A 6 $\Omega$ resistance wire is doubled on itself. What is the new resistance of the wire, assuming uniform conductivity?

Steps to Solve

1. Write the formula for resistance

The resistance $R$ of a wire is given by the formula:

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

where $\rho$ is the resistivity of the material, $L$ is the length of the wire, and $A$ is the cross-sectional area of the wire.

2. Determine the new length and area

When the wire is doubled on itself, the new length $L'$ is half of the original length, so $L' = \frac{L}{2}$. The new area $A'$ is twice the original area, so $A' = 2A$.

3. Calculate the new resistance

The new resistance $R'$ can be calculated as follows:

$$ R' = \rho \frac{L'}{A'} = \rho \frac{L/2}{2A} = \rho \frac{L}{4A} = \frac{1}{4} \rho \frac{L}{A} $$

4. Express the new resistance in terms of the original resistance

Since $R = \rho \frac{L}{A}$, we can substitute $R$ into the equation for $R'$:

$$ R' = \frac{1}{4} R $$