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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.

Short Answer

Expert verified
The maximum current and power occur at the resonance frequency, \( \omega = \frac{1}{\sqrt{LC}} \).

Step by step solution

01

Understand the Circuit

An L-R-C series circuit includes a resistor (R), an inductor (L), and a capacitor (C) in series connected to an AC source. The voltage amplitude across the circuit changes with angular frequency \( \omega \).
02

Calculate Current Amplitude Formula (Part a)

The impedance (Z) in an L-R-C series circuit is given by \( Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2} \). The current amplitude \( I \) is given by the formula \( I = \frac{V}{Z} \). Substituting for \( Z \), we have \( I = \frac{V}{\sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}} \).
03

Derive Average Power Formula (Part b)

The average power dissipated in the resistor is determined by the formula \( P = \frac{1}{2} I^2 R \). Using the expression for current amplitude from part (a), \( I = \frac{V}{\sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}} \), we have \( I^2 = \frac{V^2}{R^2 + (\omega L - \frac{1}{\omega C})^2} \). Thus, \( P = \frac{1}{2} \frac{V^2 R}{R^2 + (\omega L - \frac{1}{\omega C})^2} \).
04

Identify Resonance Frequency (Part c)

The resonance frequency for an L-R-C series circuit occurs when the reactance of the inductor equals the reactance of the capacitor, i.e., \( \omega L = \frac{1}{\omega C} \). Solving for \( \omega \), we have \( \omega = \frac{1}{\sqrt{LC}} \).
05

Prove Maximum Current and Power at Resonance

At resonance, \( \omega L = \frac{1}{\omega C} \) results in the impedance \( Z \) becoming \( R \). Thus, the current amplitude formula becomes \( I = \frac{V}{R} \), which is the maximum possible current for the circuit. Substituting \( \omega = \frac{1}{\sqrt{LC}} \) in the power formula \( P = \frac{1}{2} \frac{V^2 R}{R^2 + (\omega L - \frac{1}{\omega C})^2} \) gives maximum power, as the term depending on \( \omega \) becomes zero.

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Key Concepts

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

Resonance Frequency
The resonance frequency is a key concept in L-R-C circuit analysis. It marks the frequency at which the circuit's impedance is at its minimum, often making it a pivotal point of study. Resonance occurs when the inductive reactance (\(\omega L\)) equals the capacitive reactance (\(1/\omega C\)). This balance cancels out the reactive parts of the impedance, making only the resistance \(R\) contribute significantly to the total impedance. At resonance, the angular frequency, denoted as \(\omega\), is derived from the formula:

\[\omega = \frac{1}{\sqrt{LC}}\]
This is known as the natural frequency of the circuit. At this frequency, the circuit behaves as if it contains only the resistor, reducing power losses related to reactance. Circuits are often designed to operate at their resonance frequency, where they achieve maximum power efficiency.
Current Amplitude
The current amplitude in an \(L-R-C\) series circuit indicates the peak level of current flow when the circuit is connected to an AC source. This current amplitude is affected by the circuit's impedance, which is a combination of resistance, inductive reactance, and capacitive reactance:

  • Inductive Reactance, \(\omega L\)
  • Capacitive Reactance, \(1/\omega C\)

The formula for current amplitude \(I\) is:

\[I = \frac{V}{\sqrt{R^2 + (\omega L - 1/\omega C)^2}}\]
In this equation, \(V\) is the voltage amplitude. Notice how the current amplitude reaches its maximum at resonance frequency, where the impedance is reduced solely to \(R\). This is because the opposing effects of the inductance and capacitance cancel each other out, leaving less resistance to the flow of current.
Average Power Dissipation
Average power dissipation is a crucial aspect of circuit performance, especially in an \(L-R-C\) circuit. Power dissipation refers to the average rate at which power is lost as heat in the circuit, typically through the resistor. The formula used to determine this power is:

\[P = \frac{V^2 R / 2}{R^2 + (\omega L - 1/\omega C)^2}\]
This equation highlights that maximum power dissipation occurs at the circuit's resonance frequency. At this frequency, \(\omega L = 1/\omega C\), significantly simplifying the denominator. This simplification results in most of the input voltage dropping across the resistor, thereby converting most of the energy to heat efficiently. Understanding power dissipation helps in designing circuits that can withstand energy losses effectively, thereby prolonging their operation time and efficiency.
Impedance in Circuits
Impedance is the measure of the opposition that a circuit presents to the passage of AC current. It is a combination of resistance and reactance (from both inductors and capacitors) and determines both the current and phase differences in AC circuits. For an \(L-R-C\) series circuit, the impedance \(Z\) can be quantified as:

\[Z = \sqrt{R^2 + (\omega L - 1/\omega C)^2}\]
This expression underscores the impact that the inductor and capacitor have on the overall opposition to current flow. While resistance \(R\) is constant, the inductive and capacitive reactances change with the frequency of the source voltage. This frequency dependence affects not only the amplitude of current but also the efficiency of power transfer in the circuit. Whenever you're analyzing or designing an AC circuit, calculating the impedance accurately is essential for optimizing its performance and ensuring it operates effectively at desired frequency ranges.

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Most popular questions from this chapter

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?

A resistor, inductor, and capacitor are connected in parallel to an ac source with voltage amplitude \(V\) and angular frequency \(\omega .\) Let the source voltage be given by \(v=V \cos \omega t .\) (a) Show that the instantaneous voltages \(v_{R}, v_{L}\), and \(v_{C}\) at any instant are each equal to \(v\) and that \(i=i_{R}+i_{L}+i_{C}\), where \(i\) is the current through the source and \(i_{R}\) \(i_{L}\), and \(i_{C}\) are the currents through the resistor, the inductor, and the capacitor, respectively. (b) What are the phases of \(i_{R}, i_{L}\), and \(i_{C}\) with respect to \(v ?\) Use current phasors to represent \(i, i_{R}, i_{L}\), and \(i_{C}\) In a phasor diagram, show the phases of these four currents with respect to \(\mathrm{V}\). (c) Use the phasor diagram of part (b) to show that the current amplitude I for the current i through the source is given by \(I=\sqrt{I_{R}^{2}+\left(I_{C}-I_{L}\right)^{2}}\). (d) Show that the result of part (c) can be written as \(I=V / Z\), with \(1 / Z=\sqrt{1 / R^{2}+(\omega C-1 / \omega L)^{2}}\).

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?

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?

In an \(L-R-C\) series circuit, the source has a voltage amplitude of \(120 \mathrm{~V}, R=80.0 \Omega\), and the reactance of the capacitor is \(480 \Omega .\) The voltage amplitude across the capacitor is \(360 \mathrm{~V}\). (a) What is the current amplitude in the circuit? (b) What is the impedance? (c) What two values can the reactance of the inductor have? (d) For which of the two values found in part (c) is the angular frequency less than the resonance angular frequency? Explain.

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