Chapter 18, Problem 76GP

(a)

To determine

Power drawn by air conditioner.

Answer to Problem 76GP

  $2600W$

Explanation of Solution

Given:

Current, $I=12A$

Voltage, $V=220V$

Formula used:

  $P=VI$

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

Calculation:

Substitute the given values in the formula: $P=VI$

  $P=\left(220V\right)\left(12A\right)=2640W\approx 2600W$

Conclusion: The required power is $2600W$ .

(b)

To determine

Power dissipated in the wiring.

Answer to Problem 76GP

  $17W$

Explanation of Solution

Given:

Current, $I=12A$

Voltage, $V=220V$

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

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

Length, $L=15m$

Formula used:

Power increases if the resistance will increase.

So, $P=I^{2}R$

Where, $P$ is power, $R$ is resistance and $I$ is current.

  $R=\frac{\rho L}{A}=\frac{\rho 4L}{\pi d^2}$

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:

By using the formula of power and resistance:

  $\begin{array}{c}{P}_R=I^{2}R\\ =I^2\frac{\rho L}{A}\\ =I^2\frac{\rho L}{\pi r^2}\\ =I^2\frac{4\rho L}{\pi d^2}\end{array}$

Substitute all the given values:

  $P_R={\left(12A\right)}^2\frac{4\left(1.68\times {10}^{-8}\Omega \cdot m\right)\left(15m\right)}{\pi {\left(1.628\times {10}^{-3}m\right)}^2}=17.433W\approx 17W$

Conclusion: Power dissipated in the wiring is $17W$ .

(c)

To determine

Power dissipated with a diameter of $2.053mm$ .

Answer to Problem 76GP

  $11W$

Explanation of Solution

Given:

Current, $I=12A$

Voltage, $V=220V$

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

Diameter, $d=2.053mm=2.053\times {10}^{-3}m$

Length, $L=15m$

Formula used:

Power increases if the resistance will increase.

So, $P=I^{2}R$

Where, $P$ is power, $R$ is resistance and $I$ is current.

  $R=\frac{\rho L}{A}=\frac{\rho 4L}{\pi d^2}$

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:

By using the formula of power and resistance:

  $\begin{array}{c}{P}_R=I^{2}R\\ =I^2\frac{\rho L}{A}\\ =I^2\frac{\rho L}{\pi r^2}\\ =I^2\frac{4\rho L}{\pi d^2}\end{array}$

Substitute all the given values:

  $P_R={\left(12A\right)}^2\frac{4\left(1.68\times {10}^{-8}\Omega \cdot m\right)\left(15m\right)}{\pi {\left(2.053\times {10}^{-3}m\right)}^2}=10.962W\approx 11W$

Conclusion: Power dissipated with a diameter of $2.053mm$ is $11W$ .

(d)

To determine

Money saved per month by using no. $12$ wire.

Answer to Problem 76GP

  $\$0.2795/month\approx 28centspermonth$

Explanation of Solution

Given:

Power dissipated in the wiring is $17.433W$ .

Power dissipated with a diameter of $2.053mm$ is $10.962W$ .

The cost of electricity is $12centsperkWh$ .

Formula used:

To calculate the cost of the energy, multiply the kilowatts of power consumed by the number of hours in operation times the cost per $kWh$ .

Calculation:

According to the question,

  $Saving=\left(17.433W-10.962W\right)\left(\frac{1kW}{1000W}\right)\left(30d\right)\left(\frac{12h}{1d}\right)\left(\frac{\$0.12}{1kWh}\right)$

  $=\$0.2795/month\approx 28centspermonth$

Conclusion: The saving is $\$0.2795/month\approx 28centspermonth$