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31.32. An \(L-R-C\) series circuit consists of a source with voltage amplitude 120 \(\mathrm{V}\) and angular frequency 50.0 \(\mathrm{rad} / \mathrm{s}\) , a resistor with \(R=400 \Omega\) an inductor with \(L=9.00 \mathrm{H}\) , 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?

Short Answer

Expert verified
(a) \(C \approx 4.44 \times 10^{-5} \mathrm{F}\) (b) The voltage amplitude across the inductor is 135 V.

Step by step solution

01

Understanding Resonance in an L-R-C Circuit

In an L-R-C series circuit, the current amplitude is at its maximum when the circuit is in resonance. This occurs when the inductive reactance (\(X_L\)) equals the capacitive reactance (\(X_C\)). Thus, at resonance, \(X_L = X_C \). The reactance of the inductor is given by \(X_L = \omega L\) and the reactance of the capacitor is given by \(X_C = \frac{1}{\omega C}\), where \(\omega\) is the angular frequency.
02

Solving for the Capacitor Value at Resonance

At resonance, \(\omega L = \frac{1}{\omega C}\). Rearranging this equation to solve for \(C\), we get: \[C = \frac{1}{\omega^2 L}\]Plug in \(\omega = 50.0\, \mathrm{rad/s}\) and \(L = 9.00\, \mathrm{H}\):\[C = \frac{1}{(50.0)^2 \cdot 9.00}\]\[C = \frac{1}{22500}\]\[C \approx 4.44 \times 10^{-5} \mathrm{F}\]
03

Calculating the Voltage Amplitude Across the Inductor

With the value of \(C\) that results in resonance, the total impedance \(Z\) of the circuit is equal to the resistance \(R\), because \(X_L = X_C\) and they cancel each other out. Thus, the current amplitude \(I_0\) is:\[I_0 = \frac{V_0}{R} = \frac{120}{400} = 0.3\, \mathrm{A}\]The amplitude of the voltage across the inductor \(V_L\) is:\[V_L = I_0 \cdot X_L = 0.3 \cdot (\omega L) = 0.3 \cdot 450 = 135\, \mathrm{V}\]

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Key Concepts

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

Resonance in Circuits
Resonance in an L-R-C circuit is a fascinating event where the circuit's reactive components perfectly balance each other out. This happens when the inductive reactance and capacitive reactance are equal. At resonance, the current in the circuit reaches its peak because the reactances cancel each other, making the total impedance minimal and equal to just the ohmic resistance.
This balance results in the maximum transfer of energy between the inductor and the capacitor, causing the circuit to oscillate at its natural frequency.
  • The effect is similar to a swing moving back and forth with the maximum amplitude.
  • A resonant circuit can store energy efficiently with minimal losses.
Ultimately, understanding resonance is crucial, as it is the foundation for tuning circuits to specific frequencies in applications such as radio receivers and signal processing.
Inductive Reactance
Inductive reactance is an important property of an inductor in an AC circuit. It determines how much the inductor will oppose the change in current. Defined by the formula \(X_L = \omega L\), where \(\omega\) represents the angular frequency and \(L\) is the inductance, it shows how inductance and frequency influence the reactance.
The higher the frequency or the inductance, the higher the opposition to the current changes:
  • An inductor resists changes in current majorly due to its stored magnetic field.
  • This reactance increases with frequency, making it a key player in filtering high-frequency signals.
In resonant circuits, understanding inductive reactance helps balance the equation with capacitive reactance, achieving the desired resonance and ensuring the efficient transfer of signals.
Capacitive Reactance
Capacitive reactance, denoted as \(X_C\), measures a capacitor's opposition to changes in voltage in an AC circuit. It is given by the equation \(X_C = \frac{1}{\omega C}\), where \(\omega\) is the angular frequency and \(C\) is the capacitance. Unlike inductive reactance, which increases with higher frequency, capacitive reactance decreases as frequency increases:
  • Capacitors resist changes in voltage by storing energy in an electric field.
  • This property is instrumental in filtering low frequencies, allowing high-frequency signals to pass.
In a resonant circuit, finding the right capacitive reactance to balance with inductive reactance is critical to achieving resonance. Through this balance, circuits can be optimized for specific performance criteria, such as minimizing impedance to maximize current flow.
Resonance Frequency
Resonance frequency is an essential concept in circuit design, referring to the frequency at which resonance occurs. In the L-R-C circuit, it is determined by the equation \(\omega_0 = \frac{1}{\sqrt{LC}}\), where \(L\) represents inductance and \(C\) represents capacitance. At this frequency:
  • The inductive and capacitive reactance equalize, allowing maximal current flow.
  • The impedance of the circuit is minimized, equating to the resistance \(R\).
The resonance frequency is akin to the natural frequency of an oscillator, where a system tends to oscillate with greater amplitude.
Recognizing and using resonance frequency is crucial in designing circuits for applications such as filtering, frequency tuning, and signal amplification, ensuring they operate at maximum efficiency.

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Most popular questions from this chapter

31.62. A series circuit consists of a \(1.50-\mathrm{mH}\) inductor, a \(125-\Omega\) resistor, and a \(25.0-\mathrm{nF}\) capacitor connected across an ac source having an rms voltage of 35.0 \(\mathrm{V}\) and variable frequency. (a) At what angular frequency will the current amplitude be equal to \(\frac{1}{3}\) of its maximum possible value? (b) At the frequency in part (a) what are the current amplitude and the voltage amplitude across each of the circuit elements (including the ac source)?

31.14. You have a \(200-\Omega\) resistor, a \(0.400-\mathrm{H}\) inductor, and a \(6.00-\mu \mathrm{F}\) capacitor. Suppose you take the resistor and inductor and make a series circuit with a voltage source that has voltage amplitude 30.0 \(\mathrm{V}\) and an angular frequency of 250 \(\mathrm{rad} / \mathrm{s}\) . (a) What is the impedance of the circuit? (b) What is the current amplitude? (c) What are the voltage amplitudes across the resistor and across the inductor? (d) What is the phase angle \(\phi\) of the source voltage with respect to the current? Does the source voltage lag or lead the conrent? (e) Construct the phasor diagram.

31.47. An \(L-R-C\) series circuit consists of a \(50.0-\Omega\) resistor, a \(10.0-\mu\) F capacitor, a \(3.50-\mathrm{mH}\) inductor, and an ac voltage source of voltage anplitude 60.0 \(\mathrm{V}\) operating at 1250 \(\mathrm{Hz}\) (a) Find the current amplitude and the voltage amplitudes across the inductor, the resistor, and the capacitor. Why can the voltage amplitudes add up to more than 60.0 \(\mathrm{V} ?\) (b) If the frequency is now doubled, but nothing else is changed, which of the quantities in part (a) will change? Find the new values for those that do change.

31.25. The power of a certain CD player operating at 120 \(\mathrm{V}\) rms is 20.0 \(\mathrm{W}\) . Assuming that the CD player behaves like a pure resistance, find (a) the maximum instantancous power; (b) the rms current; (c) the resistance of this player.

31.65. An inductor, a capacitor, and a resistor are all connected in series across an ac source. If the resistance, inductance, and capacitance are all doubled, by what factor does each of the following quantities change? Indicate whether they increase or decrease: (a) the resonance angular frequency; (b) the inductive reactance; (c) the capacitive reactance. (d) Does the impedance double?

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