Parallel Resistors (special cases formula)

AI Thread Summary
The discussion revolves around calculating the resistance of individual wires in a parallel arrangement of a car's rear window defroster, which has a total resistance of 1.4 ohms from 15 strips. The formula used to find the resistance of one wire resulted in 21 ohms. When calculating the total power dissipated with a 12 V supply, the initial attempt incorrectly used the total resistance instead of the individual wire resistance, leading to a miscalculation of 102 W. The correct power dissipation is 6.86 W, calculated using the resistance of one wire, raising questions about the interpretation of the problem regarding total versus individual power dissipation. The discussion highlights the importance of using the correct resistance value for accurate power calculations in parallel resistor configurations.
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Homework Statement


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?[/B]

Homework Equations



$$ R_{T} = \frac {R}{n} $$
$$ P = \frac{V^2}{R} $$[/B]

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?[/B]
 
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If they are indeed looking for the total power dissipated then your solution is correct and theirs is wrong.
 
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