make sure its not expecting r instead of R.
     
  11. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Yeah, let us know the outcome :)
     
  12. Harald Kapp

    Harald Kapp Moderator Moderator

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    I think not. Your solution is not correct. consider this schematic:
    [​IMG]
    R3 and R4 are in series, so the total resistance is R34=R3+R4.
    R34 in turn is parallel to R2, so the total resistanceR234 is R2||R34= R2*R34/(R2+R34)
    R234 in turn is in series with R1 so the total resistance of the network is R1234 = R1+ R234

    Since R1=R2=R3=R4:
    R34 = 2R
    R234 = R*2R/(R+2R)
    R1234 = R + R*2R/(R+2R) = R* (1+2/(1+2)) = R*5/3

    If it were a tool question, we would have to know which computer tool you use to solve the equations.
     

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