Question

A $6\Omega$ resistance wire is doubled by folding. What is the new resistance?

Answer 1

Step 1: Given Data

The initial resistance on the wire $R_i=6\Omega$

Let the initial length of the wire be $L_i$.

Given that the final length of the wire $L_f=\frac{L_i}{2}$

Let the initial area of the wire be $A_i$.

Given that the final area of the wire $A_f=2A_i$

Let the resistivity of the wire be $\rho$.

Let the final resistance of the wire be $R_f$.

Step 2: Formula Used

We know that resistance is given as,

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

$\Rightarrow \rho =\frac{RA}{L}$

Step 3: Calculate new Resistance

Since the resistivity of the wire is the same in both cases, we can write

$\rho =\frac{R_i{A}_i}{L_i}=\frac{R_f{A}_f}{L_f}$

Upon substituting the values we get

$\frac{6A_i}{L_i}=\frac{4A_i{R}_f}{L_i}$

$\Rightarrow R_f=1.5\Omega$

Hence, the new resistance of the wire will be $1.5\Omega$.