Chapter 18, Problem 59GP

To determine

The diameter of the heating element of a $110V,1500W$ heater is $5.4m$ long.

Answer to Problem 59GP

  $2.9\times {10}^{-4}m$

Explanation of Solution

Given info:

Length, $L=5.4m$

Power, $P=1500W$

Voltage, $V=110V$

Formula used:

By the definition of the power, $P=VI$

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

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

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

The definition of resistance and plug in the formula for the area of a circle: $R=\frac{\rho L}{A}=\frac{\rho 4L}{\pi d^2}$

Where,

  $\rho$ is the resistivity.

  $L$ is the length of wire

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

  $\pi$ is constant and $r$ is a radius

  $d$ is a diameter.

Calculation:

Rearrange the expression, $P=\frac{V^2}{R}$ . So, $R=\frac{V^2}{P}\mathrm{...}(1)$

By using the formula of resistance, a new expression,

  $R=\frac{\rho L}{A}=\frac{\rho L}{\pi r^2}=\frac{\rho 4L}{\pi d^2}\mathrm{...}(2)$

By the equation $\left(1\right)$ and $\left(2\right)$

  $4\rho \frac{L}{\pi d^2}=\frac{V^2}{P}$

  $d=\sqrt{\frac{4\rho LP}{\pi V^2}}$

Substitute all the given values in the expression, $d=\sqrt{\frac{4\rho LP}{\pi V^2}}$

  $d=\sqrt{\frac{4\left(9.71\times {10}^{-8}\Omega \cdot m\right)\left(5.4m\right)\left(1500W\right)}{\pi {\left(110V\right)}^2}}=2.9\times {10}^{-4}m$

Conclusion: The diameter of the heating element of a $110V,1500W$ heater is $5.4m$ long: $2.9\times {10}^{-4}m$ .