Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

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

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

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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 $\$ 1080$.

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Short Answer

Step by Step Solution

Step 1: Concept Introduction

Step 2: Given data and required data.

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

Step 4: Calculation for the cost.

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Concept Introduction

Step 2: Given data and required data.

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

Step 4: Calculation for the cost.

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