Problem 66 - University Physics with Modern Physics

Question

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 \(31.5 .\)

Step-by-Step Solution

Answer

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

Step 1: 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: \[ \omega_0 = \frac{1}{\sqrt{1.80 \times 0.900 \times 10^{-6}}} \approx 787.4 \text{ rad/s} \]

Step 2: 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 \( R = 300 \Omega \): \[ I_{\text{rms-0}} = \frac{V_{\text{rms}}}{R} = \frac{60.0}{300} = 0.2 \text{ A} \]

Step 3: 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.

Step 4: 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.

Key Concepts

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 \( C = 0.900 \underline{\phantom{xxx}}\mu \text{F} \), 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.