Chapter 18, Problem 20P

To determine

The voltage-drop along a $26m$ length household no. $14$ copper.

Answer to Problem 20P

  $2.518V$

Explanation of Solution

Given info:

Resistivity of copper, $\rho =1.68\times {10}^{-8}\Omega \cdot m$

Length; $L=26.0m$

Diameter, $D=1.628mm=1.628\times {10}^{-3}m$

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.

The cross-sectional area:

  $A=\pi r^2$

Where, $\pi$ is constant and $r$ is a radius.

Formula of radius:

  $r=\frac{D}{2}$ Where, $D$ is a diameter.

Ohm’s law: $I=\frac{V}{R}$

Where,

  $I$ is the current.

  $V$ is the voltage.

  $R$ is the resistance.

Calculation:

By using the formula of resistivity, calculate the resistance:

  $R=\frac{\rho L}{\pi r^2}$

Substitute all the given values;

  $R=\frac{\left(1.68\times {10}^{-8}\Omega \cdot m\right)\left(26.0m\right)}{\left[\frac{\pi {\left(1.628\times {10}^{-3}m\right)}^2}{4}\right]}=0.210\Omega$

Now, by using ohm’s law; $V=IR$

Substitute all the given values,

  $V=12A\times 0.210=2.518V$

Conclusion: The voltage drop is $2.518V$ .