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Parallel resistanee is reciprocal of the sum?

  1. Feb 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Why is parallel resistance the reciprocal of all individual resistances?

    2. Relevant equations


    3. The attempt at a solution

    Well, since V is constant and I is different, you can write it as I=V/R, and since V won't change, you can make it I=V*(1/R1+1/R2),etc. So I get that, but why do you then have to take the reciprocal of all the resistances to get the total resistance? Wouldn't it just be the direct sum of the individual 1/Rs??
  2. jcsd
  3. Feb 6, 2009 #2


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    Because, by definition, [itex] I = V/R_{eq} [/itex]. So setting this equal to your expression, we get

    [tex] \frac{1}{R_{eq}} = \frac{1}{R_1 + R_2 + \ldots} [/tex]
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