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An \(R L C\) circuit has a resonance frequency of \(2.4 \mathrm{kHz}\). If the capacitance is \(47 \mu \mathrm{F},\) what is the inductance?

Short Answer

Expert verified
The inductance is approximately 0.1507 H.

Step by step solution

01

Understanding Resonance Frequency Equation

The resonance frequency of an RLC circuit is given by the formula:\[ f_0 = \frac{1}{2\pi\sqrt{LC}}\]where \( f_0 \) is the resonance frequency, \( L \) is the inductance, and \( C \) is the capacitance.
02

Rearranging to Solve for Inductance

To find the inductance \( L \), we need to rearrange the equation as follows:\[ L = \frac{1}{(2\pi f_0)^2 C}\]
03

Substituting Given Values

Substitute the resonance frequency \( f_0 = 2400 \text{ Hz} \) and the capacitance \( C = 47 \times 10^{-6} \text{ F} \) into the formula:\[ L = \frac{1}{(2\pi \times 2400)^2 \times 47 \times 10^{-6}}\]
04

Calculate the Inductance

Calculate the right-hand side of the equation to find \( L \):- Calculate \( (2\pi \times 2400)^2 \approx 144\pi^2 \times 10^6 \)- Simplifying further gives:\[ L \approx \frac{1}{141.37\times 47 \times 10^{-6}} \approx \frac{1}{6.64469} \approx 0.1507 \text{ H}\]

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Key Concepts

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

Resonance Frequency
The resonance frequency is a fundamental property of RLC circuits, which are composed of resistors (R), inductors (L), and capacitors (C). When a circuit resonates, it means the inductive and capacitive reactances are equal in magnitude, which allows the circuit to oscillate at its natural frequency without external forces.

This frequency is given by the formula:
  • \( f_0 = \frac{1}{2\pi\sqrt{LC}} \)
Here, \( f_0 \) represents the resonance frequency in Hertz, \( L \) is the inductance in Henries, and \( C \) is the capacitance in Farads.

At resonance, the circuit achieves maximum energy transfer and minimal impedance, making it a critical point for many electronics applications like radios and filters. The importance of understanding resonance frequency lies in its ability to optimize the performance and efficiency of the RLC circuit.
Inductance Calculation
Inductance in an RLC circuit refers to the ability of a component to store energy in a magnetic field. It is measured in Henries (H). When trying to find the inductance required for a circuit to resonate at a given frequency, you need to rearrange the resonance frequency formula.

Starting from:
  • \( f_0 = \frac{1}{2\pi\sqrt{LC}} \)
Rearrange it to solve for \( L \):
  • \( L = \frac{1}{(2\pi f_0)^2 C} \)
Inductance can then be calculated by substituting known values of \( f_0 \) and \( C \). This step-by-step approach ensures accuracy.

This calculation is vital for designing circuits to achieve specific frequencies, which is essential for applications like creating RF circuits and tuning LC oscillators.
Capacitance
Capacitance is a measure of a capacitor's ability to store electrical energy in an electric field. In an RLC circuit, the capacitance plays a crucial role in determining the resonance frequency.

For a fixed inducer, changing the capacitance would change the circuit’s resonance frequency according to the formula:
  • \( f_0 = \frac{1}{2\pi\sqrt{LC}} \)
Capacitance is measured in Farads (F), and it often appears in microfarads (\( \mu F \)) because of its relatively small values in practice.

Having a clear understanding of the role of capacitance helps in designing circuits that specifically target the desired frequency range, maximizing the efficiency and functionality of the RLC circuit.

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Most popular questions from this chapter

An \(R L C\) circuit with \(R=25.0 \Omega, L=325 \mathrm{mH},\) and \(C=\) \(45.2 \mu \mathrm{F}\) is connected to an ac generator with an \(\mathrm{rms}\) voltage of 24 V. Determine the average power delivered to this circuit when the frequency of the generator is (a) equal to the resonance frequency, (b) twice the resonance frequency, and (c) half the resonance frequency.

IP Black-Box Experiment You are given a sealed box with two electrical terminals. The box contains a \(5.00-\Omega\) resistor in series with either an inductor or a capacitor. When you attach an ac generator with an rms voltage of \(0.750 \mathrm{V}\) to the terminals of the box, you find that the current increases with increasing frequency. (a) Does the box contain an inductor or a capacitor?

IP An rms voltage of \(20.5 \mathrm{V}\) with a frequency of \(1.00 \mathrm{kHz}\) is applied to a \(0.395-\mu \mathrm{F}\) capacitor. (a) What is the rms current in this circuit? (b) By what factor does the current change if the frequency of the voltage is doubled? \((c)\) Calculate the current for a frequency of \(2.00 \mathrm{kHz}\)

IP Tuning a Radio A radio tuning circuit contains an RLC circuit with \(R=5.0 \Omega\) and \(L=2.8 \mu \mathrm{H.}\) (a) What capacitance is needed to produce a resonance frequency of \(85 \mathrm{MHz}\) ? (b) If the capacitance is increased above the value found in part (a), will the impedance increase, decrease, or stay the same? Explain. (c) Find the impedance of the circuit at reso- nance. (d) Find the impedance of the circuit when the capacitance is \(1 \%\) higher than the value found in part (a).

\(\cdot\) The rms voltage across a \(0.010-\mu F\) capacitor is \(1.8 \mathrm{V}\) at a frequency of \(52 \mathrm{Hz}\). What are (a) the rms and (b) the maximum current through the capacitor?

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