/*! 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

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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 \$0.02772.

Step by step solution

01

Calculate the power used by the lamp

The power consumed by an electrical device can be calculated using the formula P = IV, where P is the power in watts (W), I is the current in amperes (A), and V is the voltage in volts (V). For the lamp, the power P is calculated by multiplying the current (1.70 A) by the voltage (110 V).
02

Convert power to kilowatts

Electrical power is often measured in kilowatts (kW) when considering energy consumption. Since 1 kW = 1,000 W, divide the power obtained in watts by 1,000 to convert it to kilowatts.
03

Calculate the energy consumed per day

The energy consumed can be found by multiplying the power usage (in kW) by the time the lamp is operated, here assumed to be one day (24 hours). This gives the daily energy usage in kilowatt-hours (kWh).
04

Calculate the cost of energy per day

Finally, multiply the daily energy consumption (in kWh) by the cost of energy per kWh provided by the power company to find the cost of operating the lamp for one day.

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Key Concepts

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

Ohm's Law
Understanding Ohm's Law is crucial for calculating electrical power and energy consumption. It's a foundational principle in the field of electrical engineering that explains the relationship between voltage (V), current (I), and resistance (R). Ohm's Law is mathematically represented as
\( V = IR \).
It states that the voltage across a resistor is directly proportional to the current flowing through it, with the resistance being the proportionality factor. When you know any two of these variables, you can calculate the third. For example, if you have a lamp with a known resistance, and you measure the current flowing through it, you can use Ohm's Law to calculate the voltage across the lamp. Conversely, with a known voltage and resistance, the current can be deduced. It's important because it allows you to understand how the three variables interact in an electric circuit, which is critical when determining the electrical power consumption of devices.
Energy Consumption
When talking about electrical devices, energy consumption refers to the amount of electrical energy used by the device over a certain period. It's typically measured in kilowatt-hours(kWh), which represents the energy usage of an electric power of one kilowatt running for one hour.

Considering the lamp from the exercise, energy consumption is determined by the power it uses (in watts or kilowatts) and the duration it's operated (hours). To calculate daily energy usage, we multiply the power of the lamp in kilowatts by the number of hours it is kept on. If the lamp runs continuously for a full day, that would mean 24 hours. Reducing energy consumption can lead to cost savings and environmental benefits, as lower energy use can mean lesser fossil fuels burned in power plants, hence, resulting in reduced greenhouse gas emissions.
Kilowatt-Hours (kWh)
Kilowatt-hours, abbreviated as kWh, is the unit of energy commonly used to measure electricity consumption. One kilowatt-hour is equivalent to the energy used by a 1,000 watts (1 kilowatt) electric device, such as the lamp in our example, running for one hour. It's an important concept because utility companies use it to bill customers for their electricity usage.

Understanding kWh helps consumers to interpret their energy bills and make informed decisions about their energy use. For instance, knowing how many kWh a device consumes indicates how 'energy-hungry' it is. Users can opt for devices that consume less power to decrease their energy bills and their impact on the environment. Monitoring kWh usage can also encourage energy conservation and efficiency efforts in homes and businesses.
Current and Voltage Relationship
Current (I) and voltage (V) are two fundamental aspects of electricity that have a direct relationship in terms of power consumption. Using the power formula
\( P = IV \),
we see that the power (P) consumed by an electrical apparatus is the product of the voltage supplied to it and the current it draws. Voltage is the electrical force that pushes electric charges through a conductor, while current is the rate at which those charges pass through a point in the conductor.

The lamp from our exercise operates on a 110 V line and draws a current of 1.70 A, and by multiplying these two values, we determine its power usage in watts. This relationship indicates that for a given power requirement, an increase in voltage would result in a decrease in current, and vice versa. This is also why power lines operate at high voltages—to keep the current lower and reduce energy losses over long distances.

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Most popular questions from this chapter

A high-voltage transmission line with a diameter of 2.00 \(\mathrm{cm}\) and a length of 200 \(\mathrm{km}\) carries a steady current of 1000 \(\mathrm{A}\) . If the conductor is copper wire with a free charge density of \(8.49 \times 10^{28}\) electrons/m \(^{3}\) , how long does it take one electron to travel the full length of the line?

A Van de Graaff generator (see Figure 25.29\()\) is operating so that the potential difference between the high-voltage electrode \(B\) and the charging needles at \(A\) is 15.0 \(\mathrm{kV}\) . Calculate the power required to drive the belt against electrical forces at an instant when the effective current delivered to the high-voltage electrode is 500\(\mu A\) .

The current in a resistor decreases by 3.00 A when the voltage applied across the resistor decreases from 12.0 V to 6.00 V. Find the resistance of the resistor.

A small sphere that carries a charge \(q\) is whirled in a circle at the end of an insulating string. The angular frequency of rotation is \(\omega\) . What average current does this rotating charge represent?

Suppose that the current through a conductor decreases exponentially with time according to the equation \(I(t)=I_{0} e^{-t / \tau}\) where \(I_{0}\) is the initial current (at \(t=0 ),\) and \(\tau\) is a constant having dimensions of time. Consider a fixed observation point within the conductor. (a) How much charge passes this point between \(t=0\) and \(t=\tau\) . (b) How much charge passes this point between \(t=0\) and \(t=10 \tau ?\) (c) What If? How much charge passes this point between \(t=0\) and \(t=\infty\) ?

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