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Resistor equation problem

  1. Sep 26, 2015 #1
    1. The problem statement, all variables and given/known data
    The equivalent resistance of the series combination of two resistors is p.When
    they are joined in parallel, the equivalent resistance is q. If p = nq, find the
    minimum possible value of n.

    2. Relevant equations
    1/Rparallel = 1/R1+1/R2+...
    Rseries = R1+R2+...

    3. The attempt at a solution

    R1+R2=p
    R1 R2 / ( R1 + R2 ) = q

    Then, I substitute p as nq

    R1+R2=nq
    R1 R2 / ( R1 + R2 ) = q

    I divide the equation 1 and equation 2 and get
    ##n=\frac{(R_1+R_2)^{2}}{R_1R_2}##
    which I think that the minimum value is zero

    However, the textbook says that the answer is 4
    Please help
     
  2. jcsd
  3. Sep 27, 2015 #2

    ehild

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    R1+R2=nq. Resistances are positive quantities. If n=0 R1+R2 =0. Is it possible???
     
  4. Sep 27, 2015 #3
    Hmm.. Yes, you're right.. It's impossible

    So, how come we get 4 ??

    I think of using derivative to get the minimum value..
    The y-axis is n
    But, I don't know what my x-axis is.
     
  5. Sep 27, 2015 #4

    ehild

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  6. Sep 27, 2015 #5

    rude man

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    hint: fix R2 = constant, then minimize n with respect to R1 & see what if anything comes out of it.
     
  7. Sep 27, 2015 #6
    ##
    n=\frac{R_1^2+2R_1R_2+R_2^2}{R_1R_2} = 2 + \frac{R_1^2}{R_1R_2} + \frac{R_2^2}{R_1R_2}
    ##

    The AM GM of function y= ##\frac{R_1^2}{R_1R_2} + \frac{R_2^2}{R_1R_2}## is
    ##
    \frac{R_1^2+R_2^2}{R_1R_2} \geq 2\sqrt{\frac{R_1^2 R_2^2}{R_1^2 R_2^2}}
    ##
    ## \frac{R_1^2+R_2^2}{R_1R_2} \geq 2 ## The minimum value of this function is 2
    So, the minimum value of n is 2+2 = 4...
    Thanks a lot for your help !
     
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