Chapter 18, Problem 77GP

To determine

The diameter of the wire.

Answer to Problem 77GP

  $1.8\times {10}^{-4}m$

Explanation of Solution

Given:

Length, $L=3.8m$

Power, $P=95W$

Voltage, $V=120V$

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 $P=\frac{V^2}{R}$

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

It is known that resistance:

  $R=\frac{\rho L}{A}=\frac{\rho L}{\pi r^2}=\frac{\rho L}{\pi {\left(\frac{d}{2}\right)}^2}$

Where,

  $\rho$ is the resistivity.

  $L$ is the length of wire

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

  $\pi$ is constant

  $r$ is a radius.

  $d$ is a diameter.

Calculation:

By using the formula of power and resistance;

  $P=\frac{V^2}{R}\Rightarrow R=\frac{V^2}{P}$

  $R=\frac{\rho L}{A}=\frac{\rho L}{\pi {\left(\frac{1}{2}d\right)}^2}$

  $\Rightarrow \frac{V^2}{P}=\frac{\rho L}{\pi {\left(\frac{1}{2}d\right)}^2}$

  $\Rightarrow d=\sqrt{\frac{4\rho LP}{\pi V^2}}=\sqrt{\frac{4\left(100\times {10}^{-8}\right)\left(3.8m\right)\left(95W\right)}{\pi {\left(120V\right)}^2}}=1.8\times {10}^{-4}m$

Conclusion: The diameter is $1.8\times {10}^{-4}m$ .