Chapter 18, Problem 38P

To determine

Power dissipation in the cord.

Answer to Problem 38P

7.8 W

Explanation of Solution

Given:

Current, $I=15.0A$

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

Length, $L=2.7m$

Radius, $r=\frac{0.129cm}{2}=\frac{0.129\times {10}^{-2}}{2}m$

Formula used:

By the definition of the power,

  $P=VI$

Where, $V$ is voltage and $I$ is current.

By the ohm’s law, a new expression for electrical power will be $P=\frac{V^2}{R}=I^{2}R$

Where, $V$ is voltage, $I$ is current and $R$ is resistance.

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.

  $\pi$ is constant

Calculation:

By using the formula of power and resistance;

  $P=I^{2}R=I^2\rho \frac{L}{A}$

Substitute the given values

  $P={\left(15.0A\right)}^2\left(1.68\times {10}^{-8}\Omega \cdot m\right)\frac{\left(2.7m\right)}{\pi {\left(\frac{0.129}{2}\times {10}^{-2}m\right)}^2}=7.8W$

Conclusion: The required power is 7.8 W.