/*! 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 10 An inductor has an inductance of... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 23: Problem 10

An inductor has an inductance of \(0.080 \mathrm{H}\). The voltage across this inductor is \(55 \mathrm{V}\) and has a frequency of \(650 \mathrm{Hz}\). What is the current in the inductor?

Short Answer

Expert verified
The current in the inductor is approximately 0.1679 A.

Step by step solution

01

Understand the Given Values

First, identify the values given in the problem. We have an inductance \(L = 0.080 \text{ H}\), voltage \(V = 55 \text{ V}\), and frequency \(f = 650 \text{ Hz}\).
02

Formula for Inductive Reactance

The inductive reactance \(X_L\) is given by the formula \(X_L = 2\pi f L\). This is the resistance offered by the inductor to the AC current.
03

Calculate Inductive Reactance

Substitute the given values into the formula: \(X_L = 2\pi \times 650 \times 0.080\). Calculate this to find the inductive reactance.
04

Compute Inductive Reactance

Perform the multiplication within the formula: \(X_L = 2 \times 3.1416 \times 650 \times 0.080\). This gives us \(X_L \approx 327.47 \text{ ohms}\).
05

Ohm's Law for AC Circuits

In AC circuits, the current \(I\) can be found using the formula \(I = \frac{V}{X_L}\), where \(V\) is the voltage across the inductor and \(X_L\) is the inductive reactance.
06

Calculate the Current

Substitute the values into the formula: \(I = \frac{55}{327.47}\). Perform the division to find the current in the inductor.
07

Compute the Current

Calculate \(I = \frac{55}{327.47}\), which gives \(I \approx 0.1679 \text{ A}\). Thus, the current is approximately 0.1679 amperes.

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 91影视!

Key Concepts

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

Inductive Reactance
Inductive reactance is an important concept when dealing with AC circuits and inductors. It refers to the effective resistance that an inductor provides when AC current flows through it. Unlike resistance in a DC circuit, which is constant, inductive reactance changes with the frequency of the AC signal.
To calculate inductive reactance, use the formula: \(X_L = 2\pi f L\). Here, \(L\) is the inductance in henrys, and \(f\) is the frequency in hertz. This formula shows that as the frequency or the inductance increases, the reactance also increases.
Understanding how inductive reactance affects a circuit helps predict how the current will behave. High inductive reactance means the inductor will significantly oppose the current flow, while low reactance implies minimal opposition.
Ohm's Law for AC Circuits
Ohm's Law is a fundamental principle in both DC and AC circuits, but it adapts slightly when dealing with alternating current. In AC circuits, Ohm's Law is expressed as \(I = \frac{V}{X_L}\), where \(I\) is the current, \(V\) is the voltage across the inductor, and \(X_L\) is the inductive reactance. This version highlights the relationship between voltage, current, and reactance in AC circuits.
When applying Ohm's Law in AC circuits, always remember that reactance, not resistance, determines the current flow. An accurate calculation of inductive reactance allows for determining the current through an inductor by simply dividing the voltage by the reactance. Therefore, mastering this adapted form of Ohm's Law is crucial for accurately analyzing and working with AC circuits that contain inductors.
Inductor Voltage and Frequency
The relationship between inductor voltage and frequency is crucial in understanding AC circuits. Inductors behave differently based on the frequency of the voltage applied across them. The formula \(X_L = 2\pi f L\) demonstrates that inductive reactance is directly proportional to frequency.
Since inductive reactance increases with frequency, an increase in voltage frequency means the inductor will provide more opposition to the current flow. This is why inductors are often used to filter high-frequency signals in circuits. Conversely, at lower frequencies, the reactance is less, allowing more current to flow through.
The combined effects of frequency and inductance explain how inductors can selectively block or allow specific ranges of frequencies, highlighting their importance in various electronic applications.

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

Two parallel plate capacitors are filled with the same dielectric material and have the same plate area. However, the plate separation of capacitor 1 is twice that of capacitor 2 . When capacitor 1 is connected across the terminals of an ac generator, the generator delivers an rms current of 0.60 A. Concepts: (i) Which of the two capacitors has the greater capacitance? (ii) Is the equivalent capacitance of the parallel combination \(\left(C_{\mathrm{P}}\right)\) greater or smaller than the capacitance of capacitor \(1 ?\) (iii) Is the capacitive reactance of \(C_{\mathrm{P}}\) greater or smaller than for \(C_{1} ?\) (iv) When both capacitors are connected in parallel across the terminals of the generator, is the current from the generator greater or smaller than when capacitor 1 is connected alone? Calculations: What is the current delivered by the generator when both capacitors are connected in parallel across the terminals?

In a series circuit, a generator \((1350 \mathrm{Hz}, 15.0 \mathrm{V})\) is connected to a \(16.0-\Omega\) resistor, a \(4.10-\mu \mathrm{F}\) capacitor, and a \(5.30-\mathrm{mH}\) inductor. Find the voltage across each circuit element.

A capacitor (capacitance \(C_{1}\) ) is connected across the terminals of an ac generator. Without changing the voltage or frequency of the generator, a second capacitor (capacitance \(C_{2}\) ) is added in series with the first one. As a result, the current delivered by the generator decreases by a factor of three. Suppose that the second capacitor had been added in parallel with the first one, instead of in series. By what factor would the current delivered by the generator have increased?

A capacitor is connected across an ac generator whose frequency is \(750 \mathrm{Hz}\) and whose peak output voltage is \(140 \mathrm{V}\). The rms current in the circuit is 3.0 A. (a) What is the capacitance of the capacitor? (b) What is the magnitude of the maximum charge on one plate of the capacitor?

A capacitor is connected across the terminals of an ac generator that has a frequency of \(440 \mathrm{Hz}\) and supplies a voltage of \(24 \mathrm{V}\). When a second capacitor is connected in parallel with the first one, the current from the generator increases by 0.18 A. Find the capacitance of the second capacitor.
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Scientific Method Physics

Read Explanation

Electricity

Read Explanation

Work Energy and Power

Read Explanation

Particle Model of Matter

Read Explanation

Modern Physics

Read Explanation
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.