Question

A $6\Omega$ resistance wire is doubled on itself. Calculate the new resistance of the wire.

Answer 1

Step 1: Given data and formula used

Given data:

1. Resistance of the wire is $R=6\Omega$

Consider the length of the wire be $l$ and the area of the cross-section of the wire be $A$.

Formula:

The resistance of the wire is given by $R=\rho \frac{l}{A}$

Where $\rho$ is the resistivity of the material of the wire.

When the wire is doubled on itself, the length of the wire becomes $l/2$ and the area of cross-section of the wire becomes $2A$.

The new resistance of the wire becomes $R\text{'}=\rho \frac{l/2}{2A}$

Step 2: Finding the formula of new resistance of the wire

Divide $R\text{'}$ by $R$ to calculate the new resistance of the wire.

$$\begin{gathered} \frac{R\text{'}}{R}=\frac{\rho {\displaystyle \frac{l/2}{2A}}}{\rho {\displaystyle \frac{l}{A}}} \\ \frac{R\text{'}}{R}=\frac{l}{4A}\times \frac{A}{l} \\ \frac{R\text{'}}{R}=\frac{1}{4} \\ R\text{'}=\frac{R}{4} \end{gathered}$$

Step 3: Calculating the new resistance of the wire

Substituting the value of resistance into the formula to calculate $R\text{'}$.

$$\begin{gathered} R\text{'}=\frac{6\Omega }{4} \\ R\text{'}=1.5\Omega \end{gathered}$$

Thus, the new resistance of the wire is $1.5\Omega$.