/*! 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 16 An inductor used in a dc power s... [FREE SOLUTION] | 91Ó°ÊÓ

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An inductor used in a dc power supply has an inductance of 12.0 H and a resistance of 180 \(\Omega\). It carries a current of 0.500 A. (a) What is the energy stored in the magnetic field? (b) At what rate is thermal energy developed in the inductor? (c) Does your answer to part (b) mean that the magnetic-field energy is decreasing with time? Explain.

Short Answer

Expert verified
(a) 1.5 J; (b) 45 W; (c) No, magnetic-field energy is constant with constant current.

Step by step solution

01

Calculate the Energy Stored in the Magnetic Field

The energy stored in the magnetic field of an inductor can be calculated using the formula \( U = \frac{1}{2} L I^2 \), where \( L \) is the inductance and \( I \) is the current. Plugging in the known values, \( L = 12.0 \text{ H} \) and \( I = 0.500 \text{ A} \), we get: \[ U = \frac{1}{2} \times 12.0 \times (0.500)^2 = \frac{1}{2} \times 12.0 \times 0.25 \] \[ U = 1.5 \text{ J} \]. Thus, the energy stored is 1.5 joules.
02

Calculate the Rate of Thermal Energy Development

The rate at which thermal energy is developed in the resistor (due to the current flowing through it) is given by the power loss formula \( P = I^2 R \), where \( I \) is the current and \( R \) is the resistance. With \( I = 0.500 \text{ A} \) and \( R = 180 \Omega \), the power is \[ P = (0.500)^2 \times 180 = 0.25 \times 180 = 45 \text{ W} \]. Thus, the thermal energy is developed at a rate of 45 watts.
03

Analyze the Relationship Between Parts (a) and (b)

The energy stored in the magnetic field represents potential energy within the inductor, while the thermal energy developed is due to resistive losses. Part (b) refers to power loss in the resistor, not a change in magnetic-field energy. As long as the current remains constant, the stored magnetic-field energy does not decrease despite thermal losses.

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Key Concepts

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

Inductor Energy Storage
An inductor creates a magnetic field around it when current flows through its coils. This magnetic field stores energy. The energy stored in the magnetic field of an inductor can be described with the formula: \[ U = \frac{1}{2} L I^2 \] where
  • \( U \) is the energy stored in joules,
  • \( L \) is the inductance in henrys (H), and
  • \( I \) is the current in amperes (A).
Applying this formula, we see that the energy depends on both the inductance and the square of the current. This means even a small current can store significant energy if the inductance is large. For example, with an inductance of 12.0 H and a current of 0.500 A, as provided in the exercise, the energy stored equals 1.5 Joules. The presence of this energy is essential for dictating how the inductor would perform when changes occur in the current flow.
Thermal Energy Development
When current flows through a resistor, it generates heat, which is a form of thermal energy. This heat generation results in energy loss known as resistive power loss. The rate at which this energy is converted to thermal energy is given by the formula:\[ P = I^2 R \]where
  • \( P \) is the power or rate of energy conversion in watts (W),
  • \( I \) is the current in amperes (A), and
  • \( R \) is the resistance in ohms (\( \Omega \)).
In the exercise, with a current of 0.500 A and a resistance of 180 Ω, the rate of thermal energy development is calculated to be 45 watts. This indicates that the resistor causes the surrounding area to heat up as electrical energy is dissipated as heat. Understanding this concept helps in designing circuits since excessive heat can damage components.
Resistive Power Loss
Resistive power loss occurs when electrical energy is converted into heat in an inductor's resistor. This happens wherever current flows through a resistive material. Key points to remember:
  • It is not linked to the depletion of the magnetic energy stored in the inductor.
  • While the magnetic field stores potential energy, the resistive component uses some of that energy by converting it into heat.
In our exercise, even though the thermal energy is developed at 45 watts, the energy stored in the magnetic field remains unchanged as long as the current remains constant. This separation is crucial, as it highlights that resistive power loss reflects inefficiencies in energy use rather than a direct decline in magnetic energy. Understanding this helps in evaluating the efficiency and energy management of systems with resistive components.

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Most popular questions from this chapter

An \(L\)-\(C\) circuit consists of a 60.0-mH inductor and a 250-\(\mu\)F capacitor. The initial charge on the capacitor is 6.00 \(\mu\)C, and the initial current in the inductor is zero. (a) What is the maximum voltage across the capacitor? (b) What is the maximum current in the inductor? (c) What is the maximum energy stored in the inductor? (d) When the current in the inductor has half its maximum value, what is the charge on the capacitor and what is the energy stored in the inductor?

An \(L\)-\(C\) circuit containing an 80.0-mH inductor and a 1.25-nF capacitor oscillates with a maximum current of 0.750 A. Calculate: (a) the maximum charge on the capacitor and (b) the oscillation frequency of the circuit. (c) Assuming the capacitor had its maximum charge at time \(t = 0\), calculate the energy stored in the inductor after 2.50 ms of oscillation.

A coil has 400 turns and self-inductance 7.50 mH. The current in the coil varies with time according to \(i = (680 \, \mathrm{mA}) \mathrm{cos} (\pi{t}/0.0250 \, \mathrm{s})\). (a) What is the maximum emf induced in the coil? (b) What is the maximum average flux through each turn of the coil? (c) At \(t = 0.0180\) s, what is the magnitude of the induced emf?

An \(L\)-\(R\)-\(C\) series circuit has \(L = 0.600\) H and \(C = 3.00 \, \mu \mathrm{F}\). (a) Calculate the angular frequency of oscillation for the circuit when \(R =\) 0. (b) What value of \(R\) gives critical damping? (c) What is the oscillation frequency \(\omega'\) when \(R\) has half of the value that produces critical damping?

An air-filled toroidal solenoid has 300 turns of wire, a mean radius of 12.0 cm, and a cross-sectional area of 4.00 cm\(^2\). If the current is 5.00 A, calculate: (a) the magnetic field in the solenoid; (b) the self inductance of the solenoid; (c) the energy stored in the magnetic field; (d) the energy density in the magnetic field. (e) Check your answer for part (d) by dividing your answer to part (c) by the volume of the solenoid.

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