/*! 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 49 Compute the cost per day of oper... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Compute the cost per day of operating a lamp that draws a current of \(1.70 \mathrm{A}\) from a \(110-\mathrm{V}\) line. Assume the cost of energy from the power company is \(\$ 0.0600 / \mathrm{kWh}\).

Short Answer

Expert verified
The cost of operating the lamp per day is approximately $0.2693.

Step by step solution

01

Calculate Power

Calculate power using the power formula \(P = IV\) where \(I = 1.70A\) is the current and \(V = 110V\) is the voltage. \(P = (1.70A)*(110V) = 187W\).
02

Convert to Kilowatts

Convert the power from Watts to Kilowatts because the cost of energy is given in $/kWh. 1 Kilowatt = 1000 Watts, so the power in Kilowatts = \(187W / 1000 = 0.187 kW\).
03

Calculation of Daily Energy Consumption

First, calculate energy consumption for an hour by multiplying the power by 1 hr, since energy is power * time. That gives \(0.187kW * 1hr = 0.187 kWh\). Then, since there are 24 hours in a day, multiply this by 24 to get \(0.187 kWh * 24 = 4.488 kWh\). This is the total energy consumed in a day.
04

Calculate Cost

Finally, calculate the cost by multiplying the daily energy consumption by the cost of energy per kWh. That gives \(4.488 kWh * $0.0600/kWh = $0.2693\) per day.

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Key Concepts

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

Current and Voltage
Electricity is all about the movement of electric charges, and two key players in this process are current and voltage. In our exercise with the lamp, understanding these terms helps us compute how much power the lamp uses.Current, denoted as \(I\), is the flow of electric charge in a circuit measured in amperes (A). Think of it like the flow of water through a pipe. The more water that flows, the higher the current.
  • A higher current means more electrons are passing through the circuit per second.
  • In our exercise, the lamp uses a current of 1.70 A.
Voltage, denoted as \(V\), is the electric potential difference between two points in a circuit. It's like the pressure pushing the water through the pipe. The higher the voltage, the higher the potential for electric flow.
  • Voltage is measured in volts (V).
  • Our lamp is plugged into a 110 V line, which provides the necessary pressure to push the current through.
Understanding both current and voltage is essential for calculating electric power, which is the rate at which energy is used or produced. The basic formula is \(P = IV\), where \(P\) is power measured in watts (W). This tells us how much work (or energy) is done per second.
Energy Consumption
Once we know the power, understanding energy consumption helps us figure out how much energy is used over time. Typically, we measure energy consumption in kilowatt-hours (kWh), which reflects how much power (in kilowatts) is used over a specific period (in hours).In our scenario with the lamp, we first calculated the power usage as \(P = 187 W\) or \(0.187 kW\). To determine how much energy the lamp consumes daily, we multiply this power by the number of hours the lamp is used.
  • Power (in kW) \(\times\) Time (in hours) = Energy Consumption (in kWh).
  • So, for one hour: \(0.187 kW \times 1 hr = 0.187 kWh\).
  • For a full day: \(0.187 kWh \times 24 hr = 4.488 kWh\).
Energy consumption helps to determine the total energy used by an appliance and reflects how much work the appliance did over the given time. Knowing how energy consumption is calculated allows you to plan and manage your electricity usage more effectively.
Cost of Electricity
The cost of electricity is the final step in understanding how much you'll pay for using electrical devices, like our lamp, over time. The cost depends on both the energy consumption and the rate charged by the electricity provider, usually given in dollars per kilowatt-hour (\\(/kWh).To find out the daily cost:
  • First, calculate the total energy consumption in a day, which we found to be \(4.488 kWh\).
  • Then, multiply this by the cost rate: \(\\)0.0600 / kWh\).
  • The result: \(4.488 kWh \times \\(0.0600/kWh = \\)0.2693\) per day.
This simple calculation helps in budgeting and understanding how much each appliance contributes to the overall electricity bill. By knowing these costs, you can make informed decisions on the usage and efficiency of your devices. Adjusting any of the factors like reducing usage time, or using devices with better efficiency can lead to significant savings on energy costs.

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Most popular questions from this chapter

A Van de Graall generator produces a beam of \(2.00-\mathrm{MeV}\) deuterons, which are heavy hydrogen nuclei containing a proton and a neutron. (a) If the beam current is \(10.0 \mu \mathrm{A}\) how far apart are the deuterons? (b) Is the electric force of repulsion among them a significant factor in beam stability? Explain.

A certain lightbulb has a tungsten filament with a resistance of \(19.0 \Omega\) when cold and \(140 \Omega\) when hot. Assume that the resistivity of tungsten varies linearly with temperature even over the large temperature range involved here, and find the temperature of the hot filament. Assume the initial temperature is \(20.0^{\circ} \mathrm{C}\).

An electric current is given by the expression \(I(t)=\) \(100 \sin (120 \pi n, \text { where } I \text { is in amperes and } t\) is in seconds. What is the total charge carried by the current from \(t=0\) to \(t=(1 / 240) \mathrm{s} ?\)

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Batteries are rated in terms of ampere-hours \((A \cdot h) .\) For example, a battery that can produce a current of \(2.00 \mathrm{A}\) for \(3.00 \mathrm{h}\) is rated at \(6.00 \mathrm{A} \cdot \mathrm{h} .\) (a) What is the total energy, in kilowatt-hours, stored in a \(12.0-\mathrm{V}\) battery rated at \(55.0 \mathrm{A} \cdot \mathrm{h} ?\) (b) At \(\$ 0.0600\) per kilowatt-hour, what is the value of the electricity produced by this battery?

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