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An \(L-R-C\) circuit has \(L=0.450 \mathrm{H}, C=2.50 \times 10^{-5} \mathrm{F}\) and resistance \(R\) (a) What is the angular frequency of the circuit when \(R=0 ?\) (b) What value must \(R\) have to give a 5.0\(\%\) decrease in angular frequency compared to the value calculated in part (a)?

Short Answer

Expert verified
(a) 298 rad/s, (b) 84.05 ohms.

Step by step solution

01

Understanding Resonant Angular Frequency

The resonant angular frequency of an LC circuit with no resistance is given by the formula: \( \omega_0 = \frac{1}{\sqrt{LC}} \). We will use this formula to find the angular frequency when the resistance is zero.
02

Calculate Angular Frequency for R=0

Plug the values of \(L\) and \(C\) into the formula for \(\omega_0\): \[ \omega_0 = \frac{1}{\sqrt{0.450 \times 2.50 \times 10^{-5}}} \] Calculating the expression under the square root: \[ \omega_0 = \frac{1}{\sqrt{1.125 \times 10^{-5}}} = \frac{1}{3.354 \times 10^{-3}} \] Therefore, \(\omega_0 \approx 298 \text{ rad/s}\).
03

Understand Effect of Resistance on Resonant Angular Frequency

The presence of resistance decreases the angular frequency of the circuit. The altered angular frequency is given by: \( \omega = \sqrt{\frac{1}{LC} - \left( \frac{R}{2L} \right)^2} \). We want this \(\omega\) to be 5% less than \(\omega_0\).
04

Calculate Decreased Angular Frequency

5% less than \(\omega_0\) equals \(0.95 \cdot \omega_0\). So calculate: \[ \omega = 0.95 \cdot 298 = 283.1 \text{ rad/s} \]. This is the angular frequency of the circuit we want to achieve using a resistance R.
05

Solve for R

Using the equation \( \omega = \sqrt{\frac{1}{LC} - \left( \frac{R}{2L} \right)^2} \) with \(\omega = 283.1\): \[ 283.1 = \sqrt{\frac{1}{0.450 \times 2.50 \times 10^{-5}} - \left( \frac{R}{2 \times 0.450} \right)^2} \]. Square both sides to solve for \(R\): \[ 283.1^2 = \frac{1}{1.125 \times 10^{-5}} - \left( \frac{R}{0.9} \right)^2 \].
06

Calculate the Square of Both Sides

Square \(283.1\) to calculate: \[ 80169.61 = \frac{1}{1.125 \times 10^{-5}} - \frac{R^2}{0.81} \]. Now express \(\frac{1}{1.125 \times 10^{-5}}\) which equals approximately 88888.89. Set the equation to find \(R^2\): \[ 80169.61 = 88888.89 - \frac{R^2}{0.81} \].
07

Solve for R

Rearrange the equation to solve for \(R^2\): \[ \frac{R^2}{0.81} = 88888.89 - 80169.61 \]. Therefore, \(\frac{R^2}{0.81} = 8719.28\). Multiply both sides by 0.81 to find \(R^2\): \[ R^2 = 7062.6368 \]. Taking the square root of both sides gives \(R\): \[ R = \sqrt{7062.6368} \approx 84.05 \text{ ohms}\].

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Key Concepts

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

Resonant Angular Frequency
The concept of resonant angular frequency is fundamental in the study of L-R-C circuits. It is the frequency at which the circuit oscillates naturally when there is no resistance present. This phenomena occurs due to the energy exchange between the inductor (L) and capacitor (C). The resonant angular frequency, denoted as \( \omega_0 \), is calculated using the formula: \( \omega_0 = \frac{1}{\sqrt{LC}} \).
This formula emerges from the balance between the inductive and capacitive reactances, which cancel out at resonance, thus allowing the circuit to oscillate freely without any damping effect from resistance. For a circuit with given values of inductance \( L = 0.450 \) henries and capacitance \( C = 2.50 \times 10^{-5} \) farads, the calculation yields a resonant angular frequency of approximately 298 rad/s.
  • This is significant because at this frequency, the circuit ideally allows maximum energy transfer.
  • No energy is wasted in resistive components, and thus you get the pure oscillatory behavior.
Understanding resonant angular frequency helps predict how an L-R-C circuit will behave under different conditions. It gives insights into tuning radios and other devices that rely on selective frequency vibration.
Effective Resistance
In practical L-R-C circuits, resistance is present and affects circuit behavior significantly. Effective resistance, denoted by R, hinders the free oscillation by introducing energy loss in the form of heat. This energy loss due to resistance is a key factor in reducing the resonant angular frequency of the circuit.
The impact of resistance on resonant angular frequency is represented by modifying the frequency equation to:\[ \omega = \sqrt{\frac{1}{LC} - \left( \frac{R}{2L} \right)^2} \].
This equation shows how resistance affects circuit performance:
  • A higher resistance value results in a larger reduction in frequency due to increased damping.
  • In the solved problem, altering the resistance causes a decrease in angular frequency to 283.1 rad/s, which is 5% below the initial resonance frequency.
Resistance not only affects frequency but can cause the circuit to stop oscillating if it becomes too high, pushing the system into a "critically damped" or "overdamped" state where oscillation ceases.
Circuit Impedance
Circuit impedance is a comprehensive measure of opposition that a circuit offers to the flow of alternating current (AC). While resistance applies to both DC and AC currents, impedance is special because it takes both resistance and the circuit's reactance into account.
In the context of an L-R-C circuit, impedance (Z) affects how much current the circuit draws at specific frequencies. At resonance, impedance is primarily resistive because the reactive parts cancel each other out. However, when not at resonance:
  • The inductive and capacitive reactance contribute, making the impedance a complex quantity.
  • The total impedance can be calculated using: \( Z = \sqrt{R^2 + (X_L - X_C)^2} \), where \( X_L = \omega L \) and \( X_C = \frac{1}{\omega C} \).
By understanding impedance, you can better predict circuit performance over a range of frequencies, ensuring efficient design and functioning of electronic devices that rely on resonance, like filters and radio transmitters. Hence, analyzing impedance helps in selecting components to achieve desired frequency responses and energy efficiency.

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Most popular questions from this chapter

An inductor with an inductance of 2.50 \(\mathrm{H}\) and a resistance of 8.00\(\Omega\) is connected to the terminals of a battery with an emf of 6.00 \(\mathrm{V}\) and negligible intermal resistance. Find (a) the initial rate of increase of current in the circuit; (b) the rate of increase of current at the instant when the current is \(0.500 \mathrm{A} ;(\mathrm{c})\) the current 0.250 \(\mathrm{s}\) after the circuit is closed; (d) the final steady-state current.

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