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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# An air conditioner connected to a 120 V RMS ac line is equivalent to a ${\mathbf{12}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{\Omega }}$ resistance and a ${\mathbf{1}}{\mathbf{.}}{\mathbf{30}}{\mathbf{}}{\mathbf{\Omega }}$ inductive reactance in series. (a) Calculate the impedance of the air conditioner. (b) Calculate the average rate at which energy is supplied to the appliance.

1. The impedance of the air conditioner is $12.1\mathrm{\Omega }$.
2. The average rate at which energy is supplied to the appliance is $1.19×{10}^{3}\mathrm{W}$.
See the step by step solution

## Step 1: Listing the given quantities:

The resistor, $\mathrm{R}=12\mathrm{\Omega }$

The inductive reactance, ${\mathrm{X}}_{\mathrm{L}}=1.3\mathrm{\Omega }$

The RMS value, $\mathrm{\epsilon }=120\mathrm{V}$

## Step 2: Understanding the concepts of impedance and power:

The impedance of the air conditioner can be found from resistance, and inductive and capacitive reactance using the corresponding relation. Then using the formula for power, we can find the average rate at which energy is supplied to the appliance.

Formula:

$\mathrm{Z}=\sqrt{{\mathrm{R}}^{2}+{\left({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}\right)}^{2}}$ ….. (1)

Here, XC is the capacitive reactance.

## Step 3: (a) Calculations of the impedance of the air conditioner:

To find impedance $\left(\mathrm{Z}\right)$ use equation (1).

$\mathrm{Z}=\sqrt{{\mathrm{R}}^{2}+{\left({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}\right)}^{2}}$

Substitute known numerical values in the above equation.

$\begin{array}{rcl}\mathrm{Z}& =& \sqrt{{12}^{2}+{\left(1.3-0\right)}^{2}}\\ & =& \sqrt{144+1.69}\\ & =& \sqrt{145.07}\\ & =& 12.1\mathrm{\Omega }\end{array}$

Hence, the impedance of the air conditioner is $12.1\mathrm{\Omega }$.

## Step 4: (b) Calculations of the average rate at which energy supplied to the appliance:

Define the average rate of energy $\left({\mathrm{P}}_{\mathrm{avg}}\right)$ as follow.

$\begin{array}{rcl}{\mathrm{P}}_{\mathrm{avg}}& =& \frac{{\mathrm{\epsilon }}_{\mathrm{RMS}}^{2}\mathrm{R}}{{\mathrm{Z}}^{2}}\\ & =& \frac{{\left(120\right)}^{2}×12}{{\left(12.07\right)}^{2}}\\ & =& 1.19×{10}^{3}\mathrm{W}\end{array}$

The average rate at which energy supplied to the appliance is $1.19×{10}^{3}\mathrm{W}$.