Question

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

Answer 1

Step 1: Given data:

Initial resistance $R=4.25\Omega$

Length increases by 4 times,$l^{\text{'}}=4l$

Area decreased by three times, $A\text{'}=\frac{A}{3}$

Step 2: Resistance in terms of specific resistance,

The formula for finding resistance is $R=\frac{\rho l}{A}$

To find the new resistance we use the formula, $R^{\text{'}}=\frac{\rho l\text{'}}{A^{\text{'}}}$

$R^{\text{'}}=$New resistance

$l^{\text{'}}=$New length

$A^{\text{'}}=$New area

$\rho =$Specific resistance

Substituting the values in the equation:

$$\begin{gathered} R^{\text{'}}=\frac{\rho l\text{'}}{A^{\text{'}}} \\ {R}^{\text{'}}=\frac{\rho \times 4I}{A/3} \\ {R}^{\text{'}}=\frac{12\rho I}{A} \\ {R}^{\text{'}}=12R \end{gathered}$$

Step 3: Find the resistance of another wire,

Substitute the value of R from the previous step.

$$\begin{gathered} R^{\text{'}}=12\times 4.25 \\ {R}^{\text{'}}=51\Omega \end{gathered}$$

Thus, the resistance of another wire made from the same material is $51\Omega$.