# word problem

• Jun 29th 2006, 09:39 AM
kwtolley
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.
• Jun 29th 2006, 11:37 AM
earboth
Quote:

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.

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}$$\displaystyle \ =\ 27\Omega Bye EB • Jun 30th 2006, 04:33 AM topsquark Quote: Originally Posted by earboth Hi, kwtolley, calculate first the specific resistance \displaystyle \rho form: \displaystyle R=\frac{\rho\cdot l}{A} where the resistance R is measured in \displaystyle \Omega, the length l in m and the area A in square-millimeters. You'll get: \displaystyle 36\Omega=\frac{\rho \cdot 20 m}{\pi \cdot (3mm)^2} \displaystyle \Longrightarrow\ \rho=\frac{36\Omega \cdot\pi\cdot 9 mm^2}{20m} Now plug in the values you know and you'll get: \displaystyle R=\frac{\frac{36\Omega \cdot\pi\cdot 9 mm^2}{20m}\cdot 60 m}{\pi\cdot 36 mm^2}$$\displaystyle \ =\ 27\Omega$

Bye

EB

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
• Jul 2nd 2006, 08:02 AM
kwtolley
word problem
thanks eb and dan for looking over this problem with me.