/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 (a) How much power does a televi... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) How much power does a television use if it draws \(2.00 \mathrm{~A}\) on a 120 - \(\mathrm{V}\) line? (b) What energy in \(\mathrm{kWh}\) does the television use in 30 days if it is used an average of \(7.00 \mathrm{~h} /\) day? (c) Find the cost of operating the television for 30 days if the cost of energy is \(\$ 0.11 / \mathrm{kWh}\).

Short Answer

Expert verified
The television uses 240 W, consumes 50.4 kWh in 30 days, costing \$5.54.

Step by step solution

01

Determine the Power Usage of the Television

To find the power usage, use the formula for electrical power: \( P = IV \), where \( I \) is the current and \( V \) is the voltage. Here, \( I = 2.00 \; \text{A} \) and \( V = 120 \; \text{V} \). Plug these values into the formula:\[ P = 2.00 \times 120 = 240 \; \text{W} \] Thus, the television uses 240 watts of power.
02

Calculate the Energy Consumption Over 30 Days

The energy used by the television is calculated using the formula:\[ E = P \times t \]where \( P = 240 \; \text{W} \), and \( t \) is the total time in hours. The television runs for an average of \( 7.00 \; \text{hours/day} \) for 30 days:\[ t = 7.00 \times 30 = 210 \; \text{hours} \]Convert the energy into kilowatt-hours (kWh) since power companies use this unit:\[ E = 240 \; \text{W} \times 210 \; \text{hours} \times \frac{1 \; \text{kW}}{1000 \; \text{W}} = 50.4 \; \text{kWh} \] The television uses 50.4 kWh in 30 days.
03

Determine the Cost of Operating the Television

To find the cost, use the formula:\[ \text{Cost} = E \times \text{Rate} \]where \( E = 50.4 \; \text{kWh} \) and the rate is \( \\(0.11/\text{kWh} \):\[ \text{Cost} = 50.4 \times 0.11 = 5.544 \; \text{USD} \] Thus, the cost of operating the television for 30 days is approximately \( \\)5.54 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Energy Consumption
Energy consumption refers to the amount of energy used by an appliance, device, or system. When it comes to understanding energy consumption, consider the power (in watts) and the duration (in hours) for which an appliance is operated. The formula used to calculate energy consumption is \( E = P \times t \), where \( E \) represents energy in watt-hours, \( P \) denotes power in watts, and \( t \) is time in hours.
This calculation helps us understand how much energy is used over a certain period, like a day or month. For instance, if a television uses 240 watts and operates for 210 hours in a month, it consumes 50,400 watt-hours. To make this figure more manageable, we often convert it to kilowatt-hours by dividing by 1,000, which results in 50.4 kWh.
Kilowatt-Hours
Kilowatt-hours (kWh) is a unit of energy commonly used by electric utility companies to measure energy consumption. One kilowatt-hour equals 1,000 watt-hours and represents the amount of energy consumed when a 1,000-watt device runs for one hour.
By using kWh, we simplify large watt-hour values, making it easier to handle, especially for billing purposes. In the context of our television example, its monthly energy use of 50,400 watt-hours is expressed as 50.4 kWh. This conversion assists both consumers and utility providers in tracking and managing electricity usage, giving a clearer understanding of how much energy is consumed and billed.
Electricity Cost
Electricity cost is the expense associated with consuming electrical energy, usually expressed in terms of cost per kilowatt-hour (kWh). To calculate the total cost of electricity used, we multiply the total energy consumption by the cost rate. The formula for electricity cost is \[ \text{Cost} = E \times \text{Rate} \]where \( E \) is energy in kWh and the "Rate" is the price per kWh.
For instance, with a rate of \\(0.11 per kWh, operating a television that uses 50.4 kWh over a month will cost approximately \\)5.54. Understanding electricity costs allows consumers to manage their energy use better and plan their expenses effectively.
Current and Voltage
Current and voltage are fundamental concepts in understanding electricity. Current, measured in amperes (A), refers to the flow of electric charge in a circuit, while voltage, measured in volts (V), is the electric potential difference between two points.
To calculate electrical power, expressed in watts (W), we use the formula \( P = I \times V \), where \( I \) is the current and \( V \) is the voltage. In our example with the television, a current of 2.00 amperes and a voltage of 120 volts results in a power usage of 240 watts.
Understanding these concepts helps in analyzing how various appliances operate and aids in energy management and electrical safety.

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Most popular questions from this chapter

How many amperes will a \(75 \overline{0}\) -W lamp draw on a \(11 \overline{0}-\mathrm{V}\) circuit?

A coffeepot operates on \(12.0 \mathrm{~V}\). If it draws \(2.50 \mathrm{~A}\), find its resistance.

What force is exerted on a test charge of \(4.00 \times 10^{-5} \mathrm{C}\) if it is placed in an electric field of magnitude \(3.00 \times 10^{6} \mathrm{~N} / \mathrm{C} ?\)

Using the formulas for series and parallel circuits, fill in the blanks in the tables shown opposite each circuit. In the blanks across from Battery under \(V\) : Write the emf of the battery. \(I\) : Write the total current in the circuit. \(R\) : Write the equivalent or total resistance of the entire circuit. In the blanks across from \(R_{1}\) under \(V\) : Write the voltage drop across \(R_{1}\). \(I:\) Write the current flowing through \(R_{1}\). \(R\) : Write the resistance of \(R_{1}\). In the blanks across from \(R_{2}, R_{3}, \ldots\), fill in the appropriate numbers under \(V, I\), and R. (Begin by looking for key information given in the table and work from there.)

Three charges are located along the \(x\) -axis. Charge \(A(+3.00 \mu \mathrm{C})\) is located at the origin. Charge \(B(+5.50 \mu \mathrm{C})\) is located at \(x=+0.400 \mathrm{~m}\). Charge \(C(-4.60 \mu \mathrm{C})\) is located at \(x=+0.750 \mathrm{~m} .\) (a) Find the total force (and direction) on charge \(B\). (b) Find the total force (and direction) on charge \(A\). (c) Find the total force (and direction) on charge \(C\).

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