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Question:

Doubling the length and halving the diameter of a wire?

A wire has resistance 56.8 ohms. If another wire consists of the same material but has twice the length and half the diameter of the first, what is the resistance of the 2nd wire?

Answer:

The resistance of a wire is given by R = rho*L/A, L is length, A is cross-sectional area, and rho is resistivity, which is dependent on material. It will be constant here as we are dealing with the same material. Let's say that the radius of the first wire is r1. R1 = rho*L1/A1 = rho*L1/(pi*r1^2) rho = R1*A1/L1 = R1*pi*r1^2/L1 For the second wire, R2 = rho*L2/A2 = L2/(pi*r2^2) * R1*Pi*r1^2/L1 L2 is twice L1, L2=2*L1. The diameter is half the first, which means that the radius of half the first, so r2 = r1/2. R2 = 2*L1/(pi*(r1/2)^2) * R1*pi*r1^2/L1 = 8*R1 = 454 ohms.
Resistance = rho * L/A where rho = resistivity of the material, so it's the same for both wires. You've doubled the length, and you've divided the area by a factor of 2^2 = 4. Calculate rho * (2L)/(0.25A) and compare that to rho * L/A.
The two wires are composed of the same material, so we can concern ourselves solely with their respective dimensions. The second wire has half the diameter of the first, which means a cross section of the second wire has only 1/4th the cross-sectional area of the first wire (the area of a circle being proportional to the square of the diameter). The second wire is also twice as long as the first wire. Since electrical resistance is inversely proportional to the cross-sectional area of the wire and directly proportional to the length of the wire, we must have the resistance of the second wire to be 1/(1/4)*2*56.8 = 4*2*56.8 = 8*56.8 = 454.4 ohms

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