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Equivalent resistance in a parallel circuit

  1. Jul 19, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that the equivalent resistance in a parallel circuit with 2 resistors with resistance R1 and R2 is always lesser than the resistance of either resistor.

    2. Relevant equations

    Rt= (R1 x R2)/(R1 + R2)


    3. The attempt at a solution

    I tried starting from R1 + R2 > R1 , where R2 > 0
     
  2. jcsd
  3. Jul 19, 2013 #2

    gneill

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    Staff: Mentor

    If Rt is smaller than either R1 or R2, then what does that say about the ratios Rt/R1 and Rt/R2?
     
  4. Jul 19, 2013 #3

    ehild

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    Gold Member

    That is a good start. The resistances are all positive, R1, R2 and also Rt.

    You know that the reciprocal of the parallel resistances add up:

    1/Rt=1/R1+1/R2

    Multiply the equation by Rt. ...

    ehild
     
  5. Jul 19, 2013 #4
    You can simply compute the difference between R_t and R_1 or R_2 wich is easily seen to be positive

    [tex] \frac {R_1 R_2} {R_1 + R_2} - R_1 = [/tex]
     
  6. Jul 19, 2013 #5
    EDIT: I've come up to R1/(R1 + R2) > R1 , where R2 > 0

    since R1+R2>R1 , then when R1 is divided by R1+R2(a larger number), surely R1/(R1+R2) is lesser than the original number.

    now I can't have any reasoning that when I multiply R1/(R1+R2) by R2, tht this inequality still stands.

    EDIT:
    I see thanks, but isn't it the other way around?
     
    Last edited: Jul 20, 2013
  7. Jul 20, 2013 #6

    Curious3141

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    You must have meant negative, the way that difference was written.

    For the thread starter: this is a simple way to do it. Just express the second term over the same denominator, and combine terms and cancel stuff out. The end result should be obviously negative. You can then repeat it for ##R_2##.
     
  8. Jul 20, 2013 #7

    ehild

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    1=Rt/R1+Rt/R2

    As both terms on the right-hand side must be positive, both are smaller than 1.

    ehild
     
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