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The minimum capacitance of a variable capacitor in a radio is 4.18 pF. (a) What is the inductance of a coil connected to this capacitor if the oscillation frequency of the \(L\)-\(C\) circuit is \(1600 \times 10^3\) Hz, corresponding to one end of the \(\textbf{AM}\) radio broadcast band, when the capacitor is set to its minimum capacitance? (b) The frequency at the other end of the broadcast band is \(540 \times 10^3\) Hz. What is the maximum capacitance of the capacitor if the oscillation frequency is adjustable over the range of the broadcast band?

Short Answer

Expert verified
(a) The inductance is approximately 0.98 \(\mu H\). (b) The maximum capacitance is approximately 55.56 pF.

Step by step solution

01

Understand the Resonant Frequency Formula

The resonant frequency of an L-C circuit is given by the formula\[ f = \frac{1}{2\pi \sqrt{LC}} \]where \( f \) is the frequency, \( L \) is the inductance, and \( C \) is the capacitance. This formula can be used to find either the inductance or the capacitance if the frequency and one of the other variables are known.
02

Calculate the Inductance

We are given that the frequency \( f = 1600 \times 10^3 \) Hz and the capacitance \( C = 4.18 \times 10^{-12} \) F. Substitute these values into the resonant frequency formula and solve for \( L \): \[ \begin{align*}f &= \frac{1}{2\pi \sqrt{LC}} \1600 \times 10^3 &= \frac{1}{2\pi \sqrt{L \times 4.18 \times 10^{-12}}} \L &= \frac{1}{(2\pi \times 1600 \times 10^3)^2 \times 4.18 \times 10^{-12}} \L &\approx 0.98 \text{ \(\mu H\)}\end{align*} \]
03

Calculate Maximum Capacitance

For the lower frequency end at \(540 \times 10^3 \) Hz, use the same formula to solve for \( C_{max} \):\[ \begin{align*}540 \times 10^3 &= \frac{1}{2\pi \sqrt{0.98 \times 10^{-6} \times C_{max}}} \C_{max} &= \frac{1}{(2\pi \times 540 \times 10^3)^2 \times 0.98 \times 10^{-6}} \C_{max} &\approx 55.56 \text{ pF} \\end{align*} \]

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Key Concepts

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

Resonant Frequency
In electronics, the resonant frequency is the natural frequency at which an LC circuit oscillates. This frequency is determined by the circuit's inductance and capacitance. For an LC circuit, the resonant frequency is given by the formula:\[ f = \frac{1}{2\pi \sqrt{LC}} \]where
  • \( f \) is the resonant frequency,
  • \( L \) is the inductance, and
  • \( C \) is the capacitance.
To find the resonant frequency, you need to know the values of both the inductance and the capacitance. Conversely, if you have the frequency and one other factor, you can solve for the remaining unknown. Understanding this relationship is key to configuring and tuning circuits like those found in radios, where specific frequencies are targeted to receive signals.
Inductance Calculation
Inductance represents the property of a conductor by which a change in current generates an induced electromotive force, or voltage. It is measured in henrys (H). When we have a fixed frequency and a known capacitance, we can find the inductance using the resonant frequency formula. In the context of the exercise, given a frequency of \(1600 \times 10^3\) Hz, the required formula transformation allows us to solve for inductance \( L \):\[ L = \frac{1}{(2\pi f)^2 \times C} \]Plugging in the numbers:
  • \( f = 1600 \times 10^3 \),
  • \( C = 4.18 \times 10^{-12} \text{ F} \).
By calculating, we find that the inductance \( L \) is approximately \(0.98 \mu H\). This computation is crucial for designing circuits that require precise frequency control.
Capacitance Range
A variable capacitor can change its capacitance within a specific range to adjust for different frequencies. This adjustability is critical for tuning circuits like those in radios. In the given scenario, we determine the maximum capacitance needed to cover the broadcast range. Using the lower end frequency of \(540 \times 10^3\) Hz and the calculated inductance, we solve for the maximum capacitance \( C_{max} \):\[ C_{max} = \frac{1}{(2\pi f_{min})^2 \times L} \]Implementing the known values:
  • \( f_{min} = 540 \times 10^3 \text{ Hz} \)
  • \( L = 0.98 \mu H \).
The result is a capacitance of approximately \( 55.56 \text{ pF} \). This flexibility in adjusting capacitance allows the circuit to accommodate the entire range of the AM broadcast band.
LC Circuit
An LC circuit is a simple electrical circuit consisting of an inductor (L) and a capacitor (C) connected together. This circuit can store energy temporarily in two interconnected forms: the magnetic field in the inductor and the electric field in the capacitor. The significance of this circuit lies in its ability to resonate at a particular frequency, determined by the values of L and C. LC circuits play a pivotal role in tuning and filtering applications, especially in radio frequency systems, where they help in selecting and amplifying desired frequencies while rejecting others. Key attributes of LC circuits include:
  • Ability to oscillate naturally if not dampened.
  • Functioning as a simple harmonic oscillator.
  • Determining bandwidth and frequency response in radio.
Understanding LC circuits is fundamental to effectively controlling frequencies in electronic systems like radios and signal processors.
Radio Frequencies
Radio frequencies (RF) refer to the range of electromagnetic frequencies from about 3 kHz to 300 GHz used in wireless communication systems. They allow devices like radios to transmit and receive sound, data, and images wirelessly. Working with RF involves a deep understanding of how frequencies are generated, modulated, and transmitted. Radio frequencies are systematically organized in bands, with the AM broadcast band extending from 540 kHz to 1600 kHz. Each channel or frequency in this band can be tuned using an LC circuit with a variable capacitor. Essentials of RF include:
  • Frequency Adjustability: Crucial for tuning different channels.
  • Modulation Techniques: How information is encoded onto carrier waves.
  • Signal Strength and Clarity: Importance of minimizing interference for clear reception.
Comprehending these frequencies and their manipulation allows the development and operation of advanced wireless communication technology.

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Most popular questions from this chapter

Two coils are wound around the same cylindrical form, like the coils in Example 30.1. When the current in the first coil is decreasing at a rate of -0.242 A/s, the induced emf in the second coil has magnitude \(1.65 \times 10^{-3}\) V. (a) What is the mutual inductance of the pair of coils? (b) If the second coil has 25 turns, what is the flux through each turn when the current in the first coil equals 1.20 A? (c) If the current in the second coil increases at a rate of 0.360 A/s, what is the magnitude of the induced emf in the first coil?

An \(L\)-\(C\) circuit containing an 80.0-mH inductor and a 1.25-nF capacitor oscillates with a maximum current of 0.750 A. Calculate: (a) the maximum charge on the capacitor and (b) the oscillation frequency of the circuit. (c) Assuming the capacitor had its maximum charge at time \(t = 0\), calculate the energy stored in the inductor after 2.50 ms of oscillation.

An inductor is connected to the terminals of a battery that has an emf of 16.0 V and negligible internal resistance. The current is 4.86 mA at 0.940 ms after the connection is completed. After a long time, the current is 6.45 mA. What are (a) the resistance \(R\) of the inductor and (b) the inductance \(L\) of the inductor?

At the instant when the current in an inductor is increasing at a rate of 0.0640 A/s, the magnitude of the self-induced emf is 0.0160 V. (a) What is the inductance of the inductor? (b) If the inductor is a solenoid with 400 turns, what is the average magnetic flux through each turn when the current is 0.720 A?

A toroidal solenoid has 500 turns, cross-sectional area 6.25 cm\(^2\), and mean radius 4.00 cm. (a) Calculate the coil's selfinductance. (b) If the current decreases uniformly from 5.00 A to 2.00 A in 3.00 ms, calculate the self- induced emf in the coil. (c) The current is directed from terminal \(a\) of the coil to terminal \(b\). Is the direction of the induced emf from \(a\) to \(b\) or from \(b\) to \(a\)?

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