Newbie question on equations

Discussion in 'General Electronics Chat' started by OBIE, Feb 28, 2012.

  1. OBIE

    Thread Starter New Member

    Feb 28, 2012
    11
    0
    Hello,
    I've been going thru the worksheets and came across some algebra manipulations that were beyond me.
    The question is:
    Manipulate this equation to solve for resistor value R1, given the values of R2 and Rparallel:

    Rparallel = R1 R2
    R1 + R2The answer is

    R1 = R2 Rparallel/ R2 − Rparallel


    Appologies for the bad paste job
    The question is
    How do you get from the first equation to solve for R1?
     
  2. MrChips

    Moderator

    Oct 2, 2009
    12,442
    3,361
    The answer you are looking for is

    \frac{1}{R}  = \frac{1}{R1}  +  \frac{1}{R2}

    Hence to solve for R1

    \frac{1}{R1}  = \frac{1}{R}  -  \frac{1}{R2}

    \frac{1}{R1}  = \frac{R2 - R}{R . R2}

    R1  = \frac{R . R2}{R2 - R}
     
  3. OBIE

    Thread Starter New Member

    Feb 28, 2012
    11
    0
    Hello Mr. Chips
    Thank you for your reply.
    I am familiar with summing the reciprocals for a parallel circuit and can solve those without any problem.
    The equation I am dealing with is an alternate, and it said that it is only used for 2 resistors in parallel.

    Rparallel= R1xR2/R1+R2
    The question is: solve for R1
    The answer given is
    R1=R2xRparallel/R2-Rparallel
    The answer is plain algebraic manipulation, but this one escapes me and I'd appreciate a step by step procedure
    Thank you
    Obie (John)
     
  4. MrChips

    Moderator

    Oct 2, 2009
    12,442
    3,361
    Then what is your question?
     
  5. crutschow

    Expert

    Mar 14, 2008
    13,014
    3,233
    Rp = R1R2 / (R1+R2)

    Rp (R1+R2) = R1R2

    RpR1 + RpR2 = R1R2

    R1R2 - RpR1 = RpR2

    R1(R2 - Rp) = RpR2

    R1 = RpR2 / (R2 - Rp)
     
  6. OBIE

    Thread Starter New Member

    Feb 28, 2012
    11
    0
    Hello Crutshow,
    Beautiful!
    Now that I see it, it's so simple. I hope with a bit more practice that I'll be able to handle most equations.
    Many thanks,
    Obie
     
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