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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. (d) Graph \(P\) as a function of \(\omega\) for \(V=100 \mathrm{V}, R=200 \Omega, L=2.0 \mathrm{H},\) and \(C=0.50 \mu \mathrm{F}\) . Compare to the light purple curve in Fig. \(31.19 .\) Discuss the behavior of \(I\) and \(P\) in the limits \(\omega=0\) and \(\omega \rightarrow \infty\) .

Short Answer

Expert verified
The current amplitude is maximum at resonance \(\omega = \frac{1}{\sqrt{LC}}\), where power is also maximized. Impedance increases at \(\omega = 0\) and \(\omega \to \infty\).

Step by step solution

01

Setting Up the Circuit Equations

In an L-R-C series circuit, the impedance is given by the formula:\[ Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2} \].The current amplitude through the circuit can be determined using Ohm's Law for AC circuits, which states that the current amplitude \( I \) is the voltage amplitude \( V \) divided by the impedance \( Z \).
02

Deriving the Current Amplitude

Substitute the expression for impedance into Ohm’s Law to find the current amplitude:\[I = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}}. \]
03

Finding Average Power Dissipation

The average power dissipated in the resistor is given by the formula:\[ P = \frac{1}{2} I^2 R. \]Substitute the expression for \( I \) from the previous step:\[ P = \frac{1}{2} \left( \frac{V}{\sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}} \right)^2 R.\]Simplify to obtain:\[ P = \frac{V^2 R / 2}{R^2 + (\omega L - \frac{1}{\omega C})^2}. \]
04

Maximizing the Current and Power

The expression \((\omega L - \frac{1}{\omega C})\) becomes zero when \(\omega L = \frac{1}{\omega C}\), leading to \(\omega = \frac{1}{\sqrt{LC}}\).At this resonance frequency, both the impedance and thus current amplitude \(I\), and the power \(P\) dissipated become maximized, confirming resonant conditions.
05

Analyzing Behavior at Limits

For \(\omega = 0\), the circuit becomes inductive, and the impedance tends toward infinity, leading to minimal current and power. As \(\omega \to \infty\), the circuit becomes capacitive, and again the impedance increases, reducing current and power. This illustrates the dependence on frequency noted earlier.

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Key Concepts

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

Current Amplitude in RLC Circuits
The current amplitude, often represented by the symbol \( I \), is an important concept when analyzing RLC circuits. In AC circuits, current amplitude refers to the peak current value that flows through the circuit. For an RLC circuit connected to an AC source, the current amplitude can be determined by Ohm's Law, adapted for AC circuits. Ohm's Law states that \( I = \frac{V}{Z} \), where \( V \) is the voltage amplitude and \( Z \) is the impedance of the circuit.

Impedance in an RLC circuit is given by the expression \( Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2} \). By substituting this into Ohm's Law, we find that the current amplitude in an RLC circuit is:

\[ I = \frac{V}{\sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}} \]

This equation shows how the current amplitude varies with the angular frequency \( \omega \) of the AC source. The impedance affects the current amplitude, causing it to change as the frequency changes. The current amplitude is maximal at resonance frequency when impedance is minimized.
Average Power Dissipation
In electrical circuits, power dissipation refers to the conversion of electrical energy into heat. This occurs mainly in resistors, as they impede the flow of current. The average power dissipated in a resistor within an AC circuit can be calculated using the following formula:

\[ P = \frac{1}{2} I^2 R \]

For RLC circuits, substituting the expression for current amplitude into this formula gives us an equation for the average power dissipated:
  • Start from: \( I = \frac{V}{\sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}} \)
  • Substitute into power formula: \( P = \frac{1}{2} \left( \frac{V}{\sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}} \right)^2 R \)
  • Simplify to: \( P = \frac{V^2 R / 2}{R^2 + (\omega L - \frac{1}{\omega C})^2} \)

Average power dissipation is crucial for understanding energy losses in a circuit, which affect efficiency. At resonance frequency, this power reaches its maximum because the circuit optimally delivers energy through the resistor.
Resonance Frequency in RLC Circuits
The resonance frequency is a key feature in RLC circuits. It is the frequency at which the impedance of the circuit is at its minimum, allowing maximum current flow and power dissipation. This occurs because the inductive and capacitive reactances cancel each other out. At resonance, the angular frequency \( \omega \) is given by:

\[ \omega = \frac{1}{\sqrt{LC}} \]

When \( \omega \) equals this resonance frequency, the terms \( \omega L \) and \( \frac{1}{\omega C} \) are equal, and thus cancel out in the expression for impedance. Therefore, the impedance is reduced to just the resistance \( R \), simplifying greatly:
  • Resonance condition: \( Z = R \)
  • Minimum impedance leads to maximal current: \( I_{\text{max}} = \frac{V}{R} \)
  • Maximal power dissipation: \( P_{\text{max}} = \frac{V^2}{2R} \)

Understanding resonance is essential for designing circuits that require precise frequency responses, such as filters and oscillators.
Impedance in AC Circuits
Impedance is a fundamental concept in AC circuit analysis. Unlike resistance, which is constant, impedance in AC circuits varies with frequency. It combines both resistive elements and reactance, which is the opposition offered by capacitors and inductors to alternating current.

For an RLC circuit, impedance is expressed as:

\[ Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2} \]

Here, \( R \) is the resistance, \( \omega L \) represents the inductive reactance, and \( \frac{1}{\omega C} \) represents the capacitive reactance. The interplay between these components is crucial:
  • Inductive reactance increases with frequency \( \omega \), making the circuit behave like an inductor at higher frequencies.
  • Capacitive reactance decreases with increasing frequency, making the circuit behave more like a capacitor at lower frequencies.
  • Impedance is minimized at the resonance frequency, where the reactances cancel.

This property of impedance determines how AC components interact and affects the current flow and power dissipation in the circuit, making it a vital aspect of AC circuit design and analysis.

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Most popular questions from this chapter

In an \(L-R-C\) series circuit, \(L=0.280 \mathrm{H}\) and \(C=\) 4.00\(\mu \mathrm{F}\) . The voltage amplitude of the source is 120 \(\mathrm{V}\) . (a) What is the resonance angular frequency of the circuit? (b) When the source operates at the resonance angular frequency, the current amplitude in the circuit is 1.70 A. What is the resistance \(R\) of the resistor? (c) At the resonance angular frequency, what are the peak voltages across the inductor, the capacitor, and the resistor?

In an \(L-R-C\) series circuit, the components have the following values: \(L=20.0 \mathrm{mH}, C=140 \mathrm{nF},\) and \(R=350 \Omega .\) The generator has an rms voltage of 120 \(\mathrm{V}\) and a frequency of 1.25 \(\mathrm{kHz}\) . Determine (a) the power supplied by the generator and (b) the power dissipated in the resistor.

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 current? (e) Construct the phasor diagram.

A Step-Down Transformer. A transformer connected to a \(120-\mathrm{V}(\mathrm{rms})\) ac line is to supply 12.0 \(\mathrm{V}(\mathrm{rms})\) to a portable electronic device. The load resistance in the secondary is 5.00\(\Omega .\) (a) What should the ratio of primary to secondary turns of the transformer be? (b) What rms current must the secondary supply? (c) What average power is delivered to the load? (d) What resistance connected directly across the \(120-\mathrm{V}\) line would draw the same power as the transformer? Show that this is equal to 5.00\(\Omega\) times the square of the ratio of primary to secondary turns.

A resistor with \(R=300 \Omega\) and an inductor are connected in series across an ac source that has voltage amplitude 500 \(\mathrm{V}\) . The rate at which electrical energy is dissipated in the resistor is 216 \(\mathrm{W}\) (a) What is the impedance \(Z\) of the circuit? (b) What is the amplitude of the voltage across the inductor? (c) What is the power factor?

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