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

Short Answer

Expert verified
Resonance frequency \( \omega_0 \approx 787.4 \text{ rad/s} \); RMS current at resonance is \( 0.2 \text{ A} \).

Step by step solution

01

Calculate the Resonance Angular Frequency

The resonance angular frequency \( \omega_0 \) is calculated using the formula: \[ \omega_0 = \frac{1}{\sqrt{LC}} \] where \( L = 1.80 \text{ H} \) and \( C = 0.900 \times 10^{-6} \text{ F} \). Substituting the given values: \[ \omega_0 = \frac{1}{\sqrt{1.80 \times 0.900 \times 10^{-6}}} \approx 787.4 \text{ rad/s} \]
02

Calculate the RMS Current at Resonance

At resonance, the impedance of the circuit is purely resistive, equal to the resistance \( R \). The resonance RMS current \( I_{\text{rms-0}} \) is calculated using \( V_{\text{rms}} = 60.0 \text{ V} \) and \( R = 300 \Omega \): \[ I_{\text{rms-0}} = \frac{V_{\text{rms}}}{R} = \frac{60.0}{300} = 0.2 \text{ A} \]
03

Find the Frequencies where Current is Half the Resonance Value

To find the frequencies \( \omega_1 \) and \( \omega_2 \) where the RMS current is half the value at resonance, i.e., \( I_{\text{rms}} = \frac{0.2}{2} = 0.1 \text{ A} \), we use the half-power bandwidth concept. This involves solving \[ \frac{V_{\text{rms}}}{\sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}} = 0.1 \] iteratively or using appropriate approximations for an LRC circuit's frequency response.
04

Calculate Resonance Width and Compare Results

The resonance width is given by \( |\omega_1 - \omega_2| \). Repeat the calculations for different resistances (\( R = 300 \Omega, 30 \Omega, \) and \( 3 \Omega \)), observing that a lower \( R \) results in a narrower width as higher \( R \) dampens less. Compare these results to the theoretical explanation of damping and resonance width discussed in Section 31.5.

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Key Concepts

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

Resonance Angular Frequency
In an L-R-C circuit, resonance occurs when the reactive components of the circuit cancel each other out. This happens at a specific angular frequency, known as the resonance angular frequency, denoted by \( \omega_0 \). You can find this frequency using the formula:
  • \( \omega_0 = \frac{1}{\sqrt{LC}} \)
where \( L \) is the inductance and \( C \) is the capacitance. It is crucial to understand that at this frequency, the inductive reactance \( \omega L \) and the capacitive reactance \( \frac{1}{\omega C} \) are equal in magnitude but opposite in sign, thus cancelling each other out entirely. Therefore, the circuit behaves purely resistively, minimizing the impedance and allowing for maximum current flow. For example, in our problem, with \( L = 1.80 \text{ H} \) and \( C = 0.900 \text{ }\mu \text{F} \), the resonance angular frequency comes out to be approximately \( 787.4 \text{ rad/s} \). This is crucial in tuning circuits, such as radio receivers, to receive the desired frequency signals.
L-R-C Series Circuit
An L-R-C series circuit consists of three components: an Inductor (L), a Resistor (R), and a Capacitor (C), all in series along a single path. The series configuration means that the same current flows through each component, but the impedance (total resistance to current flow) varies with frequency.
  • At low frequencies, the capacitor dominates.
  • At high frequencies, the inductor dominates.
  • At resonance, the impedance is minimal and purely resistive, determined solely by \( R \).
This specific circuit configuration is significant because it exhibits the characteristic of resonant frequency, where the current reaches its maximum and the system oscillates at its natural frequency. Understanding the behavior of each component and how they interact in this circuit helps control and manage alternating current (AC) systems efficiently. In the provided example, with \( R = 300 \Omega \), the resonance results in the current being \( I_{\text{rms-0}} = 0.2 \text{ A} \) because the circuit offers minimal opposition to the flow of AC current at resonance.
Resonance Width
The resonance width measures how narrowly defined the resonant peak is in terms of frequency. In simpler terms, it defines the range of frequencies over which the circuit can effectively operate near its peak performance. Specifically, this width is denoted by the distance between two frequencies, \( \omega_1 \) and \( \omega_2 \), where the current is half its resonance value:
  • The resonance width \( |\omega_1 - \omega_2| \) provides insight into the circuit's selectivity.
  • A smaller width indicates a sharper peak and better selectivity, which means the circuit is more sensitive to changes in frequency.
  • Larger values of resistance \( R \) generally increase the resonance width, leading to reduced selectivity and a broader resonance peak.
In our exercise example, using different resistances such as \( 300 \Omega, 30 \Omega, \) and \( 3 \Omega \) shows how damping - which is linked to resistance - affects the resonance width. More significant resistance results in broader peaks, meaning the range over which the circuit remains effective widens. This behavior aligns with the theoretical expectations discussed in academic resources on damping and resonance.

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Most popular questions from this chapter

An \(L_{-} R-C\) series circuit is connected to a \(120-\mathrm{Hz}\) ac source that has \(V_{\mathrm{rms}}=80.0 \mathrm{V} .\) The circuit has a resistance of 75.0\(\Omega\) and an impedance at this frequency of 105\(\Omega .\) What average power is delivered to the circuit by the source?

In an \(L-R-C\) series circuit the current is given by \(i=I \cos \omega t .\) The voltage amplitudes for the resistor, inductor, and capacitor are \(V_{R}, \quad V_{L},\) and \(V_{C}\) . (a) Show that the instantaneous power into the resistor is \(p_{R}=V_{R} I \cos ^{2} \omega t=\frac{1}{2} V_{R} I(1+\cos 2 \omega t)\) What does this expression give for the average power into the resistor? (b) Show that the instantaneous power into the inductor is \(p_{L}=-V_{L} I \sin \omega t \cos \omega t=-\frac{1}{2} V_{L} I \sin 2 \omega t .\) What does this expression give for the average power into the inductor? (c) Show that the instantaneous power into the capacitor is \(p_{C}=\) \(V_{C} I \sin \omega t \cos \omega t=\frac{1}{2} V_{C} I \sin 2 \omega t .\) What does this expression give for the average power into the capacitor? (d) The instantaneous power delivered by the source is shown in Section 31.4 to be \(p=V I \cos \omega t(\cos \phi \cos \omega t-\sin \phi \sin \omega t) .\) Show that \(p_{R}+p_{L}+\) \(p_{C}\) equals \(p\) at each instant of time.

At a frequency \(\omega_{1}\) the reactance of a certain capacitor equals that of a certain inductor. (a) If the frequency is changed to \(\omega_{2}=2 \omega_{1},\) what is the ratio of the reactance of the inductor to that of the capacitor? Which reactance is larger? (b) If the frequency is changed to \(\omega_{3}=\omega_{1} / 3,\) what is the ratio of the reactance of the inductor to that of the capacitor? Which reactance is larger? (c) If the capacitor and inductor are placed in series with a resistor of resistance \(R\) to form an \(L-R-C\) series circuit, what will be the resonance angular frequency of the circuit?

An \(L-R-C\) series circuit with \(L=0.120 \mathrm{H}, R=240 \Omega\) and \(C=7.30 \mu\) F carries an rms current of 0.450 \(\mathrm{A}\) with a frequency of 400 \(\mathrm{Hz}\) (a) What are the phase angle and power factor for this circuit? (b) What is the impedance of the circuit? (c) What is the rms voltage of the source? (d) What average power is delivered by the source? (e) What is the average rate at which electrical energy is converted to thermal energy in the resistor? (f) What is the average rate at which electrical energy is dissipated (converted to other forms) in the capacitor? (g) In the inductor?

An \(L-R-C\) series circuit is connected to an ac source of constant voltage amplitude \(V\) and variable angular frequency \(\omega\) (a) Show that the current amplitude, as a function of \(\omega,\) is $$I=\frac{V}{\sqrt{R^{2}+(\omega L-1 / \omega C)^{2}}}$$ (b) Show that the average power dissipated in the resistor is $$P=\frac{V^{2} R / 2}{R^{2}+(\omega L-1 / \omega C)^{2}}$$ (c) Show that \(I\) and \(P\) are both maximum when \(\omega=1 / \sqrt{L C},\) the resonance frequency of the circuit. (d) Graph \(P\) as a function of \(\omega\) for \(V=100 \mathrm{V}, R=200 \Omega, L=2.0 \mathrm{H},\) and \(C=0.50 \mu \mathrm{F}\) . Compare to the light purple curve in Fig. \(31.19 .\) Discuss the behavior of \(I\) and \(P\) in the limits \(\omega=0\) and \(\omega \rightarrow \infty\) .

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