Chapter 18, Problem 16P

To determine

The resistance of each piece.

Answer to Problem 16P

  $R_1=2\Omega$ and $R_2=8\Omega$

Explanation of Solution

Given info:

Resistance of a copper wire $=10.0\Omega$

The resistance of one piece is $4.0$ times the resistance of the other.

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:

Formula of resistivity, $R_0=\frac{\rho L_0}{A}$

By using the condition that resistance of one wire is four times the resistance of the other wire: $R_1+4R_1=R_0\to R_1=\frac{R_0}{5}$

It is known that the cross-sectional area and resistivity remains the same. The resistance of the length will be changed. If one wire has four times the resistance of the other, it must be four times as long. The shorter wire must be exactly $\frac{1}{5}$ the original length.

  $L_1=\frac{L_0}{5};L_2=L_0-L_1=L_0-L_0/5=\frac{4L_0}{5}$

It is known that the resistance is depended only on the length, the resistances are $\frac{1}{5}$ and $\frac{4}{5}$ of the original resistance $10\Omega$ . So,

  $R_1=\frac{R_0}{5}=\frac{10\Omega }{5}=2\Omega$

  $R_2=\frac{4R_0}{5}=\frac{4\times 10\Omega }{5}=\frac{40\Omega }{5}=8\Omega$

Conclusion:

The wire must be cut at a point $\frac{1}{5}$ of its total length. The resistances of the smaller piece will be $2\Omega$ and larger piece will be $8\Omega$ .