• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q78PE

Expert-verified
Found in: Page 736

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# A small office-building air conditioner operates on $$408 - V$$ AC and consumes $$50.0{\rm{ }}kW$$ .(a) What is its effective resistance? (b) What is the cost of running the air conditioner during a hot summer month when it is on $$8.00{\rm{ }}h{\rm{ }}per{\rm{ }}day$$ for$$30$$days and electricity costs $$9.00{\rm{ }}cents/kW / h$$?

1. Value of the effective resistance is $$R = 3.33{\rm{ }}\Omega$$.
2. Cost of running the air conditioner during a hot summer month is $${\mathop{\rm Cos}\nolimits} t = \ 1080$$.
See the step by step solution

## Step 1: Concept Introduction

The power is defined by the formula

$$P = I\Delta V$$

Where$$I$$ is the current and $$\Delta V$$is potential difference.

The power, or rate at which energy is delivered to a circuit element, is $$P = I\Delta V$$ if a potential difference $$\Delta V$$ is maintained across the element.

We can represent the power applied to a resistor as \begin{align}{c}P = {I^2}R\\P = \frac{{{{(\Delta V)}^2}}}{R}\end{align} since the potential difference across a resistor is given by $$\Delta V = IR$$.

A resistor's inherent energy manifests as the energy transmitted to it through electrical transmission.

The power of a process is the amount of some type of energy converted into a different type divided by the time interval $$\Delta t$$ in which the process occurred:

$$P = \frac{{\Delta E}}{{\Delta t}}$$

The SI unit of power is the watt $$(W)$$.

$$1$$ Watt is $$1$$ joule/second$$(1\;W = 1\;J/s)$$.

## Step 2: Given data and required data.

The potential difference across the air conditioner is:

$$\Delta V = 408\;V$$

The power consumption of the air conditioner is:

\begin{align}{}P& = (50.0\;kW)\left( {\frac{{1000\;W}}{{1\;kW}}} \right)\\P &= 50.0 \times {10^3}\;W\end{align}

The cost of electricity is:

$$9.00{\rm{ }}cents/kW / h$$

## Step 3: Calculation of power consumption of the air conditioner.

(a)

The power consumption of the air conditioner is found from equation $$(1)$$:

$$P = \frac{{{{(\Delta V)}^2}}}{R}$$

Solving for $$R$$, we get:

$${\rm{R = }}\frac{{{{{\rm{(\Delta V)}}}^{\rm{2}}}}}{{\rm{P}}}$$

Entering the values for $$\Delta V$$ and $$P$$, it gives:

\begin{align}{}R &= \frac{{{{(408\;V)}^2}}}{{50.0 \times {{10}^3}\;W}}\\R &= 3.33{\rm{ }}\Omega \end{align}

Therefore, the resistance value is obtained as $$3.33{\rm{ }}\Omega$$.

## Step 4: Calculation for the cost.

(b)

The energy consumed by the air conditioner if it runs $$8.00\;h$$ per day for $$30$$days is found from equation $$(2)$$:

\begin{align}{}E &= P\Delta t\\E &= (50.0\;kW)\left( {\frac{{8.00\;h}}{{day}}} \right)(30.0{\rm{ }}days{\rm{ }})\\E &= 1.20 \times {10^4}\;kW \times h\end{align}

The cost of this energy is found by multiplying it by the cost of a single kilowatt-hour we are given:

\begin{align}{}{\mathop{\rm Cos}\nolimits} t &= \left( {1.20 \times {{10}^4}\;kW \times h} \right)(9.00{\rm{ }}cents{\rm{ }}/kW \times h)\\{\mathop{\rm Cos}\nolimits} t &= 1.08 \times {10^5}{\rm{ }}cents{\rm{ }}\\{\mathop{\rm Cos}\nolimits} t &= \left( {1.08 \times {{10}^5}{\rm{ }}cents{\rm{ }}} \right)\left( {\frac{{\ 1}}{{100{\rm{ }}cents{\rm{ }}}}} \right)\\{\mathop{\rm Cos}\nolimits} t &= \ 1080\end{align}

Therefore, the cost value is obtained as $$\ 1080$$.