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College Physics (Urone)
Found in: Page 736

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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 \(9.00{\rm{ }}cents/kW / h\)?

  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 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.


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.


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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