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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 0.1684 A.

Step by step solution

01

Understand the Problem

We need to find the current through an inductor with given voltage, inductance, and frequency. Inductive reactance will help us calculate the current using Ohm's Law for AC circuits.
02

Calculate Inductive Reactance

The formula for inductive reactance \(X_L\) is \( X_L = 2\pi f L \), where \( f \) is frequency and \( L \) is inductance. Substitute the values: \( X_L = 2\pi (650 \text{ Hz})(0.080 \text{ H}) \).
03

Perform Calculations for Reactance

Calculate \( X_L \): \[ X_L = 2\pi \times 650 \times 0.080 \approx 326.73 \text{ ohms} \]
04

Use Ohm's Law to Find Current

Ohm's Law for AC circuits is \( I = \frac{V}{X_L} \), where \( V \) is voltage. Substitute the values to find the current: \[ I = \frac{55}{326.73} \approx 0.1684 \text{ A} \]
05

Conclude the Solution

The current through the inductor is approximately \(0.1684 \text{ Amperes}\).

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Key Concepts

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

Inductive Reactance
In alternating current (AC) circuits, inductors play a key role in resisting changes in current. This is measured by a property known as inductive reactance, represented as \( X_L \). Inductive reactance is not a fixed value; it varies with the frequency of the AC signal and the inductance of the coil. The formula to calculate it is: \[ X_L = 2\pi f L \]where:
  • \( f \) is the frequency of the AC source,
  • \( L \) is the inductance of the coil.

This formula shows that as the frequency or the inductance increases, the reactance increases, making it harder for the current to pass through the inductor. In our problem, by substituting the given values, we calculated the inductive reactance as approximately \( 326.73 \) ohms. This resistive behavior is crucial to designing AC circuits, as it affects the current flow.
AC Ohm's Law
Ohm's Law is a fundamental principle used to find the current in both DC and AC circuits. In AC circuits, however, we must consider reactance instead of just resistance. This variation is known as AC Ohm's Law, which states:\[ I = \frac{V}{X_L} \]where:
  • \( I \) is the current through the inductor,
  • \( V \) is the voltage across the inductor,
  • \( X_L \) is the inductive reactance.

This formula helps us understand how much current will flow through an inductor at a given reactance and voltage. It's a handy tool for circuit analysis and helps provide a clear picture of an inductive circuit's behavior. In our example, using AC Ohm's Law allowed us to determine that the current is approximately \( 0.1684 \) Amperes.
Frequency and Inductance
Frequency and inductance are two foundational elements that heavily influence AC circuits. Frequency, measured in hertz (Hz), indicates how fast the current alternates per second. Inductance, measured in henries (H), represents the inductor's ability to store energy in a magnetic field. Both these factors together affect the inductive reactance:
  • As frequency increases, the reactance increases, making it more difficult for AC to pass through the inductor.
  • Higher inductance equates to more reactance, acting like a higher resistance in the circuit.
Increasing either the frequency or inductance increases the opposition to current flow. Understanding the interplay between these elements allows for the intentional design of filters, oscillators, and other components in electronics. In this exercise, given a frequency of \( 650 \) Hz and inductance of \( 0.080 \) H, we see a significant effect on reactance and thus on the current through the circuit.

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Most popular questions from this chapter

A series \(\mathrm{RCL}\) circuit contains a \(47.0-\Omega\) resistor, a \(2.00-\mu \mathrm{F}\) capacitor, and a \(4.00-\mathrm{mH}\) inductor. When the frequency is \(2550 \mathrm{Hz},\) what is the power factor of the circuit?

A tank circuit in a radio transmitter is a series RCL circuit connected to an antenna. The antenna broadcasts radio signals at the resonant frequency of the tank circuit. Suppose that a certain tank circuit in a shortwave radio transmitter has a fixed capacitance of \(1.8 \times 10^{-11} \mathrm{F}\) and a variable inductance. If the antenna is intended to broadcast radio signals ranging in frequency from 4.0 MHz to 9.0 MHz, find the (a) minimum and (b) maximum inductance of the tank circuit.

A circuit consists of an \(85-\Omega\) resistor in series with a \(4.0-\mu \mathrm{F}\) capacitor, and the two are connected between the terminals of an ac generator. The voltage of the generator is fixed. At what frequency is the current in the circuit one-half the value that exists when the frequency is very large?

A series RCL circuit has a resonant frequency of \(690 \mathrm{kHz}\). If the value of the capacitance is \(2.0 \times 10^{-9} \mathrm{F},\) what is the value of the inductance?

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?
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