Question

A wire of resistivity ? must be replaced in a circuit by a wire of the same material but 4 times as long. If, however, the resistance of the new wire is to be the same as the resistance of the original wire, the diameter of the new wire must beAnswer4 times the diameter of the original wire.1/2 the diameter of the original wire.1/4 the diameter of the original wire.the same as the diameter of the original wire.2 times the diameter of the original wire.

Answer 1

Answer:

2 times the diameter of the original wire.

Explanation:

Let's call $\rho$ the resistivity of the wire. The original resistance of the wire is given by

$R=\rho \frac{L}{A}$

where L is the length of the wire and A is the cross-sectional area. Writing the area as

$A=\pi (\frac{d}{2})^2$

where d is the diameter and d/2 corresponds to the radius, we can rewrite the resistance of the wire as

$R=\rho \frac{L}{\pi (d/2)^2}=4\frac{\rho L}{\pi d^2}$

Which can be rewritten solving for d, the diameter:

$d=\sqrt{\frac{4\rho L}{\pi R}}$

Now, we know that the wire must be replaced by a wire with same material (so, same resistivity $\rho$), but with 4 times long, which means the new length is $L'=4L$. Substituting this into the formula, and keeping in mind that the resistance R must remain the same, we find the new diameter:

$d'=\sqrt{\frac{4 \rho L' }{\pi R}}=\sqrt{\frac{4 \rho (4L) }{\pi R}}=2\sqrt{\frac{4 \rho L' }{\pi R}}=2d$

so, the diameter must be

2 times the diameter of the original wire.