A wire 10 m long with a diameter of 1.0 mm has a resistance of 5.0 Ω. The resistance of a second wire made of the same material, but 3.0 m long with a diameter of 4.0 mm, is approximately

Question

Quizplus · Accepted Answer

We can use the formula for resistance of a wire:

$R = ho \frac{l}{A}$

where $R$ is resistance, $ho$ is resistivity (a constant for the material), $l$ is length, and $A$ is cross-sectional area.

For the first wire, we have:

$5.0\ \Omega = ho \frac{10\ m}{(\pi/4)(1\ mm)^2}$

Solving for $ho$, we get:

$ho = \frac{5.0\ \Omega \cdot (\pi/4)(1\ mm)^2}{10\ m} \approx 6.37 	imes 10^{-8}\ \Omega\cdot m$

Now we can use this resistivity value for the second wire. The only difference is that the length and cross-sectional area are different. The cross-sectional area is proportional to the square of the diameter, so we have:

$A_2 = \frac{(4\ mm)^2}{(1\ mm)^2}A_1 = 16A_1$

For the length, we have $l_2 = 3.0\ m$.

Putting it all together, we get:

$R_2 = ho \frac{l_2}{A_2} = \frac{ho}{l_2}\frac{1}{16A_1}l_2 = \frac{1}{16}\frac{R_1}{l_2} = \frac{1}{16}\frac{5.0\ \Omega}{3.0\ m} \approx 0.094\ \Omega$

A Wire 10 M Long with a Diameter of 1