/*! 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 43 Energy in a typical inductor. (a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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?

Short Answer

Expert verified
(a) 0.00675 J; (b) 14.0 A, which is a high and less practical current for typical lab circuits.

Step by step solution

01

Recall the Formula for Inductance

The energy stored in an inductor can be calculated using the formula \( E = \frac{1}{2} L I^2 \) where \( E \) is the energy in joules, \( L \) is the inductance in henrys (\(H\)), and \( I \) is the current through the inductor in amperes (\(A\)).
02

Calculate Energy for Given Current

Given \( L = 10.2 \) millihenrys (or \( 0.0102 \) henrys) and \( I = 1.15 \) amperes, substitute these values into the energy formula: \[ E = \frac{1}{2} \times 0.0102 \times (1.15)^2 \]Calculate it to find the energy stored:\[ E = \frac{1}{2} \times 0.0102 \times 1.3225 \approx 0.00675 \text{ joules} \]
03

Find Current for Given Energy

To find the current that would store 1 joule of energy in the same inductor, rearrange the formula to solve for \( I \):\[ I = \sqrt{\frac{2E}{L}} \]Substitute \( E = 1.0 \) joule and \( L = 0.0102 \) henrys:\[ I = \sqrt{\frac{2 \times 1.0}{0.0102}} \approx \sqrt{196.08} \approx 14.0 \text{ amperes} \]
04

Evaluate Reasonability of Current

Consider the 14.0 amperes of current and determine if this is a common value for typical laboratory circuits. Such high currents (14.0A) are considered relatively large and could be impractical for small-scale laboratory equipment due to increased risk of overheating and the need for larger components.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Inductance Formula
The inductance formula is key to understanding how energy is stored in an inductor. It is expressed as \( E = \frac{1}{2} L I^2 \), where \( E \) is the energy in joules, \( L \) is the inductance in henrys (H), and \( I \) is the current in amperes (A). This formula reveals that the energy storage capability of an inductor depends on both its inductance and the current passing through it.

When you double the current, the energy stored increases by four times, demonstrating a non-linear relationship. This highlights the importance of both inductance and current in evaluating the energy storage capacity of inductors.

In practical terms, it means that small variations in current can lead to significant changes in the energy stored. Understanding this formula is crucial for designing and analyzing circuits that involve inductors.
Current Calculation
In the exercise, calculating the current required to store a specific amount of energy is essential. To do this, we rearrange the inductance formula to solve for current, derived as \( I = \sqrt{\frac{2E}{L}} \).

This approach allows us to predict the current needed for achieving a certain energy level in the inductor. For instance, if we want to store 1 joule of energy in a 10.2 mH inductor, we use the formula:

  • Given the energy \( E = 1.0 \) joule
  • Inductance \( L = 0.0102 \) henrys
  • The calculation becomes \( I = \sqrt{\frac{2 \times 1.0}{0.0102}} \approx 14.0 \) amperes
This shows that substantial current is required for storing even a modest amount of energy, which is crucial for engineers to consider in design to ensure circuit safety and practicality.
Energy Storage in Inductors
Energy storage in inductors is a critical mechanism in electromagnetics. Inductors store energy in the form of a magnetic field when current flows through them. This stored energy can then be released when the current decreases, making inductors useful in applications like transformers, filters, and energy storage systems.

Furthermore, the energy storage capacity of an inductor is directly affected by its inductance value and the amount of current flowing through it.

When you have a high inductance value and large current, the inductor can store significant energy. However, this comes with challenges such as heat dissipation and component sizing, especially in smaller circuits where managing space and thermal conditions is crucial.

Engineers must balance these factors to optimize energy storage while maintaining circuit efficiency and reliability.
Laboratory Circuit Safety
When dealing with inductors in laboratory settings, safety must be a priority. High currents can pose risks like overheating, which can damage components or cause injury. It's important to use appropriate protective measures such as:
  • Using properly rated components to handle the current
  • Implementing overload protection mechanisms like fuses or circuit breakers
  • Ensuring adequate ventilation and spacing to avoid heat build-up
Design considerations should also include the practicality of the current levels.

As the exercise demonstrates, currents as high as 14.0 amperes are uncommon for typical lab circuits, indicating the need for careful assessment of the circuit design. Safe handling practices and appropriate component choice ensure that experiments and circuits are conducted safely and effectively.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Measuring blood flow. Blood contains positive and negative ions and therefore is a conductor. A blood vessel, therefore, can be viewed as an electrical wire. We can even picture the flowing blood as a series of parallel conducting slabs whose thickness is the diameter \(d\) of the vessel moving with speed \(v\) .(See Figure \(21.62 . )\) (a) If the blood vessel is placed in a magnetic field \(B\) perpendicular to the vessel, as in the figure, show that the motional potential difference induced across it is \(\mathcal{E}=v B d .\) (b) If you expect that the blood will be flowing at 15 \(\mathrm{cm} / \mathrm{s}\) for a vessel 5.0 \(\mathrm{mm}\) in diameter, what strength of magnetic field will you need to produce a potential difference of 1.0 \(\mathrm{mV} ?\) (c) Show that the volume rate of flow \((R)\) of theblood is equal to \(R=\pi \mathcal{E} d / 4 B .\) (Note: Although the method developed here is useful in measuring the rate of blood flow in a vessel, it is limited to use in surgery because measurement of the potential \(\mathcal{E}\) must be made directly across the vessel.)

A step-up transformer. A transformer connected to a 120 \(\mathrm{V}\) (rms) ac line is to supply \(13,000 \mathrm{V}\) (rms) for a neon sign. To reduce the shock hazard, a fuse is to be inserted in the primary circuit and is to blow when the rms current in the secondary circuit exceeds 8.50 \(\mathrm{mA}\) (a) What is the ratio of secondary to primary turns of the transformer? (b) What power must be supplied to the transformer when the rms secondary current is 8.50 \(\mathrm{mA}\) ? (c) What current rating should the fuse in the primary circuit have?

Self-inductance of a solenoid. A long, straight solenoid has \(N\) turns, a uniform cross-sectional area \(A,\) and length \(l .\) Use the definition of self- inductance expressed by Equation 21.13 to show that the inductance of this solenoid is given approximately by the equation \(L=\mu_{0} A N^{2} / l .\) Assume that the magnetic field is uniform inside the solenoid and zero outside. (Your answer is approximate because \(B\) is actually smaller at the ends than at the center of the solenoid. For this reason, your answer is actually an upper limit on the inductance.)

When a certain inductor carries a current \(I,\) it stores 3.0 \(\mathrm{mJ}\) of magnetic energy. How much current (in terms of \(I )\) would it have to carry to store 9.0 \(\mathrm{mJ}\) of energy?

A closely wound rectangular coil of 80 turns has dimensions of 25.0 \(\mathrm{cm}\) by 40.0 \(\mathrm{cm}\) . The plane of the coil is rotated from a position where it makes an angle of \(37.0^{\circ}\) with a magnetic field of 1.10 \(\mathrm{T}\) to a position perpendicular to the field. The rotation takes 0.0600 s. What is the average emf induced in the coil?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.