Chapter 18, Problem 63GP

To determine

To compare: The resistance of new combination to the resistance of the original wire.

Answer to Problem 63GP

The resistance of this new combination $\frac{1}{4}$ to the resistance of the original wire.

Explanation of Solution

Given:

A length of wire is cut in half.

Two lengths are wrapped together.

Formula used:

It is known that resistance:

  $R=\frac{\rho L}{A}$

Where,

  $\rho$ is the resistivity.

  $L$ is the length of wire

  $A$ is the cross-sectional area of the wire.

Calculation:

The equation relating resistance $R$ to the resistivity $\rho$ , length $L$ and cross-sectional area $A$ of the wire. By using the formula of resistivity:

  $R_{old}=\frac{\rho L}{A}$

It is given that a piece of wire having resistance $R$ is cut into half. The resistance of the one part will be $\frac{R}{2}$ because $R\alpha \mathcal{l}$ .

Two part are placed together side by side, the equivalent resistance will be

  $\frac{1}{\mathrm{Re}q}=\frac{1}{R/2}+\frac{1}{R/2}$

  $\frac{1}{\mathrm{Re}q}=\frac{4}{R}$

  $\mathrm{Re}q=\frac{1}{4}$

Thus, the new resistance is $\frac{1}{4}$ the original resistance.

Conclusion: The new resistance is $\frac{1}{4}$ the original resistance.