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An ac series \(R-L-C\) circuit contains a \(120 \Omega\) resistor, a \(2.0 \mu \mathrm{F}\) capacitor, and a \(5.0 \mathrm{mH}\) inductor. Find (a) the resonance angular frequency and (b) the length of time that each cycle lasts at the resonance angular frequency.

Short Answer

Expert verified
(a) The resonance angular frequency is \( 10^{4} \text{ rad/s} \). (b) Each cycle lasts approximately \( 6.2832 \times 10^{-4} \text{ s} \).

Step by step solution

01

Identify the Given Values

We are provided with the values for the resistance \( R = 120 \ \Omega \), the capacitance \( C = 2.0 \ \mu F \), and the inductance \( L = 5.0 \ mH \). We need to find the resonance angular frequency and the time period per cycle at resonance.
02

Use the Formula for Resonance Angular Frequency

The formula for the resonance angular frequency \( \omega_0 \) in an \( R-L-C \) circuit is \( \omega_0 = \frac{1}{\sqrt{LC}} \).
03

Convert Units

Convert \( L = 5.0 \ \mathrm{mH} \) to \( \mathrm{H} \): \( 5.0 \ \mathrm{mH} = 5.0 \times 10^{-3} \ \mathrm{H} \). Convert \( C = 2.0 \ \mu F \) to \( \mathrm{F} \): \( 2.0 \ \mu F = 2.0 \times 10^{-6} \ \mathrm{F} \).
04

Calculate the Resonance Angular Frequency

Substitute the values into the formula: \[ \omega_0 = \frac{1}{\sqrt{(5.0 \times 10^{-3})(2.0 \times 10^{-6})}} \]Calculate: \[ \omega_0 = \frac{1}{\sqrt{10^{-8}}} = \frac{1}{10^{-4}} = 10^{4} \ \text{rad/s} \]
05

Calculate the Period of Each Cycle

The period \( T \) of a cycle is the reciprocal of the frequency. The frequency \( f \) is related to angular frequency \( \omega_0 \) by \( f = \frac{\omega_0}{2\pi} \). Thus, \( T = \frac{1}{f} = \frac{2\pi}{\omega_0} \).Substitute \( \omega_0 = 10^{4} \ \text{rad/s} \) into the period formula:\[ T = \frac{2\pi}{10^{4}} = \frac{2\pi}{10^{4}} \, \text{s} \]
06

Simplify the Period Expression

\[ T = \frac{2\times 3.1416}{10^{4}} \approx 6.2832 \times 10^{-4} \ \text{s} \]

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Key Concepts

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

Resonance Angular Frequency
In an RLC circuit, the resonance angular frequency is a key characteristic that defines the frequency at which the impedance of the circuit is minimized, and thus, the circuit can potentially store and transfer energy most efficiently. This frequency is important because, at resonance, the inductive reactance and the capacitive reactance are equal in magnitude but opposite in phase. Hence, they cancel each other's effects.
To calculate the resonance angular frequency \( \omega_0 \) in a series RLC circuit, we use the formula:
  • \( \omega_0 = \frac{1}{\sqrt{LC}} \)
where:
  • \( L \) is the inductance in henries (H)
  • \( C \) is the capacitance in farads (F)
By substituting the given values of \( L = 5.0 \times 10^{-3} \) H and \( C = 2.0 \times 10^{-6} \) F into the formula, we find the resonance angular frequency to be \( \omega_0 = 10^4 \text{ rad/s} \). This indicates the circuit's natural frequency of oscillation.
Inductive Reactance
Inductive reactance is a measure of the opposition that an inductor offers to the change in current in an AC circuit. It is denoted by \( X_L \) and is calculated using the formula:
  • \( X_L = \omega L \)
where:
  • \( \omega \) is the angular frequency in rad/s
  • \( L \) is the inductance in henries (H)
The unit for inductive reactance is ohms (\( \Omega \)).
Inductive reactance increases linearly with frequency; the higher the frequency, the greater the opposition to the AC current.
In the context of resonance, at the resonance angular frequency, the inductive reactance becomes exactly equal to the capacitive reactance, allowing for minimal impedance overall in the circuit.
Capacitive Reactance
Similar to inductive reactance, capacitive reactance quantifies how much a capacitor resists changes in voltage in an AC circuit. It is represented by \( X_C \) and can be calculated using:
  • \( X_C = \frac{1}{\omega C} \)
where:
  • \( \omega \) is the angular frequency in rad/s
  • \( C \) is the capacitance in farads (F)
Like inductive reactance, capacitive reactance is measured in ohms (\( \Omega \)). However, unlike inductive reactance, capacitive reactance decreases with increasing frequency. This inverse relationship means that, in an AC circuit with higher frequency, a capacitor will let more current through.
At the resonance angular frequency, the capacitive reactance balances out the inductive reactance, thus lowering the circuit's impedance to a minimum.
Alternating Current Circuits
Alternating Current (AC) circuits are those where the current flows back and forth periodically, as opposed to in a single direction like Direct Current (DC) circuits. These circuits are used extensively in power systems due to their efficiency in transmitting electrical power over long distances.
In AC circuits, components like resistors, inductors, and capacitors respond to changes in voltage and current differently compared to DC circuits. Each has its unique effect on the circuit's behavior, especially with respect to voltage and current.
  • Resistors: They provide consistent opposition to current, regardless of frequency.
  • Inductors: They oppose changes in current and their reactance increases with frequency.
  • Capacitors: They oppose changes in voltage and their reactance decreases with frequency.
Understanding these dynamics is crucial when analyzing circuits like the RLC circuit. The interplay between resistance, inductive reactance, and capacitive reactance determines overall circuit behavior, especially around the resonance point.

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Most popular questions from this chapter

In an \(R-L-C\) series circuit, \(R=300 \Omega, L=0.400 \mathrm{H},\) and \(C=6.00 \times 10^{-8} \mathrm{~F}\). When the ac source operates at the resonance frequency of the circuit, the current amplitude is 0.500 A. (a) What is the voltage amplitude of the source? (b) What is the amplitude of the voltage across the resistor, across the inductor, and across the capacitor? (c) What is the average power supplied by the source?

You have a special light bulb with a very delicate wire filament. The wire will break if the current in it ever exceeds \(1.50 \mathrm{~A},\) even for an instant. What is the largest root-mean-square current you can run through this bulb?

A series \(R-L-C\) circuit is connected to a \(120 \mathrm{~Hz}\) ac source that has \(V_{\mathrm{rms}}=80.0 \mathrm{~V}\). The circuit has a resistance of \(75.0 \Omega\) and an impedance of \(105 \Omega\) at this frequency. What average power is delivered to the circuit by the source?

(a) At what angular frequency will a \(5.00 \mu \mathrm{F}\) capacitor have the same reactance as a \(10.0 \mathrm{mH}\) inductor? (b) If the capacitor and inductor in part (a) are connected in an \(L-C\) circuit, what will be the resonance angular frequency of that circuit?

(a) Compute the reactance of a \(0.450 \mathrm{H}\) inductor at frequencies of \(60.0 \mathrm{~Hz}\) and \(600 \mathrm{~Hz}\). (b) Compute the reactance of a \(2.50 \mu \mathrm{F}\) capacitor at the same frequencies. (c) At what frequency is the reactance of a \(0.450 \mathrm{H}\) inductor equal to that of a \(2.50 \mu \mathrm{F}\) capacitor?

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