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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?

Short Answer

Expert verified
(a) Ratio is 4; inductor's reactance is larger. (b) Ratio is 1/9; capacitor's reactance is larger. (c) Resonance frequency is \(\omega_1\).

Step by step solution

01

Understanding Initial Reactance Equality

We know that at frequency \( \omega_1 \), the reactance of the capacitor \( X_C \) equals that of the inductor \( X_L \). For a capacitor, \( X_C = \frac{1}{\omega_1 C} \), and for an inductor, \( X_L = \omega_1 L \). Thus, at \( \omega_1 \), \( \frac{1}{\omega_1 C} = \omega_1 L \). This simplifies to \( \omega_1^2 = \frac{1}{LC} \).
02

Reactance Ratio at Frequency \(\omega_2 = 2\omega_1\)

The reactance of the capacitor at \( \omega_2 \) is \( X_C = \frac{1}{\omega_2 C} = \frac{1}{2\omega_1 C} \), and that of the inductor is \( X_L = \omega_2 L = 2\omega_1 L \). The ratio of the inductor to capacitor reactance is \( \frac{X_L}{X_C} = \frac{2\omega_1 L}{\frac{1}{2\omega_1 C}} = 4\omega_1^2 LC = 4 \). The inductor reactance is larger.
03

Reactance Ratio at Frequency \(\omega_3 = \frac{\omega_1}{3}\)

The reactance of the capacitor at \( \omega_3 \) is \( X_C = \frac{1}{\omega_3 C} = 3 \omega_1 C \), and that of the inductor is \( X_L = \omega_3 L = \frac{\omega_1 L}{3} \). The ratio of the inductor to capacitor reactance is \( \frac{X_L}{X_C} = \frac{\frac{\omega_1 L}{3}}{3\omega_1 C} = \frac{1}{9} \). The capacitor reactance is larger.
04

Finding Resonance Frequency for L-R-C Circuit

For an L-R-C series circuit, the resonance frequency occurs where the reactances cancel out, i.e., \( \omega^2 = \frac{1}{LC} \). This gives the resonant angular frequency as \( \omega_0 = \omega_1 \), since \( \omega_1^2 = \frac{1}{LC} \) was established in step 1.

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Key Concepts

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

Capacitor Reactance
Capacitor reactance, represented as \( X_C \), is the opposition offered by a capacitor to the flow of alternating current (AC) in a circuit. It is inversely proportional to the frequency of the AC signal and the capacitance of the capacitor. This relationship is described by the formula: \[ X_C = \frac{1}{\omega C} \]Where:
  • \( X_C \) is the capacitor reactance.
  • \( \omega \) is the angular frequency of the AC signal.
  • \( C \) is the capacitance.
The inversely proportional relationship means that as the frequency of the AC signal increases, the reactance decreases, allowing more current to flow through the capacitor. Conversely, a decrease in frequency leads to an increase in reactance.
This concept is crucial in determining how capacitors behave in different circuit conditions.
Inductor Reactance
Inductor reactance, denoted as \( X_L \), is the opposition that an inductor presents to the flow of AC current. It is directly proportional to both the inductance of the inductor and the frequency of the AC signal. The formula for inductor reactance is:\[ X_L = \omega L \]Where:
  • \( X_L \) is the inductor reactance.
  • \( \omega \) is the angular frequency.
  • \( L \) is the inductance.
As the frequency increases, the inductor will react more strongly by increasing its reactance, thus opposing the current. This characteristic is used in designing circuits that suppress high-frequency noise or allow certain frequency ranges to pass.
L-R-C Circuit
An L-R-C circuit is a series or parallel circuit composed of an inductor \( L \), a resistor \( R \), and a capacitor \( C \). This setup is quintessential in AC circuit analysis due to its ability to exhibit resonance. In resonance, the circuit can oscillate at a particular frequency, known as the resonant frequency.The resonance in an L-R-C series circuit occurs when the total reactance is zero, meaning the inductive and capacitive reactances cancel each other out:\[ \omega^2 = \frac{1}{LC} \]At the resonant frequency, the impedance of the circuit is purely resistive, and it allows maximum current to flow through. This property makes L-R-C circuits useful in applications like radio tuning and frequency filters.
Frequency Ratio
The frequency ratio in the context of reactance refers to how changes in frequency affect the reactance of capacitors and inductors. When the frequency doubles (\( \omega_2 = 2\omega_1 \)) or is reduced to a third (\( \omega_3 = \frac{\omega_1}{3} \)), these ratios impact the comparative reactance between capacitors and inductors.
  • At double the frequency: The reactance of a capacitor decreases, and the inductor's reactance increases, making the inductor reactance larger.
  • At one-third of the frequency: The opposite occurs, where the capacitor reactance becomes larger as its reactance increases with lowering frequency.
Understanding frequency ratios helps in predicting and tuning the behavior of circuits in varying frequency environments.

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Most popular questions from this chapter

You want to double the resonance angular frequency of an \(L-R-C\) series circuit by changing only the pertinent circuit elements all by the same factor. (a) Which ones should you change? (b) By what factor should you change them?

In an \(L-R-C\) series circuit the source is operated at its resonant angular frequency. At this frequency, the reactance \(X_{C}\) of the capacitor is \(200 \Omega\) and the voltage amplitude across the capacitor is \(600 \mathrm{~V}\). The circuit has \(R=300 \Omega\). What is the voltage amplitude of the source?

An A.C. source of angular frequency \(\omega\) is fed across a resistor \(R\) and a capacitor \(C\) in series, The current registered is \(I .\) If now the frequency of source is changed to \(\omega / 3\) (but maintanining the same voltage), the current in the circuit is found to be halved. Calculate the ratio ofreactance to resistance at the original frequency \(\omega\).

An inductor with \(L=10 \mathrm{mH}\) is connected across an ac source that has voltage amplitude \(44 \mathrm{~V}\). (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 \(4 \mathrm{~A}\) ?

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 quanti- 1 ties change? Indicate whether they increase or decrease: (a) the resonance angular frequency; (b) the inductive reactance; \((\mathrm{c})\) the capacitive reactance. (d) Does the impedance double?

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