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Chapter 30: Problem 18

It has been proposed to use large inductors as energy storage devices. (a) Ilow much electrical energy is converied to light and thermal energy by a \(150 \mathrm{~W}\) light bulb 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
The energy converted by a 150 W light bulb in one day is \( \approx 1.296 脳 10^{7} \) Joules. The inductance needed to store this amount of energy with a current of 80.0 A is \( \approx 406.25\) Henry.

Step by step solution

01

Calculate the energy converted by a light bulb

The electrical energy converted to light and heat by a light bulb can be calculated using the power formula \(P = \frac{E}{t}\), where \(P\) is power, \(E\) is energy and \(t\) is time. Given that the power is 150 W and the time is one day (which we convert to seconds for consistency of units), we can isolate \(E\) to get the formula \(E = Pt\). In this step, it's important to ensure that the units of power and time are consistent. Here, both should be in their standard units: Watts for power and seconds for time.
02

Find the energy

Using the formula from step 1, replace \(P\) with 150 W and \(t\) with 86,400 seconds (24 hours converted to seconds). Calculate the result to find the energy \(E\).
03

Find the inductance

Once the energy \(E\) has been found, this can be placed into the formula for energy stored in an inductor \(E = \frac{1}{2}LI^{2}\), where \(L\) is inductance and \(I\) is current. The current is given as 80.0 A. Re-arrange the formula to isolate \(L\) to get \(L = \frac{2E}{I^{2}}\) and substitute the values of \(E\) and \(I\) into the formula to calculate \(L\) which is the inductance.

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Key Concepts

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

Energy Conversion
Energy conversion is the process of changing one form of energy into another. In everyday life, we often convert electrical energy into more usable forms such as light or heat.
In this exercise, we are focusing on an ordinary light bulb which converts 150 watts of electrical energy into both light and thermal energy throughout a day. To clearly understand the conversion, we use the power formula, which makes it easy to calculate how much energy is used.
* Electrical energy is initially in the form of current and voltage.
* When Circuit is operational, energy is supplied to the light bulb.
* The bulb's filament heats up and emits light.
Energy consumption of electrical devices like light bulbs is often measured in watt-hours. Here, maintaining consistent units is key: power in watts and time in seconds. For example, 150 watts over 24 hours equates to a tremendous quantity of energy that can illuminate and provide warmth for quite some time.
Inductor Energy Storage
Inductors play a crucial role as energy storage devices in electrical circuits. They are components capable of storing energy in a magnetic field when electric current passes through them. For someone trying to save energy from a light bulb scenario, understanding how inductors work is essential.
The energy stored in an inductor comes from the current flowing through it, which creates a magnetic field around the coil.
In mathematical terms, the energy in an inductor is calculated using the formula: \[ E = \frac{1}{2} L I^2 \] Where:
  • \(E\) is the energy,
  • \(L\) is the inductance, and
  • \(I\) is the current passing through the inductor.
This formula shows how even a small amount of inductance combined with a substantial current can lead to a significant amount of energy storage.
Inductors are often used in applications where temporary energy storage and controlled release are necessary, indicating their pivotal role beyond simple energy storage.
Power Formula
The power formula is a fundamental concept in physics and electronics, used to easily calculate the energy used by electrical devices. In simple terms, power is the rate at which energy is used or converted in a system. If you've ever tried to figure out how much energy a device uses, you've likely used the power formula.
To find energy consumption, we use the relationship: \[ P = \frac{E}{t} \] Which is rearranged to solve for energy: \[ E = P \times t \] Here:
  • \(P\) represents power measured in watts,
  • \(E\) denotes energy in joules, and
  • \(t\) is the time in seconds over which power is utilized.
This formula emphasizes the importance of unit consistency鈥攚atts for power, seconds for time鈥攖o ensure correct energy calculations. Understanding and applying the power formula is crucial for analyzing how electrical devices operate and determining their energy efficiency.

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Most popular questions from this chapter

If part of the magnet develops resistance and liquid helium boils away, rendering more and more of the magnct nonsuperconducting. how will this quench affect the time for the current to drop to half of its initial value? (a) The time will be shorter bccause the resistance will increase; (b) the time will be longer because the resistance will increase: (c) the time will be the same; (d) not enough information is given.

\(A 15.0 \mu F\) capacitor is charged by a 150.0 V power supply, then disconnected from the power and connected in scries with a \(0.280 \mathrm{mH}\) inductor. Calculate: (a) the oscillation frequency of the circuit: (b) the energy stored in the capacitor at time \(t=0 \mathrm{~ms}\) (the moment of connection with the inductor); (c) the energy stored in the inductor at \(t=1.30 \mathrm{~ms}\)

A solenoidal coil with 25 turns of wire is wound tightly around another coil with 300 turns (see Example 30.1 ). The inner solenoid is \(25.0 \mathrm{~cm}\) long and has a diameter of \(2.00 \mathrm{~cm}\). At a certain time, the current in the inner solenoid is \(0.120 \mathrm{~A}\) and is increasing at a rate of \(1.75 \times 10^{3} \mathrm{~A} / \mathrm{s}\). For this time, calculate: (a) the average magnetic flux through each turn of the inner solenoid; (b) the mutual inductance of the two solenoids; (c) the emf induced in the outer solenoid by the changing current in the inner solenoid.

. In Fig. \(30.11 .\) suppose that \(\varepsilon=60.0 \mathrm{~V}, R=240 \Omega .\) and \(L=0.160 \mathrm{H}\). With switch \(S_{2}\) open, switch \(S_{1}\) is left closed until a constant currcnt is cstablishcd. Then \(S_{2}\) is closcd and \(S_{1}\) opcned, taking the battery out of the circuit. (a) What is the initial current in the resistor, just after \(S_{2}\) is closed and \(S_{1}\) is opened? (b) What is the current in the resistor at \(t=4.00 \times 10^{-4} \mathrm{~s} ?\) (c) What is the potcntial differcnoe betwecn points \(b\) and \(c\) at \(t=4.00 \times 10^{-4} \mathrm{~s}\) ? Which point is at a higher potcntial? (d) How long does it take the current to decrease to half its initial value?

An air-filled toroidal solenoid has a mean radius of \(15.0 \mathrm{~cm}\) and a cross-sectional area of \(5.00 \mathrm{~cm}^{2}\). When the current is \(12.0 \mathrm{~A}\), the encrgy stored is \(0.390 \mathrm{~J}\). How many turns does the winding have?
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