Question

A resistance wire made from German silver has a resistance of 4.25 Ω. Calculate the resistance of another wire, made from the same material, such that its length increases by 4 times and area of cross-section decreases by three times.

Answer 1

Case 1: R = 4.25 Ω

Let \(\text{‘}l\text{’}\) be length and \(\text{‘}a\text{’}\) the area of cross-section

∴ \[R=\rho\frac{l}{a}\]

\[4.25=\rho\frac{l}{a}\qquad\text{...(1)}\]

Case 2: \[R_{1}=?;\ \text{length}=4l;\ \text{area}=\frac{a}{3}\]

∴ \[R_{1}=\rho\frac{4l}{a/3}\qquad\text{...(2)}\]

Dividing (2) by (1)

\[\frac{R_{1}}{4.25}=\frac{\rho\cdot4l}{a/3}\times\frac{a}{\rho\cdot l}\]

∴ \[R_{1}=4.25\times12=51\ \Omega\]