If a wire of resistance R, resistivity ñ "roe", length L, and cross sectional area A, had its length doubled and radius halved, what would be the ratio of R'/R. Where R' is the new resistance of the wire.
I thought it would be 4 times R, giving the ratio of 4R:R'
However, I plugged theoretical numbers in to check and it did not match what I thought. Can someone please explain the ratio to me?
R1=ρ•L1/A1
R2= ρ•L2/A2= ρ•2•L1•2/A1=4• ρ•L1/A1=4•R1
To find the new resistance of the wire, we need to consider the relationship between resistance, resistivity, length, and cross-sectional area. The resistance of a wire is given by the formula:
R = (ρ * L) / A
where R is the resistance, ρ (rho) is the resistivity, L is the length of the wire, and A is the cross-sectional area of the wire.
In this case, we have two scenarios: the initial wire and the changed wire. Let's call the initial wire's length L1, resistivity ρ1, and cross-sectional area A1, and the changed wire's length L2, resistivity ρ2, and cross-sectional area A2.
According to the problem, the length is doubled, so L2 = 2L1. And the radius is halved, which means the cross-sectional area is one-fourth of the initial area, so A2 = (1/4) * A1.
Now, let's calculate the new resistance R' using these values:
R' = (ρ2 * L2) / A2
Substituting our given values:
R' = (ρ2 * 2L1) / ((1/4) * A1)
Simplifying, we can cancel out the 2's and the 1/4:
R' = (2 * ρ2 * L1) / A1
Now, we can compare the ratio R'/R by dividing R' by R:
(R'/R) = [(2 * ρ2 * L1) / A1] / [(ρ1 * L1) / A1]
Simplifying, we can cancel out the A1's and L1's:
(R'/R) = (2 * ρ2) / ρ1
So, the ratio of the new resistance R' to the initial resistance R is simply:
(R'/R) = 2 * ρ2 / ρ1
In summary, the ratio (R'/R) depends on the ratio of the resistivity of the new wire ρ2 to the resistivity of the initial wire ρ1, and this ratio is multiplied by 2. Therefore, the ratio (R'/R) is not necessarily equal to 4 times R, as you initially thought, but rather depends on the resistivity of the materials used in the wires.