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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.

Short Answer

Expert verified
(a) 0.75 A; (b) 160 Ω; (c) 618.56 Ω, 341.44 Ω; (d) 341.44 Ω yields a smaller angular frequency.

Step by step solution

01

Calculate Current Amplitude

Given the voltage across the capacitor (\( V_C = 360 \, \text{V} \)) and the capacitive reactance (\( X_C = 480 \, \Omega \)), use the relation \( V_C = I X_C \) to solve for the current amplitude (\( I \)):\[ I = \frac{V_C}{X_C} = \frac{360 \, \text{V}}{480 \, \Omega} = 0.75 \, \text{A} \]
02

Calculate the Impedance

The impedance (\( Z \)) of an L-R-C circuit is calculated using \( Z = \sqrt{R^2 + (X_L - X_C)^2} \), where \( X_L \) is the inductive reactance. First, solve for \( Z \) using the current amplitude (\( I \)) formula \( V = IZ \) where \( V \) is the voltage amplitude:\[ 120 = 0.75 Z \]\[ Z = \frac{120}{0.75} = 160 \, \Omega \]
03

Find Reactance of the Inductor

To find \( X_L \), use \( Z = \sqrt{R^2 + (X_L - X_C)^2} \) as follows:\[ 160 = \sqrt{80^2 + (X_L - 480)^2} \]Square both sides and solve for \( X_L \):\[ 160^2 = 80^2 + (X_L - 480)^2 \]\[ 25600 = 6400 + (X_L - 480)^2 \]\[ 19200 = (X_L - 480)^2 \]\[ \sqrt{19200} = |X_L - 480| \]\[ \approx 138.56 = |X_L - 480| \]Thus, the two possible values are \( X_L = 480 + 138.56 = 618.56 \, \Omega \) and \( X_L = 480 - 138.56 = 341.44 \, \Omega \).
04

Determine Angular Frequency's Relationship to Resonance

Resonance occurs when \( X_L = X_C \). For the calculated values, resonance would be closer to \( 480 \, \Omega \).The resonance angular frequency (\( \omega_0 \)) is \( \omega_0 = \frac{1}{\sqrt{LC}} \). A smaller \( X_L \) compared to \( X_C \) implies a smaller angular frequency according to \( X_L = \omega L \).Since \( 341.44 \, \Omega < 480 \, \Omega \), it indicates that the angular frequency is lower for \( X_L = 341.44 \, \Omega \).

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Key Concepts

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

Impedance
In an L-R-C series circuit, impedance is a crucial factor that represents the circuit’s opposition to the flow of alternating current (AC). Unlike simple resistance, impedance combines resistance (R) and reactance (X) in a complex manner. It is expressed as a single value and calculated using the relation:
  • \( Z = \sqrt{R^2 + (X_L - X_C)^2} \)
Where:
  • \( R \) is the resistance
  • \( X_L \) is the inductive reactance
  • \( X_C \) is the capacitive reactance
The impedance is given in ohms (\( \Omega \)) and affects how easily current flows through the circuit.
In the example provided, the circuit's impedance was calculated to be \( 160 \, \Omega \). It's important to note that this value is necessary for determining other parameters such as current amplitude in the circuit.
Current Amplitude
Current amplitude in an L-R-C series circuit relates to the maximum value of alternating current flowing through it. It is influenced by the total voltage and the impedance of the circuit, given by the formula:
  • \( I = \frac{V}{Z} \)
Where:
  • \( I \) is the current amplitude
  • \( V \) is the total voltage provided by the source
  • \( Z \) is the impedance
This is deeply connected to Ohm’s law, extended for AC circuits. For the exercise example, the current amplitude was found using:
  • \( I = \frac{360 \, \text{V}}{480 \, \Omega} = 0.75 \, \text{A} \)
Knowing the current amplitude is vital for assessing how the circuit behaves, especially in terms of energy delivery and efficiency.
Inductive Reactance
Inductive reactance is a measure of a coil's opposition to changes in current due to its inductance in AC circuits. It is calculated with:
  • \( X_L = \omega L \)
Where:
  • \( X_L \) is the inductive reactance
  • \( \omega \) is the angular frequency
  • \( L \) is the inductance
Inductive reactance causes the current to lag behind the voltage, which can affect the phase difference in the circuit.
In the exercise, potential inductive reactance values were calculated as \( 618.56 \, \Omega \) and \( 341.44 \, \Omega \), showing how changing parameters can shift circuit dynamics.
Resonance Angular Frequency
The resonance angular frequency is the frequency at which an L-R-C series circuit naturally oscillates, reaching a point of minimum impedance and maximum current amplitude. This frequency occurs when:
  • \( X_L = X_C \)
And it can be expressed as:
  • \( \omega_0 = \frac{1}{\sqrt{LC}} \)
Resonance is a special condition where energy exchange between inductive and capacitive elements is maximized without net loss, leading to greater vibrational energy. It's pivotal for applications requiring stable frequencies, such as radio and TV tunings.
When examining the exercise's findings, the angular frequency for \( X_L = 341.44 \, \Omega \) was identified to be below resonance, highlighting the critical role of precise tuning in maintaining desired circuit performance.

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Most popular questions from this chapter

A coil has a resistance of 48.0\(\Omega .\) At a frequency of 80.0 \(\mathrm{Hz}\) the voltage across the coil leads the current in it by \(52.3^{\circ} .\) Determine the inductance of the coil.

You want to double the resonance angular frequency of an L-R-C series circuit by changing only the pertinent circuit elements all by the same factor. (a) Which ones should you change? (b) By what factor should you change them?

Analyzing an \(L \cdot R-C\) Circuit. You have a \(200-\Omega\) resistor, a \(0.400-\mathrm{H}\) inductor, a \(5.00-\mu \mathrm{F}\) capacitor, and a variable-frequency ac source with an amplitude of 3.00 \(\mathrm{V} .\) You connect all four elements together to form a series circuit. (a) At what frequency will the current in the circuit be greatest? What will be the current amplitude at this frequency? (b) What will be the current amplitude at an angular frequency of 400 \(\mathrm{rad} / \mathrm{s} ?\) At this frequency, will the source voltage lead or lag the current?

In an \(L-R-C\) series circuit the current is given by \(i=I \cos \omega t .\) The voltage amplitudes for the resistor, inductor, and capacitor are \(V_{R}, \quad V_{L},\) and \(V_{C}\) . (a) Show that the instantaneous power into the resistor is \(p_{R}=V_{R} I \cos ^{2} \omega t=\frac{1}{2} V_{R} I(1+\cos 2 \omega t)\) What does this expression give for the average power into the resistor? (b) Show that the instantaneous power into the inductor is \(p_{L}=-V_{L} I \sin \omega t \cos \omega t=-\frac{1}{2} V_{L} I \sin 2 \omega t .\) What does this expression give for the average power into the inductor? (c) Show that the instantaneous power into the capacitor is \(p_{C}=\) \(V_{C} I \sin \omega t \cos \omega t=\frac{1}{2} V_{C} I \sin 2 \omega t .\) What does this expression give for the average power into the capacitor? (d) The instantaneous power delivered by the source is shown in Section 31.4 to be \(p=V I \cos \omega t(\cos \phi \cos \omega t-\sin \phi \sin \omega t) .\) Show that \(p_{R}+p_{L}+\) \(p_{C}\) equals \(p\) at each instant of time.

In an \(L-R-C\) series circuit the magnitude of the phase angle is \(54.0^{\circ},\) with the source voltage lagging the current. The reactance of the capacitor is \(350 \Omega,\) and the resistor resistance is 180\(\Omega .\) The average power delivered by the source is 140 \(\mathrm{W} .\) Find (a) the reactance of the inductor; (b) the rms current; (c) the rms voltage of the source.

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