If the length of a conductor is increased by 20% and cross-sectional area is decreased by 4% ,then calculate the percentage change in the resistance of the conductor.
The resistance \(R\) of a conductor is given by the formula:
\[
R = \rho \frac{L}{A}
\]
where:
- \(R\) is the resistance,
- \(\rho\) is the resistivity (which remains constant for the same material),
- \(L\) is the length of the conductor,
- \(A\) is the cross-sectional area of the conductor.
We are told that the length of the conductor is increased by 20%, and the cross-sectional area is decreased by 4%. Let's analyze how these changes affect the resistance.
Step 1: Find the new length and area.
- New length \(L'\) = \(L \times (1 + 0.20) = 1.2L\)
- New area \(A'\) = \(A \times (1 - 0.04) = 0.96A\)
Step 2: Calculate the new resistance.
The new resistance \(R'\) is given by:
\[
R' = \rho \frac{L'}{A'} = \rho \frac{1.2L}{0.96A}
\]
Simplifying:
\[
R' = \frac{1.2}{0.96} \times \rho \frac{L}{A} = 1.25 \times R
\]
Step 3: Find the percentage change in resistance.
The percentage change in resistance is:
\[
\text{Percentage change} = \left(\frac{R' - R}{R}\right) \times 100
\]
Substituting \(R' = 1.25R\):
\[
\text{Percentage change} = \left(\frac{1.25R - R}{R}\right) \times 100 = 0.25 \times 100 = 25\%
\]
Final Answer:
The percentage change in the resistance of the conductor is an increase of 25%.
