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Parallel Resistors (special cases formula)

  1. Oct 30, 2014 #1

    zug

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    1. The problem statement, all variables and given/known data
    Problem:
    A car's rear window defroster uses 15 strips of resistive wire in a parallel arrangement. If the total resistance is 1.4 ohms, what is the resistance of one wire?
    Solution: rearrange formula to $$ R nR_{T} = (15)(1.4 \Omega) = 21 \Omega $$

    Question: What is the total power dissipated in the defroster if 12 V is applied to it?


    2. Relevant equations

    $$ R_{T} = \frac {R}{n} $$
    $$ P = \frac{V^2}{R} $$




    3. The attempt at a solution
    To get the total power dissipated I assumed I would use the formula $$ P = \frac{V^2}{R} = \frac{12^2}{1.4 \Omega} = 102W $$

    I was wrong though, the answer is 6.86 W.

    $$ P = \frac{V^2}{R} = \frac{12^2}{21 \Omega} = 6.86 W $$


    What I'm guessing is they used the resistance value of one wire (21 ohms) for the resistance. My question is, why did they use the resistance of one wire and not the total resistance of the wires? Especially since they were asking for the total power dissipated?
     
  2. jcsd
  3. Oct 30, 2014 #2

    gneill

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    Staff: Mentor

    If they are indeed looking for the total power dissipated then your solution is correct and theirs is wrong.
     
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