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Consider an \(L-R-C\) series circuit with a 1.80-H inductor, a 0.900-\(\mu\)F capacitor, and a 300-\(\Omega\) resistor. The source has terminal rms voltage V\(_{rms}\) = 60.0 V and variable angular frequency \(\omega\). (a) What is the resonance angular frequency \(\omega_0\) of the circuit? (b) What is the rms current through the circuit at resonance, I\(_{rms}\)-0? (c) For what two values of angular frequency, \(\omega\)1 and \(\omega\)2, is the rms current half the resonance value? (d) The quantity \(\omega\)1 - \(omega\)2 defines the resonance \(width\). Calculate I\(_{rms}\)-0 and the resonance width for R = 300 \(\Omega\), 30.0 \(\Omega\), and 3.00 \(\Omega\). Describe how your results compare to the discussion in Section 31.5. the

Short Answer

Expert verified
a) \(\omega_0 = 787.6\, \text{rad/s}\), b) \(I_{\text{rms}-0} = 0.20\, \text{A}\), c) \(\omega_1 = 772.1\, \text{rad/s}\), \(\omega_2 = 803.1\, \text{rad/s}\), d) As \( R \) decreases, the resonance width decreases.

Step by step solution

01

Calculate the resonance angular frequency (蠅鈧)

The resonance angular frequency, \( \omega_0 \), of a series \( L-R-C \) circuit is given by the formula:\[ \omega_0 = \frac{1}{\sqrt{LC}} \]Given \( L = 1.80 \ \text{H} \) and \( C = 0.900 \times 10^{-6} \ \text{F} \). Substitute these values into the formula:\[ \omega_0 = \frac{1}{\sqrt{1.8 \times 0.900 \times 10^{-6}}} \approx 787.6 \ \text{rad/s} \].
02

Determine rms current at resonance, I鈧欌個

At resonance, the impedance is purely resistive and equals the resistance \( R \), so the rms current can be calculated using Ohm's Law:\[ I_{\text{rms}-0} = \frac{V_{\text{rms}}}{R} \]Substitute \( V_{\text{rms}} = 60.0 \ \text{V} \) and \( R = 300 \ \Omega \):\[ I_{\text{rms}-0} = \frac{60.0}{300} = 0.20 \ \text{A} \].
03

Calculate half-resonant frequencies 蠅鈧 and 蠅鈧

The half-resonant frequencies occur when the impedance is \( \sqrt{2} \) times the resistance \( R \). The formula is given by:\[ \omega_1, \omega_2 = \omega_0 \sqrt{1 \pm \frac{1}{2Q^2}} \]Where \( Q = \frac{\omega_0 L}{R} \). First, calculate \( Q \):\[ Q = \frac{787.6 \times 1.80}{300} \approx 4.732 \]Now find \( \omega_1 \) and \( \omega_2 \):\[ \omega_1 = 787.6 \sqrt{1 - \frac{1}{2 \times 4.732^2}} \approx 772.1 \ \text{rad/s} \]\[ \omega_2 = 787.6 \sqrt{1 + \frac{1}{2 \times 4.732^2}} \approx 803.1 \ \text{rad/s} \].
04

Evaluate the resonance width

The resonance width \( \omega_2 - \omega_1 \) is as follows:\[ \text{Width} = 803.1 - 772.1 = 31.0 \ \text{rad/s} \].
05

Calculate I鈧欌個 and resonance width for varied resistance

Repeat the above calculations for \( R = 30 \ \Omega \), and \( R = 3 \ \Omega \):- For \( R = 30 \ \Omega \): - \( I_{\text{rms}-0} = 2.0 \ \text{A} \) - New \( Q = \frac{787.6 \times 1.80}{30} \approx 47.32 \) - Resonance width \( \approx 3.1 \ \text{rad/s} \).- For \( R = 3 \ \Omega \): - \( I_{\text{rms}-0} = 20.0 \ \text{A} \) - New \( Q = \frac{787.6 \times 1.80}{3} \approx 473.2 \) - Resonance width \( \approx 0.31 \ \text{rad/s} \).As \( R \) decreases, and \( Q \) increases, the resonance width decreases, illustrating a sharper resonance.

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Key Concepts

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

Resonance Frequency
In an RLC circuit, the resonance frequency is the frequency at which the circuit naturally oscillates with the least resistance. At resonance, the reactive effects of the inductor and capacitor cancel each other out, leaving only the resistive component of the impedance. This frequency is crucial because it minimizes energy loss in the circuit and maximizes current.

Mathematically, the resonance angular frequency, \( \omega_0 \), for a series RLC circuit is given by the formula: \[ \omega_0 = \frac{1}{\sqrt{LC}} \]where \( L \) and \( C \) are the inductance and capacitance, respectively.

In our exercise, using \( L = 1.80 \, \text{H} \) and \( C = 0.900 \times 10^{-6} \, \text{F} \), the resonance angular frequency is calculated to be approximately \( 787.6 \, \text{rad/s} \). Understanding the resonance frequency helps in designing circuits that can effectively filter or select desired frequencies while ignoring others.
Inductor
An inductor is a passive electronic component that stores energy in a magnetic field when electric current passes through it. In the RLC circuit, the inductor's reactance changes with frequency, influencing how the circuit responds to alternating current (AC).

This reactance, \( X_L \), is given by:\[ X_L = \omega L \]where \( \omega \) is the angular frequency and \( L \) is the inductance.

As the frequency increases, the inductor's impedance also increases, which affects the overall impedance of the series circuit. In the exercise, an inductor with \( L = 1.80 \, \text{H} \) was used, influencing the resonance and tuning characteristics of the circuit. These characteristics are vital when tuning radios or designing filters and oscillators.
Capacitor
A capacitor is another crucial component in an RLC circuit, storing energy in an electric field. Its primary role is to oppose changes in voltage and work in tandem with the inductor to determine the circuit's resonant frequency.

The reactance of a capacitor, \( X_C \), varies inversely with frequency:\[ X_C = \frac{1}{\omega C} \]where \( \omega \) is the angular frequency, and \( C \) is the capacitance.

In our example, the circuit uses a capacitor with \( C = 0.900 \times 10^{-6} \, \text{F} \). When the frequency changes, the capacitive reactance adjusts, impacting the circuit's impedance. Capacitors are essential for managing oscillations and storing charge, making them vital for timing applications and signal processing.
Resistor
In a series RLC circuit, the resistor governs energy dissipation. Unlike inductors and capacitors, which exchange energy but ideally don't dissipate it, a resistor converts electrical energy into heat.

The resistance, \( R \), influences the quality factor (\( Q \)), which describes the sharpness of the resonance peak. The formula for \( Q \) is:\[ Q = \frac{\omega_0 L}{R} \]where \( \omega_0 \) is the resonance angular frequency, and \( L \) is the inductance.

Different resistances produce different \( Q \)-factors, affecting the resonance width and the peak current at resonance. Thus, as shown in the exercise, substituting different resistances like 300 \( \Omega \), 30 \( \Omega \), and 3 \( \Omega \) altered the resonance behavior and current through the circuit. Resistors are key for controlling current flow and stabilizing circuits.
RMS Voltage
Root Mean Square (RMS) voltage is a measure of the effective value of an alternating current (AC) voltage. It provides an equivalent value to direct current (DC) voltage, which would produce the same power dissipation in a resistor.

In AC circuits, the RMS voltage, \( V_{\text{rms}} \), is crucial for calculating power and current as it accounts for the time-averaged value of the sinusoidal input voltage.

Given RMS voltage and resistance, the current through the circuit at resonance can be determined by Ohm's Law:\[ I_{\text{rms}} = \frac{V_{\text{rms}}}{R} \]For our exercise, \( V_{\text{rms}} = 60.0 \, \text{V} \) was used to determine how much current flows through the circuit, especially important at resonance where the impedance is at its minimum. This concept is foundational in understanding power within AC systems.
Q-Factor
The Q-Factor, or quality factor, is a dimensionless parameter that describes how underdamped an oscillator or resonator is, which correlates to the sharpness of the resonance peak.

In an RLC circuit, a higher Q-Factor means the circuit has a narrow and sharp resonance peak.
  • High \( Q \)-Factor: Implies low energy dissipation and sharper frequency selectivity.
  • Low \( Q \)-Factor: Implies higher energy dissipation and a wider frequency band.
The Q-Factor is calculated using:\[ Q = \frac{\omega_0 L}{R} \]where \( \omega_0 \) is the resonance frequency, \( L \) is the inductance, and \( R \) is the resistance.

In the exercise, as resistance decreased from 300 \( \Omega \) to 3 \( \Omega \), the Q-Factor increased, indicating a sharper resonance and reduced resonance width. Understanding the Q-Factor helps in designing circuits with precise frequency selection, like filters and oscillators.

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Most popular questions from this chapter

A resistor with \(R = 300\space \Omega\) and an inductor are connected in series across an ac source that has voltage amplitude 500 V. The rate at which electrical energy is dissipated in the resistor is 286 W. What is (a) the impedance Z of the circuit; (b) the amplitude of the voltage across the inductor; (c) the power factor?

An inductor with \(L\) = 9.50 mH is connected across an ac source that has voltage amplitude 45.0 V. (a) What is the phase angle \(\phi\) for the source voltage relative to the current? Does the source voltage lag or lead the current? (b) What value for the frequency of the source results in a current amplitude of 3.90 A?

In an \(L-R-C\) series circuit, R = 400 \(\Omega\), \(L\) = 0.350 H, and \(C\) = 0.0120 \(\mu\)F. (a) What is the resonance angular frequency of the circuit? (b) The capacitor can withstand a peak voltage of 670 V. If the voltage source operates at the resonance frequency, what maximum voltage amplitude can it have if the maximum capacitor voltage is not exceeded?

An \(L-R-C\) series circuit consists of a source with voltage amplitude 120 V and angular frequency 50.0 rad/s, a resistor with R = 400 \(\Omega\), an inductor with \(L\) = 3.00 H, and a capacitor with capacitance \(C\). (a) For what value of C will the current amplitude in the circuit be a maximum? (b) When \(C\) has the value calculated in part (a), what is the amplitude of the voltage across the inductor?

A coil has a resistance of 48.0 \(\Omega\). At a frequency of 80.0 Hz the voltage across the coil leads the current in it by 52.3\(^\circ\). Determine the inductance of the coil.

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