/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 An \(L-R-C\) series circuit is c... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An \(L-R-C\) series circuit is constructed using a \(175-\Omega\) resistor, a \(12.5-\mu \mathrm{F}\) capacitor, and an \(8.00-\mathrm{mH}\) inductor, all connected across an ac source having a variable frequency and a voltage amplitude of 25.0 \(\mathrm{V}\) (a) At what angular frequency will the impedance be smallest, and what is the impedance at this frequency? (b) At the angular frequency in part (a), what is the maximum current through the inductor? (c) At the angular frequency in part (a), find the potential difference across the ac source, the resistor, the capacitor, and the inductor at the instant that the current is equal to one-half its greatest positive value. (d) In part \((\mathrm{c}),\) how are the potential differences across the resistor, inductor, and capacitor related to the potential difference across the ac source?

Short Answer

Expert verified
(a) At \( \omega \approx 1000 \, \text{rad/s} \), impedance is 175 \( \Omega \). (b) Max current is 0.143 A. (c) Instant voltages: source 25 V, resistor 12.51 V, capacitor 5.72 V, inductor 5.72 V. (d) Source voltage equals resistor voltage due to cancellation of inductor and capacitor voltages.

Step by step solution

01

Identify conditions for minimum impedance

For the impedance of the circuit to be smallest, the circuit must be at resonance. In an L-R-C series circuit, resonance occurs when the inductive reactance equals the capacitive reactance: \(X_L = X_C\), where \(X_L = \omega L\) and \(X_C = \frac{1}{\omega C}\).
02

Calculate the resonant angular frequency

Setting \(X_L = X_C\), we have \(\omega L = \frac{1}{\omega C}\). Solving for \(\omega\), we get: \[ \omega = \frac{1}{\sqrt{LC}} \] Substituting \(L = 8.00 \times 10^{-3} \) H and \(C = 12.5 \times 10^{-6} \) F, \[ \omega = \frac{1}{\sqrt{8.00 \times 10^{-3} \cdot 12.5 \times 10^{-6}}} \approx 1000 \, \text{rad/s} \]
03

Calculate the impedance at resonance

At resonance, the impedance is purely resistive and equals the resistance \( R \). Therefore: \[ Z = R = 175 \, \Omega \]
04

Calculate maximum current at resonance

The maximum current \( I_{max} \) can be calculated using Ohm's law: \( I_{max} = \frac{V}{Z} \) where \( V = 25.0 \, \text{V} \) and \( Z = 175 \, \Omega \). \[ I_{max} = \frac{25.0}{175} \approx 0.143 \, \text{A} \]
05

Calculate potential difference across components when current is halved

At the instant when the current is half its maximum value, \( I = 0.0715 \, \text{A} \). The potential difference across each component is: - Resistor: \( V_R = IR = 0.0715 \times 175 = 12.51 \, \text{V} \)- Capacitor: \( V_C = \frac{I}{\omega C} = \frac{0.0715}{1000 \times 12.5 \times 10^{-6}} = 5.72 \, \text{V} \)- Inductor: \( V_L = I \omega L = 0.0715 \times 1000 \times 8.00 \times 10^{-3} = 5.72 \, \text{V} \)- Total potential: \( V_{source} = \sqrt{(V_R)^2 + (V_L - V_C)^2} \approx 25.0 \, \text{V} \)
06

Explain the relationship of potential differences

At resonance: - The potential difference across the source equals the potential difference across the resistor: \( V_{source} = V_R \).- The voltages across the inductor and capacitor are equal in magnitude but opposite in phase, thus canceling each other. Hence, \( V_{source}^2 = V_R^2 + (V_L - V_C)^2 \approx V_R^2\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Angular Frequency
In an RLC circuit, angular frequency is crucial for understanding resonance, where the system vibrates at its natural frequency. Angular frequency, denoted as \( \omega \), is defined as the rate of change of the phase of a sinusoidal waveform, with respect to time. It's measured in radians per second.
For the circuit to be at resonance, the inductive reactance \( X_L \) must be equal to the capacitive reactance \( X_C \). This means \( X_L = X_C \) or \( \omega L = \frac{1}{\omega C} \).
  • \( X_L = \omega L \) represents the opposition to the current in an inductor. It depends on the inductance \( L \) and angular frequency \( \omega \).
  • \( X_C = \frac{1}{\omega C} \) represents the opposition to the current in a capacitor. It depends on the capacitance \( C \) and angular frequency \( \omega \).
At resonance, the resonant angular frequency \( \omega_0 \) is essential because it minimizes the impedance of the circuit. It can be found using: \[ \omega = \frac{1}{\sqrt{LC}} \]This equation shows that the resonant frequency is determined solely by the inductance \( L \) and capacitance \( C \) of the circuit, leading to an optimal energy transfer.
Impedance
Impedance, denoted as \( Z \), is a measure of how much a circuit resists the flow of alternating current (AC). It combines the effects of resistance, inductive reactance, and capacitive reactance into one term. In an RLC circuit, impedance is key to understanding how the circuit will behave at different frequencies.
  • At any point, the total impedance can be calculated as: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
  • At resonance, the circuit has its minimum impedance because \( X_L = X_C \). This results in the inductive and capacitive reactances cancelling each other out. Hence, \( Z \) simplifies to the resistance \( R \), which in this case is 175 \( \Omega \).
  • This is significant because it indicates that the circuit will allow maximum current flow when the impedance is at its lowest.
Understanding impedance helps in designing circuits that function efficiently under particular frequency constraints, leading to improved performance of electrical devices.
Maximum Current
Maximum current in a resonant RLC circuit is achieved when impedance is at its minimum, meaning the circuit is purely resistive. The expression for maximum current, \( I_{max} \), relates directly to Ohm’s Law, which states:
  • \( I_{max} = \frac{V}{Z} \)
Where \( V \) is the voltage amplitude and \( Z \) is the impedance. At resonance, since \( Z = R \), this simplifies to:
  • \( I_{max} = \frac{25.0}{175} \approx 0.143 \text{ A} \)
This demonstrates that the circuit can achieve the greatest current through its purely resistive path under resonance, emphasizing the importance of resonant frequency. Furthermore, half of the maximum current yields potential differences across each component, enabling deeper insight into the circuit’s behavior at given frequencies. Understanding this concept is vital for optimizing the function of many practical electrical systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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

In an \(L-R-C\) series circuit, the rms voltage across the resistor is \(30.0 \mathrm{V},\) across the capacitor it is \(90.0 \mathrm{V},\) and across the inductor it is 50.0 \(\mathrm{V} .\) What is the rms voltage of the source?

In an \(L-R-C\) series circuit, \(R=150 \Omega, L=0.750 \mathrm{H},\) and \(C=0.0180 \mu \mathrm{F} .\) The source has voltage amplitude \(V=150 \mathrm{V}\) and a frequency equal to the resonance frequency of the circuit. (a) What is the power factor? (b) What is the average power delivered by the source? (c) The capacitor is replaced by one with \(C=\) 0.0360\(\mu \mathrm{F}\) and the source frequency is adjusted to the new resonance value. Then what is the average power delivered by the source?

(a) What is the reactance of a \(3.00-\mathrm{H}\) inductor at a frequency of 80.0 \(\mathrm{Hz}\) ? (b) What is the inductance of an inductor whose reactance is 120\(\Omega\) at 80.0 \(\mathrm{Hz}\) ? (c) What is the reactance of a \(4.00-\mu \mathrm{F}\) capacitor at a frequency of 80.0 \(\mathrm{Hz}\) ? (d) What is the capacitance of a capacitor whose reactance is 120\(\Omega\) at 80.0 \(\mathrm{Hz} ?\)

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?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.