Resistance Calculation for Wire Dimensions Change

The resistance (\(R\)) of a wire depends on its material resistivity (\(\rho\)), length (\(L\)), and cross-sectional area (\(A\)). The formula is:

\( R = \rho \frac{L}{A} \)

For a cylindrical wire, the cross-sectional area is \(A = \pi r^2\), where \(r\) is the radius. Substituting this, the resistance becomes:

\( R = \rho \frac{L}{\pi r^2} \)

Initial Wire Resistance

Let the initial resistance be \(R_1\). Given:

• Length = \(L_1 = L\)
• Radius = \(r_1 = r\)
• Resistance = \(R_1 = R\)

So, the initial resistance is:

\( R = \rho \frac{L}{\pi r^1}^2 \)

New Wire Resistance Calculation

Let the new resistance be \(R_2\). The problem states the new wire has:

• Length = \(L_2 = L/2\)
• Radius = \(r_2 = r/2\)
• Material resistivity = \(\rho\) (same as initial)

First, calculate the new cross-sectional area (\(A_2\)):

\( A_2 = \pi r_2^2 = \pi \left(\frac{r}{2}\right)^2 = \pi \frac{r^2}{4} \)

Now, calculate the new resistance (\(R_2\)) using the resistance formula:

\( R_2 = \rho \frac{L_2}{A_2} \)

Substitute the values for \(L_2\) and \(A_2\):

\( R_2 = \rho \frac{L/2}{\pi r^2/4} \)

Simplify the expression:

\( R_2 = \rho \frac{L}{2} \times \frac{4}{\pi r^2} = \rho \frac{4L}{2\pi r^2} = \rho \frac{2L}{\pi r^2} \)

Relating New Resistance to Initial Resistance

We can rewrite \(R_2\) by factoring out the expression for the initial resistance (\(R\)):

\( R_2 = 2 \left( \rho \frac{L}{\pi r^2} \right) \)

Since \(R = \rho \frac{L}{\pi r^2}\), we have:

\( R_2 = 2R \)

Therefore, the resistance of the new wire is \(2R\).