Chapter 18, Problem 80GP

(a)

To determine

The power wasted in copper wire.

Answer to Problem 80GP

  $28.0W$

Explanation of Solution

Given:

Power $P=2250W$

Voltage $V=120V$

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

Length, $L=25.0m$

Diameter, $d=0.259cm=2.59\times {10}^{-3}m$

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

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

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 dissipated and a resistance;

  $P_R=I^{2}R=\frac{P^2}{V^2}$

  $R=\frac{P^2}{V^2}\frac{\rho L}{A}=\frac{P^2}{V^2}\frac{\rho L}{\pi r^2}=\frac{P^2}{V^2}\frac{\rho L}{\pi {\left(\frac{d}{2}\right)}^2}=\frac{P^2}{V^2}\frac{4\rho L}{\pi d^2}$

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

Substitute the given values:

  $R=\frac{{\left(2250W\right)}^{2}4\left(1.68\times {10}^{-8}\right)\left(25.0m\right)}{{\left(120V\right)}^2\pi {\left(2.59\times {10}^{-3}m\right)}^2}=28.0W$

Conclusion:

The power wasted in $25.0m$ of copper wire is $28.0W$ .

(b)

To determine

The power wasted in copper wire.

Answer to Problem 80GP

The answer is $11.1W$

Explanation of Solution

Given:

Power $P=2250W$

Voltage $V=120V$

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

Length, $L=25.0m$

Diameter, $d=0.412cm=4.12\times {10}^{-3}m$

Formula used:

Power dissipated in a resistance: $P=I^{2}R$

Where, $P$ is power, $I$ 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 dissipated and a resistance;

  $P_R=I^{2}R=\frac{P^2}{V^2}$

  $R=\frac{P^2}{V^2}\frac{\rho L}{A}=\frac{P^2}{V^2}\frac{\rho L}{\pi r^2}=\frac{P^2}{V^2}\frac{\rho L}{\pi {\left(\frac{d}{2}\right)}^2}=\frac{P^2}{V^2}\frac{4\rho L}{\pi d^2}$

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

Substitute the given values:

  $R=\frac{{\left(2250W\right)}^{2}4\left(1.68\times {10}^{-8}\right)\left(25.0m\right)}{{\left(120V\right)}^2\pi {\left(4.12\times {10}^{-3}m\right)}^2}=11.1W$

Conclusion: The required power is $11.1W$ .