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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?

Short Answer

Expert verified
(a) \(2998 \; \mathrm{rad/s}\); (b) \(70.59 \; \Omega\); (c) \(V_L = 1427 \; \mathrm{V}, V_C = 1427 \; \mathrm{V}, V_R = 120 \; \mathrm{V}\).

Step by step solution

01

Identify Known Variables

We are given that the inductance \( L = 0.280 \; \mathrm{H} \), the capacitance \( C = 4.00 \; \mu \mathrm{F} = 4.00 \times 10^{-6} \; \mathrm{F} \), and the voltage amplitude \( V_0 = 120 \; \mathrm{V} \). Furthermore, the current amplitude at resonance is \( I_0 = 1.70 \; \mathrm{A} \).
02

Calculate Resonance Angular Frequency

Use the formula for the resonance angular frequency: \( \omega_0 = \frac{1}{\sqrt{LC}} \).Substitute \( L = 0.280 \; \mathrm{H} \) and \( C = 4.00 \times 10^{-6} \; \mathrm{F} \) into the formula:\[\omega_0 = \frac{1}{\sqrt{0.280 \times 4.00 \times 10^{-6}}} = \frac{1}{\sqrt{1.12 \times 10^{-6}}} \approx 2998 \; \mathrm{rad/s}.\]
03

Determine Resistance at Resonance

At resonance, the impedance is equal to the resistance, \( R \). Use the relation \( V_0 = I_0 R \) to find \( R \):\[R = \frac{V_0}{I_0} = \frac{120}{1.70} \approx 70.59 \; \Omega.\]
04

Calculate Peak Voltage Across Inductor and Capacitor

At resonance, the peak voltage across the inductor \( V_L \) is given by \( V_L = I_0 \cdot \omega_0 \cdot L \) and across capacitor \( V_C \) by \( V_C = \frac{I_0}{\omega_0 \cdot C} \).For inductor:\[V_L = 1.70 \times 2998 \times 0.280 \approx 1427 \; \mathrm{V}.\]For capacitor:\[V_C = \frac{1.70}{2998 \times 4.00 \times 10^{-6}} \approx 1427 \; \mathrm{V}.\]
05

Calculate Peak Voltage Across Resistor

At resonance, the peak voltage across the resistor \( V_R \) is simply \( V_R = I_0 \cdot R \):\[V_R = 1.70 \times 70.59 \approx 120 \; \mathrm{V}.\]This matches the source voltage, confirming our calculations.

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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 LRC circuit, the resonance angular frequency, represented as \( \omega_0 \), is a crucial concept. It determines the frequency at which the circuit naturally prefers to oscillate. Understanding this enables us to predict the circuit's behavior under various conditions.
At resonance, the impedance of the circuit is minimized, typically resulting in the maximum possible current flow. The formula to calculate the resonance angular frequency is:
  • \( \omega_0 = \frac{1}{\sqrt{LC}} \)
where \( L \) is the inductance in henries (H) and \( C \) is the capacitance in farads (F).
Using our exercise values, \( L = 0.280 \, \mathrm{H} \) and \( C = 4.00 \times 10^{-6} \, \mathrm{F} \), we find \( \omega_0 \) by substituting the known values into the formula. After calculation, \( \omega_0 \approx 2998 \, \mathrm{rad/s} \).
This frequency is where the inductor and capacitor store and release energy maximally, leading to a significant current with minor resistance influence.
Impedance
Impedance in an LRC circuit, denoted as \( Z \), is a measure of the circuit's opposition to the current when voltage is applied. It is more than simple resistance as it considers the effects of inductance and capacitance.
At resonance, however, the impedance simplifies greatly.
  • In a series LRC circuit at resonance, the inductive reactance \( \omega L \) and capacitive reactance \( \frac{1}{\omega C} \) cancel each other out.
  • Thus, the impedance \( Z \) equals the resistance \( R \).
This simplification allows us to calculate \( R \) directly from the peak current and voltage values using the equation:
\[ R = \frac{V_0}{I_0} \]
where \( V_0 \) is the source's voltage amplitude and \( I_0 \) is the current amplitude. From the given exercise, using \( V_0 = 120 \, \mathrm{V} \) and \( I_0 = 1.70 \, \mathrm{A} \), we find that \( R \approx 70.59 \, \Omega \). This resistance is crucial as it dictates the circuit's energy loss at resonance.
Peak Voltage Calculation
In an LRC circuit at resonance, understanding the distribution of voltage across each component is vital. The peak voltage across the inductor \( V_L \), the capacitor \( V_C \), and the resistor \( V_R \) can be calculated as follows:
For the inductor:
  • \( V_L = I_0 \cdot \omega_0 \cdot L \)
Substituting the known values leads to \( V_L \approx 1427 \, \mathrm{V} \).
For the capacitor:
  • \( V_C = \frac{I_0}{\omega_0 \cdot C} \)
This also results in \( V_C \approx 1427 \, \mathrm{V} \).
The equality of \( V_L \) and \( V_C \) at resonance implies energy is perfectly exchanged between the inductor and capacitor.
Finally, for the resistor:
  • \( V_R = I_0 \cdot R \)
From calculations, \( V_R \approx 120 \, \mathrm{V} \), matching the source, which confirms our understanding of the voltage distribution at resonance. This helps recognize the separate roles each component plays in the circuit.

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Most popular questions from this chapter

An \(L-R-C\) series circuit has \(R\) = 60.0 \(\Omega\), \(L\) = 0.800 H, and \(C\) = 3.00 \(\times\) 10\(^{-4}\) F. The ac source has voltage amplitude 90.0 V and angular frequency 120 rad/s. (a) What is the maximum energy stored in the inductor? (b) When the energy stored in the inductor is a maximum, how much energy is stored in the capacitor? (c) What is the maximum energy stored in the capacitor?

A toroidal solenoid has 2900 closely wound turns, cross-sectional area 0.450 cm\(^2\), mean radius 9.00 cm, and resistance \(R\) = 2.80 \(\Omega\). Ignore the variation of the magnetic field across the cross section of the solenoid. What is the amplitude of the current in the solenoid if it is connected to an ac source that has voltage amplitude 24.0 V and frequency 495 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 \(\omega3\)= \(\omega_1\)/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?

A capacitance \(C\) and an inductance \(L\) are operated at the same angular frequency. (a) At what angular frequency will they have the same reactance? (b) If \(L\) = 5.00 mH and \(C\) = 3.50 \(\mu\)F, what is the numerical value of the angular frequency in part (a), and what is the reactance of each element?

When a solenoid is connected to a 48.0-V dc battery that has negligible internal resistance, the current in the solenoid is 5.50 A. When this solenoid is connected to an ac source that has voltage amplitude 48.0 V and angular frequency 20.0 rad/s, the current in the solenoid is 3.60 A. What is the inductance of this solenoid?

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