Wires 1 and 2 are made of the same metal. Wire 2 has twice the length and twice the diameter of wire 1. What is the ratio of the resistances of the two wires?

Answer 1

Wire 1 has radius R, and wire 2 has radius 2R. Wire 1 has length L and wire 2 has length 2L. Since resistance is inversely proportional to cross sectional area and proportional to length resistance of wire 1 is:

c*L/pi*R^2, where c is a constant based on the material.

The resistance of wire 2 is:

c * 2L/pi *(2R)^2, or squaring the 2R term we have c * 2L/pi * 4R^2, and simplifying we have: c*L/2pi*R^2.

Now, to get the ratio of wire 1 to wire 2 we divide equation 1 by equation 2 and simplify:

c*L/pi*R^2 / c*L/2pi * R^2. When you divide a fraction by another fraction you invert and multiply, so this becomes

c*L *2piR^2 / c*L *pi * R^2. Each of the variables cancel leaving the ratio of resistance of wire 1 to wire 2 is 2:1, or wire 1 has twice the resistance of wire 2 even though wire 2 is twice as long.

An Alternative Answer

Resistance is directly proportional to length, and inversely proportional to cross sectional area. The cross-sectional area of a circle is proportional to the square of the diameter. So, conductor 2 is twice as long, so you can double its resistance; but its diameter is doubled, so its resistance is then reduced by a quarter. So the resistance of conductor 2 is half that of conductor 1!