word problem

The electrical resistance R of a wire varies dirctly as its length L and inversely as the square of its diameter. A wire 20 meters long and 0.6 centimeters in diameter made from a certain alloy has a resistance of 36 ohms. What is the resistance of a piece of wire 60 meters long and 1.2 centimeters in diameter made from the same material? My answer is 24 ohms because of wire size and length. Thanks for checking my answer.

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Originally Posted by kwtolley

The electrical resistance R of a wire varies dirctly as its length L and inversely as the square of its diameter. A wire 20 meters long and 0.6 centimeters in diameter made from a certain alloy has a resistance of 36 ohms. What is the resistance of a piece of wire 60 meters long and 1.2 centimeters in diameter made from the same material? My answer is 24 ohms because of wire size and length. Thanks for checking my answer.

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Hi, kwtolley,

calculate first the specific resistance $\displaystyle \rho$ form:

$\displaystyle R=\frac{\rho\cdot l}{A}$ where the resistance R is measured in $\displaystyle \Omega$, the length l in m and the area A in square-millimeters.

You'll get:

$\displaystyle 36\Omega=\frac{\rho \cdot 20 m}{\pi \cdot (3mm)^2}$ $\displaystyle \Longrightarrow\ \rho=\frac{36\Omega \cdot\pi\cdot 9 mm^2}{20m}$

Now plug in the values you know and you'll get:

$\displaystyle R=\frac{\frac{36\Omega \cdot\pi\cdot 9 mm^2}{20m}\cdot 60 m}{\pi\cdot 36 mm^2}$$\displaystyle \ =\ 27\Omega$

Bye

EB

Quote:

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Originally Posted by earboth

Hi, kwtolley,

calculate first the specific resistance $\displaystyle \rho$ form:

$\displaystyle R=\frac{\rho\cdot l}{A}$ where the resistance R is measured in $\displaystyle \Omega$, the length l in m and the area A in square-millimeters.

You'll get:

$\displaystyle 36\Omega=\frac{\rho \cdot 20 m}{\pi \cdot (3mm)^2}$ $\displaystyle \Longrightarrow\ \rho=\frac{36\Omega \cdot\pi\cdot 9 mm^2}{20m}$

Now plug in the values you know and you'll get:

$\displaystyle R=\frac{\frac{36\Omega \cdot\pi\cdot 9 mm^2}{20m}\cdot 60 m}{\pi\cdot 36 mm^2}$$\displaystyle \ =\ 27\Omega$

Bye

EB

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Just a thought for future consideration. There's nothing wrong with what earboth did here, but I would recommend using a single unit for length. Specifically, use m for the length and m^2 for the cross-sectional area. In this particular problem the different units end up cancelling so there's no problem, but this won't happen in general.

-Dan

thanks eb and dan for looking over this problem with me.