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In an \(L-R-C\) series circuit, \(L\) = 0.280 H and C = 4.00 \(\mu\)F. The voltage amplitude of the source is 120 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) Approx. 2982 rad/s. (b) Approx. 70.6 ohms. (c) Peak voltages: 1420 V for inductor and capacitor; 120 V for resistor.

Step by step solution

01

Understanding Resonance Angular Frequency

In an L-R-C circuit, the resonance angular frequency is given by \( \omega_0 = \frac{1}{\sqrt{LC}} \). This formula is derived from the condition where the circuit's inductive reactance is equal to its capacitive reactance, minimizing impedance.
02

Calculate Resonance Angular Frequency

Using the given values, \( L = 0.280 \text{ H} \) and \( C = 4.00 \mu\text{F} = 4.00 \times 10^{-6} \text{ F} \), calculate \( \omega_0 \):\[\omega_0 = \frac{1}{\sqrt{0.280 \times 4.00 \times 10^{-6}}}\]Solving this gives:\[\omega_0 \approx 2982 \text{ rad/s}\]
03

Calculate Resistance at Resonance

At resonance, the current amplitude \( I \) is given by the voltage amplitude \( V \) divided by the resistance \( R \):\[I = \frac{V}{R} \Rightarrow R = \frac{V}{I}\]Given \( V = 120 \text{ V} \) and \( I = 1.70 \text{ A} \), calculate:\[R = \frac{120}{1.70} \approx 70.6 \text{ ohms}\]
04

Calculate Peak Voltage Across Components at Resonance

At resonance, the peak voltages across the inductor \( V_L \) and the capacitor \( V_C \) are given by \( V_L = V_C = \omega_0 L I \) where \( I \) is the current at resonance.Substituting the known values:\[V_L = \omega_0 L I = 2982 \times 0.280 \times 1.70 \approx 1420 \text{ V}\]And similarly, \( V_C = V_L \), hence \( V_C \approx 1420 \text{ V} \).The peak voltage across the resistor \( V_R \) is simply \( IR \):\[V_R = 1.70 \times 70.6 \approx 120 \text{ V}\]

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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 L-R-C circuit, resonance occurs when the inductive reactance and capacitive reactance are equal, allowing the circuit to oscillate at its natural frequency with minimal impedance. This special frequency is known as the resonance angular frequency, denoted by \( \omega_0 \). The formula for calculating \( \omega_0 \) is \( \omega_0 = \frac{1}{\sqrt{LC}} \), where \( L \) is the inductance and \( C \) is the capacitance. At this frequency, the energy stored in the inductor and capacitor is perfectly exchanged with each cycle, resulting in maximum current flow.
  • Minimizes impedance, allowing maximum current flow.
  • Expression: \( \omega_0 = \frac{1}{\sqrt{LC}} \)
Using this concept, you can plug in the provided values for \( L = 0.280 \text{ H} \) and \( C = 4.00 \times 10^{-6} \text{ F} \) to calculate the resonance frequency of the circuit as approximately 2982 rad/s.
Impedance
Impedance in an L-R-C circuit refers to the total opposition to the current flow, combining both resistive and reactive elements. Unlike pure resistance, impedance depends on frequency. It reaches its minimum value at the resonance angular frequency, where the circuit's reactance components cancel each other out. At resonance, the impedance is purely resistive and equals the resistance of the circuit.
  • Combines effects of resistance, inductance, and capacitance.
  • At resonance, impedance equals the circuit's resistance \( R \).
For the given problem, when \( \omega_0 \approx 2982 \text{ rad/s} \), the impedance is minimized, allowing us to use the relation \( R = \frac{V}{I} \) to find the resistance, which is approximately 70.6 ohms.
Inductive Reactance
Inductive reactance, represented by \( X_L \), is the opposition to current flow due to the inductor in an AC circuit. It varies with frequency, proportional to both the inductance \( L \) and the angular frequency \( \omega \). The mathematical expression for inductive reactance is \( X_L = \omega L \).
  • Increases with higher frequencies.
  • Expression: \( X_L = \omega L \).
At resonance, \( X_L \) and the capacitive reactance \( X_C \) balance out, resulting in their effects canceling. For resonance frequency, \( \omega = \omega_0 \), calculated as approximately 2982 rad/s, the voltage across the inductor can be found using \( V_L = \omega_0 L I \), yielding about 1420 V.
Capacitive Reactance
Capacitive reactance \( X_C \) operates much like inductive reactance but is related to capacitors in AC circuits. It describes how much a capacitor opposes the flow of alternating current, decreasing as the frequency increases. The formula to compute \( X_C \) is \( X_C = \frac{1}{\omega C} \).
  • Decreases with higher frequencies.
  • Expression: \( X_C = \frac{1}{\omega C} \)
In our L-R-C circuit at resonance, \( X_C \) becomes equal to \( X_L \), nullifying their opposing influences, thereby reducing impedance. At resonance, the voltage across the capacitor is calculated similarly to the inductor using the formula \( V_C = \omega_0 L I \), which also results in approximately 1420 V.

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Most popular questions from this chapter

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

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?

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?

A sinusoidal current \(i = I\space cos\space \omega\)t has an rms value \(I_{rms} =\)2.10 A. (a) What is the current amplitude? (b) The current is passed through a full-wave rectifier circuit. What is the rectified average current? (c) Which is larger: \(I_{rms}\) or \(I_{rav}\)? Explain, using graphs of \(i^2\) and of the rectified current.

(a) Show that for an \(L-R-C\) series circuit the power factor is equal to R/Z. (b) An \(L-R-C\) series circuit has phase angle -31.5\(^\circ\). The voltage amplitude of the source is 90.0 V. What is the voltage amplitude across the resistor?

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