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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?

Short Answer

Expert verified
Change both inductance \(L\) and capacitance \(C\) by a factor of \(\frac{1}{4}\).

Step by step solution

01

Understand the Resonance Angular Frequency

The resonance angular frequency \(\omega_0\) for an \(L-R-C\) series circuit is given by the formula: \[ \omega_0 = \frac{1}{\sqrt{LC}} \] where \(L\) is inductance and \(C\) is capacitance. Notice that \(R\) does not appear in the formula, indicating it doesn't affect the resonance angular frequency.
02

Identify the Pertinent Circuit Elements

Since resistance \(R\) does not influence \(\omega_0\), the pertinent elements here are \(L\) and \(C\). To double \(\omega_0\), we need to modify \(L\) and/or \(C\).
03

Derive the Required Change Factor

If we want to double \(\omega_0\), we set the new frequency \(\omega_0'\) as: \[ \omega_0' = 2\times\omega_0 = \frac{1}{\sqrt{L'C'}} \] We can equate and solve: \[ 2\sqrt{LC} = \frac{1}{\sqrt{L'C'}} \] Thus, we find that \(L'C' = \frac{1}{4}LC\). Therefore, both \(L\) and \(C\) should be changed by a factor of \(\frac{1}{4}\).
04

Conclusion

To double the resonance angular frequency, change the inductance \(L\) and the capacitance \(C\) by a factor of \(\frac{1}{4}\). Resistance \(R\) is not changed.

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Key Concepts

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

L-R-C Circuit
In the world of electronics, an L-R-C circuit, also known as a series RLC circuit, is a common configuration that consists of an inductor (L), a resistor (R), and a capacitor (C) connected in series. Each of these components plays a unique role within the circuit:
  • **Inductor (L)**: Stores energy in a magnetic field when current flows through it.
  • **Resistor (R)**: Controls the flow of current and dissipates energy as heat.
  • **Capacitor (C)**: Stores energy in an electric field, opposing voltage changes.
The behavior of an L-R-C circuit is characterized particularly by its ability to resonate at a specific angular frequency, known as the resonance angular frequency. At this frequency, the effects of the inductor and capacitor cancel each other out to the greatest extent possible, allowing the circuit to oscillate with minimal energy loss. Understanding this resonance is essential for manipulating circuit behaviors, such as achieving desired frequencies for radio transmissions or filtering specific signal frequencies.
Inductance
Inductance, represented by the letter 'L', is a property of an electrical conductor by which a change in current induces an electromotive force, either in the conductor itself or in nearby conductors. Essentially, it is the measure of an inductor's ability to store energy in a magnetic field.The unit of inductance is the henry (H), and the fundamental relationship governing inductance is given by:\[ V = L \frac{dI}{dt} \]where:
  • **V** is the induced voltage,
  • **L** is the inductance,
  • **dI/dt** is the rate of change of current through the inductor.
In the context of an L-R-C circuit, the inductance directly affects the circuit's resonance angular frequency. By changing the inductance, you can control the frequency at which the circuit naturally oscillates. To double the resonance frequency, the inductance should be changed by a factor determined through the specific needs of the circuit's design goals.
Capacitance
Capacitance, denoted as 'C', is the ability of a component or circuit to store an electric charge. It is measured in farads (F). The primary function of a capacitor is to store energy in an electric field, releasing it as needed to maintain the voltage across the circuit components.The formula relating charge and capacitance is:\[ Q = CV \]where:
  • **Q** is the electric charge,
  • **C** is the capacitance,
  • **V** is the voltage across the capacitor.
In an L-R-C circuit, the capacitance influences the resonance angular frequency along with the inductance. To achieve specific frequency goals, such as doubling the original resonance frequency, adjusting the capacitance by the exact same calculated factor as the inductance ensures the circuit's performance is maintained or improved according to design requirements.
Frequency Doubling
Frequency doubling in an electrical circuit refers to the process of increasing the circuit's resonance angular frequency by a factor of two. In an L-R-C circuit, this involves adjusting specific components to ensure the desired frequency change.The resonance frequency \( \omega_0 \) is expressed as:\[ \omega_0 = \frac{1}{\sqrt{LC}} \]To double this frequency, the relationship is altered by changing both the inductance (L) and capacitance (C) equally by a factor. In this scenario:\[ L' = \frac{L}{4} \quad \text{and} \quad C' = \frac{C}{4} \]This proportional change in both L and C reduces the product \( LC \) by a factor of 4, effectively doubling the original resonance frequency. This method ensures that the frequency goal is achieved uniformly across the circuit without affecting other circuit characteristics such as total impedance or energy storage capabilities.

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Most popular questions from this chapter

An \(L-R-C\) series circuit has \(C=5 \mu \mathrm{F}, L=0.8 \mathrm{H}\),and source voltage amplitude \(V-56.0 \mathrm{~V} .\) The source is operated at the resonance frequency of the circuit. If the voltage across the capacitor has amplitude \(80.0 \mathrm{~V}\), what is the value of \(R\) for the resistor in the circuit?

A \(250-\Omega\) resistor is connected in series with a \(5-\mu \mathrm{F}\) capacitor and an ac source. The voltage across the capacitor is \(v_{C}=(7.5 \mathrm{~V}) \sin [(120 \mathrm{rad} / \mathrm{s}) t]\) (a) Determine the capacitive reactance of the capacitor. (b) Derive an expression for the voltage \(\mathrm{V}_{\mathrm{R}}\) across the resistor.

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.

A series \(\mathrm{RC}\) circuit with \(R=10 \Omega\) has an impendance with phase angle of \(45^{\circ}\) at frequency \(f=500 \mathrm{~Hz}\). Find the frequency for which the magnitude of the impedance is (a) twice that off, (b) one-half that at \(f\).

An LCR series circuit with \(100 \Omega\) resistance is connected to an \(A C\) source of \(200 V\) and angular frequency 300 radians/sec. When only the capacitance is removed The current lags behind the voltage by \(60^{\circ} .\) When only the inductance is removed the current leads the voltage \(60^{\circ}\). Calculate the current and the power dissipated in the LCR circuit.

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