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Large inductors have been proposed as energy-storage devices. (a) How much electrical energy is converted to light and thermal energy by a 200 \(\mathrm{W}\) lightbulb in one day? (b) If the amount of energy calculated in part (a) is stored in an inductor in which the current is \(80.0 \mathrm{A},\) what is the inductance?

Short Answer

Expert verified
17,280,000 J; 5,400 H

Step by step solution

01

Understanding the Problem

To solve the problem, we need to find the amount of energy used by the lightbulb in one day and then calculate the inductance required to store the same amount of energy with a given current.
02

Calculate Energy Consumed by Lightbulb

The power of the lightbulb is given as 200 W. Power is the rate of energy consumption, which means energy can be calculated by multiplying power by time. One day is equal to 24 hours or 86,400 seconds. Thus, the energy consumed is \( E = P \times t = 200 \text{ W} \times 86,400 \text{ s} = 17,280,000 \text{ J} \).
03

Use Energy Formula for Inductor

The energy stored in an inductor is given by the formula \( E = \frac{1}{2} L I^2 \), where \( L \) is the inductance and \( I \) is the current. We know the energy \( E = 17,280,000 \text{ J} \) and current \( I = 80.0 \text{ A} \). We need to solve for \( L \).
04

Solve for Inductance

Rearrange the energy formula: \( L = \frac{2E}{I^2} \). Substitute the known values: \( L = \frac{2 \times 17,280,000 \text{ J}}{(80.0 \text{ A})^2} = \frac{34,560,000}{6,400} = 5,400 \text{ H} \).
05

Conclusion

The energy consumed by the lightbulb in one day is 17,280,000 J, and the required inductance to store this energy with a current of 80.0 A is 5,400 H.

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Key Concepts

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

Electrical Energy Conversion
Electrical energy conversion is a fundamental concept in physics and engineering, focusing on how electrical energy transforms into other forms of energy. In this exercise, a 200-watt lightbulb serves as an example of converting electrical energy into both light and heat.
  • Electrical devices like lightbulbs use electricity to produce light and sometimes heat.
  • The rate of energy usage is quantified in watts, which measures how much energy a device consumes per second.
  • For a 200-watt bulb, this means it converts 200 joules of electrical energy per second to other forms.
To find out how much energy is consumed over a day, we multiply the power (200 W) by the total seconds in 24 hours (86,400 seconds). Hence, the energy conversion calculates to 17,280,000 joules. This figure represents the total energy turned from electrical energy to light and heat by the bulb in a full day.
Inductance Calculation
Inductance is a property of an electrical component that resists changes in current, and it plays a crucial role in energy storage within inductors. An inductor can store and release energy in a magnetic field, which makes it a potential candidate for energy storage devices.
  • Inductance, denoted by the letter \( L \), is measured in henrys (H).
  • The energy stored in an inductor is described by the formula: \( E = \frac{1}{2} L I^2 \).
  • This equation shows the relationship between energy \( E \), inductance \( L \), and current \( I \).
For the given problem, where the energy to be stored equals 17,280,000 joules with a current of 80.0 amperes, we need to find the inductance \( L \). By rearranging the equation, we get \( L = \frac{2E}{I^2} \). Substituting the known values results in: \( L = \frac{2 \times 17,280,000}{80^2} = 5,400 \text{ H} \). Hence, the inductor's inductance required to store this energy at the given current is 5,400 henrys.
Power Consumption
Power consumption is a critical idea in both understanding and managing energy efficiency in electrical devices. It represents the amount of energy used by an appliance over time and can be calculated if both the power and the duration of use are known.
  • Power is measured in watts (W), indicating how much energy a device uses per second.
  • Energy consumed over time can be determined by multiplying the power by the number of seconds used.
In the context of the lightbulb, it draws 200 watts of power continuously. Over the span of an entire day, this results in a substantial energy consumption, calculated in step-by-step detail to be 17,280,000 joules. Understanding these principles aids in assessing the efficiency and cost-effectiveness of using such devices over extended periods.

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Most popular questions from this chapter

Energy in a typical inductor. (a) How much energy is stored in a 10.2 \(\mathrm{mH}\) inductor carrying a 1.15 A current? (b) How much current would such an inductor have to carry to store 1.0 \(\mathrm{J}\) of energy? Is this a reasonable amount of current for ordinary laboratory circuit elements?

A very thin 15.0 \(\mathrm{cm}\) copper bar is aligned horizontally along the east-west direction. If it moves horizontally from south to north at 11.5 \(\mathrm{m} / \mathrm{s}\) in a vertically upward magnetic field of \(1.22 \mathrm{T},\) (a) what potential difference is induced across its ends, and (b) which end (east or west) is at a higher potential? (c) What would be the potential difference if the bar moved from east to west instead?

A toroidal solenoid has a mean radius of 10.0 \(\mathrm{cm}\) and a cross- sectional area of 4.00 \(\mathrm{cm}^{2}\) and is wound uniformly with 100 turns. A second coil with 500 turns is wound uniformly on top of the first. What is the mutual inductance of these coils?

A circular area with a radius of 6.50 \(\mathrm{cm}\) lies in the \(x\) -y plane. What is the magnitude of the magnetic flux through this circle due to a uniform magnetic field \(B=0.230 \mathrm{T}\) that points (a) in the \(+z\) direction? (b) at an angle of \(53.1^{\circ}\) from the \(+z\) direction? (c) in the \(+y\) direction?

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