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A standard flashlight battery can deliver about \(2.0 \mathrm{~W} \cdot \mathrm{h}\) of energy before it runs down. (a) If a battery costs US\$0.80, what is the cost of operating a \(100 \mathrm{~W}\) lamp for \(8.0 \mathrm{~h}\) using batteries? (b) What is the cost if energy is provided at the rate of US\$0.06 per kilowatt-hour?

Short Answer

Expert verified
(a) US$320; (b) US$0.048.

Step by step solution

01

Energy required for the lamp

Calculate the total energy required to operate a 100 W lamp for 8 hours. The energy required is given by the formula: \[ \text{Energy} = \text{Power} \times \text{Time} \]\[ \text{Energy} = 100 \text{ W} \times 8 \text{ h} = 800 \text{ Watt-hours} \]
02

Determine the number of batteries needed

Each battery can deliver 2.0 Wh of energy. To find the number of batteries required:\[ \text{Number of batteries} = \frac{800 \text{ Wh}}{2.0 \text{ Wh/battery}} = 400 \text{ batteries} \]
03

Calculate the cost using batteries

Each battery costs US\(0.80. Thus, the total cost for 400 batteries is:\[ \text{Cost} = 400 \text{ batteries} \times 0.80 \text{ US\\)} = 320 \text{ US\$} \]
04

Calculate the cost using electricity rate

Energy cost is provided at the rate of US\(0.06 per kilowatt-hour (kWh). First, convert the energy from Wh to kWh:\[ 800 \text{ Wh} = 0.8 \text{ kWh} \]Now calculate the cost:\[ \text{Cost} = 0.8 \text{ kWh} \times 0.06 \frac{\text{US\\)}}{\text{kWh}} = 0.048 \text{ US\$} \]

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Key Concepts

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

Battery Energy
To understand battery energy, let's think about how much energy a battery can give before it runs out. This energy is often measured in Watt-hours (Wh), which combines power (Watt) and time (hour). A standard flashlight battery typically delivers about 2.0 Wh of energy. This means the battery can power a device using 2 Watts for 1 hour, or equivalently, 1 Watt for 2 hours. This is quite useful for small devices.
In the example exercise, knowing that a battery can deliver 2.0 Wh is key to determining how many batteries would be needed to power a larger device, like a 100 W lamp, for longer periods.
Power Consumption
Power consumption refers to the amount of energy used by a device over time, and it's measured in Watts (W). When you know the power consumption and the time a device will be used, you can calculate the total energy required using the formula:
  • Energy = Power × Time
In the flashlight example, a 100 W lamp running for 8 hours means that it uses a total of 800 Watt-hours (Wh) of energy. To find how much energy a device consumes is crucial when planning how to power it, especially if you're deciding between using batteries or plugging into an electrical outlet.
This calculation helps you ensure you have enough power supplies and can compare costs, as done in the problem.
Electrical Efficiency
Electrical efficiency is about using energy wisely and minimizing waste. It's significant when comparing how power is supplied to a device, like batteries versus direct electrical supply from the grid. One way to think about efficiency is considering how much it costs to provide the same amount of energy using different methods.
For instance, if you use batteries to power a device, you might end up paying more per unit of energy than if you used electricity from the grid. In the lamp example, using batteries to supply 800 Wh costs US$320, while the same energy from the grid only costs US$0.048. This vast difference illustrates how much more efficient it can be to use grid electricity for high power devices and long running times.
Understanding electrical efficiency helps make more informed choices about energy sources, contributing to both cost savings and environmental conservation.

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Most popular questions from this chapter

A simple ohmmeter is made by connecting a \(1.50 \mathrm{~V}\) flashlight battery in series with a resistance \(R\) and an ammeter that reads from 0 to \(1.00 \mathrm{~mA},\) as shown in Fig. \(27-59 .\) Resistance \(R\) is adjusted so that when the clip leads are shorted together, the meter deflects to its full- scale value of \(1.00 \mathrm{~mA}\). What external resistance across the leads results in a deflection of (a) \(10.0 \%,(\mathrm{~b}) 50.0 \%\) and (c) \(90.0 \%\) of full scale? (d) If the ammeter has a resistance of \(20.0 \Omega\) and the internal resistance of the battery is negligible, what is the value of \(R ?\)

A \(1.0 \mu \mathrm{F}\) capacitor with an initial stored energy of \(0.50 \mathrm{~J}\) is discharged through a \(1.0 \mathrm{M} \Omega\) resistor. (a) What is the initial charge on the capacitor? (b) What is the current through the resistor when the discharge starts? Find an expression that gives, as a function of time \(t,\) (c) the potential difference \(V_{C}\) across the capacitor, (d) the potential difference \(V_{R}\) across the resistor, and (e) the rate at which thermal energy is produced in the resistor.

A \(120 \mathrm{~V}\) power line is protected by a 15 A fuse. What is the maximum number of 500 W lamps that can be simultaneously operated in parallel on this line without "blowing" the fuse because of an excess of current?

Four \(18.0 \Omega\) resistors are connected in parallel across a \(25.0 \mathrm{~V}\) ideal battery. What is the current through the battery?

A \(15.0 \mathrm{k} \Omega\) resistor and a capacitor are connected in series, and then a \(12.0 \mathrm{~V}\) potential difference is suddenly applied across them. The potential difference across the capacitor rises to \(5.00 \mathrm{~V}\) in \(1.30 \mu \mathrm{s}\). (a) Calculate the time constant of the circuit. (b) Find the capacitance of the capacitor.

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