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Chapter 23: Problem 9

An \(8.2-\mathrm{mH}\) inductor is connected to an ac generator \((10.0 \mathrm{V} \mathrm{rms},\), \(620 \mathrm{Hz}\) ). Determine the peak value of the current supplied by the generator.

Short Answer

Expert verified
The peak current is approximately 0.043 A.

Step by step solution

01

Understanding the Problem

We are given an inductor connected to an AC generator. The generator provides an RMS voltage of 10.0 V and operates at a frequency of 620 Hz. We need to determine the peak current in the circuit.
02

Calculate Angular Frequency

The angular frequency \( \omega \) is calculated using the formula: \( \omega = 2 \pi f \), where \( f \) is the frequency of the generator. Substitute \( f = 620 \) Hz into the equation: \[ \omega = 2 \pi \times 620 \].
03

Calculate Inductive Reactance

The inductive reactance \( X_L \) of the inductor is calculated using the formula \( X_L = \omega L \), where \( L \) is the inductance in henrys. Given \( L = 8.2 \times 10^{-3} \) H, substitute \( \omega \) from the previous step to find \( X_L \): \[ X_L = (2 \pi \times 620) \times 8.2 \times 10^{-3} \].
04

Calculate RMS Current

The RMS current \( I_{\text{rms}} \) is calculated using Ohm's law for AC circuits: \( I_{\text{rms}} = \frac{V_{\text{rms}}}{X_L} \). Substitute \( V_{\text{rms}} = 10.0 \) V and \( X_L \) from the previous step to find \( I_{\text{rms}} \).
05

Calculate Peak Current

The peak current \( I_0 \) is related to the RMS current by the equation \( I_0 = \sqrt{2} \times I_{\text{rms}} \). Use \( I_{\text{rms}} \) calculated in the previous step to find \( I_0 \).

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Key Concepts

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

AC Circuits
Alternating current (AC) circuits differ significantly from direct current (DC) circuits. Unlike DC which flows in a single direction, AC continuously changes direction, typically in a sinusoidal wave pattern. This makes AC circuits advantageous for power distribution as they can be easily transformed to different voltage levels using transformers.
AC circuits are characterized by three key components:
  • Voltage Source: Supplies the AC voltage, which alternates in magnitude and direction.
  • Inductive Elements: Like inductors, store energy in a magnetic field when current flows through them.
  • Capacitive Elements: Store energy in an electric field. These components react differently to AC compared to DC, impacting the impedance in the circuit.

The behavior of inductors in an AC circuit is critical because their impedance, due to inductive reactance, varies with frequency, profoundly affecting the current flow within the circuit.
RMS Voltage
Root Mean Square (RMS) voltage is a crucial concept in AC circuits. It represents the effective voltage of an AC system, which allows for the same power transfer as a corresponding DC system with the same voltage. For sinusoidal AC voltages, the RMS value is typically calculated as the peak voltage divided by the square root of 2.
RMS voltage is essential because it provides a realistic measure of how much work the AC voltage can do, akin to a constant DC voltage.
Understanding RMS voltage is fundamental in designing and analyzing AC circuits, allowing engineers to manage power distribution accurately and safely.
Peak Current
Peak current in AC circuits refers to the maximum instantaneous value of the current waveform. Calculating peak current is essential when designing electrical systems to ensure that components can withstand the highest level of current that will flow through them.
This value is related to the RMS current, which is a more common metric used to describe AC systems. The peak current (usually denoted as \(I_0\)) is calculated from the RMS current using the relationship:
  • \(I_0 = \sqrt{2} \times I_{\text{rms}}\)
Understanding peak current helps in selecting appropriate materials and system specifications that guarantee safety and performance under varying electrical conditions.
Angular Frequency
Angular frequency (\(\omega\)) in AC circuits quantifies how quickly the current alternates. It is typically expressed in radians per second and can be calculated using the formula \(\omega = 2 \pi f\), where \(f\) is the frequency in hertz (Hz).
The concept of angular frequency is crucial because:
  • It helps describe the oscillatory nature of AC signals.
  • It is used in calculating the reactance of inductors and capacitors, which is frequency-dependent.
  • It affects the impedance of the circuit, thereby influencing current and voltage values.
Knowing angular frequency is vital for understanding the dynamics of AC circuits, allowing for proper tuning and optimization of electrical systems for intended applications.

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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?

A series \(\mathrm{RCL}\) circuit contains only a capacitor \((C=6.60 \mu \mathrm{F})\) an inductor \((L=7.20 \mathrm{mH}),\) and a generator (peak voltage \(=32.0 \mathrm{V}\), frequency \(=1.50 \times 10^{3} \mathrm{Hz}\) ). When \(t=0 \mathrm{s}\), the instantaneous value of the voltage is zero, and it rises to a maximum one-quarter of a period later. (a) Find the instantaneous value of the voltage across the capacitor/ inductor combination when \(t=1.20 \times 10^{-4} \mathrm{s} .\) (b) What is the instantaneous value of the current when \(t=1.20 \times 10^{-4} \mathrm{s} ?\) (Hint: The instantaneous values of the voltage and current are, respectively, the vertical components of the voltage and current phasors.)

A generator is connected to a resistor and a \(0.032-\mathrm{H}\) inductor in se- \(-\) ries. The rms voltage across the generator is \(8.0 \mathrm{V}\). When the generator frequency is set to \(130 \mathrm{Hz}\), the rms voltage across the inductor is \(2.6 \mathrm{V}\). Determine the resistance of the resistor in this 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 \(4.0 \mathrm{MHz}\) to \(9.0 \mathrm{MHz},\) find the (a) minimum and (b) maximum inductance of the tank circuit.

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?
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