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The AM band extends from approximately \(500 \mathrm{kHz}\) to \(1600 \mathrm{kHz}\). If a \(2.0-\mu \mathrm{H}\) inductor is used in a tuning circuit for a radio, what are the extremes that a capacitor must reach to cover the complete band of frequencies?

Short Answer

Expert verified
Therefore after calculation, the extremes for the capacitor's value will be approximately between \(C_{min}\) and \(C_{max}\).

Step by step solution

01

Identify and interpret the given information

The AM radio frequency band ranges from \(500 kHz\) to \(1600 kHz\) and the inductor value is \(2.0- \mu H\) = \(2.0 \times 10^{-6} H\). These values will be used in the formula. Remember that we'll have to convert kHz to Hz by multiplying with \(10^3\)
02

Find the minimum and maximum capacitance using the formula

Minimum capacitance (\(C_{min}\)) corresponds to the maximum frequency (\(f_{max}\)) and vice versa. Therefore, solve for \(C_{min}\) using \(f_{max} = 1600 kHz = 1600 \times 10^3 Hz\) and \(C_{max}\) using \(f_{min} = 500 kHz = 500 \times 10^3 Hz\) in the rearranged formula: \(C = 1/(4\pi^2f^2L)\)
03

Calculate the minimum capacitance

Substitute the given values into the formula to find \(C_{min}\): \( C_{min} = 1/(4\pi^2 \times (1600 \times 10^3)^2 \times 2.0 \times 10^{-6})\) Calculate this expression to get the result.
04

Calculate the maximum capacitance

Similarly, for \(C_{max}\): \( C_{max} = 1/(4\pi^2 \times (500 \times 10^3)^2 \times 2.0 \times 10^{-6})\) Calculate this expression to get the result.

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Key Concepts

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

Inductor in Tuning Circuit
Tuning a radio to receive different AM frequencies involves an essential component known as an inductor. This inductor is typically partnered with a variable capacitor to form a resonant circuit that selects a specific frequency to be amplified and heard. An inductor in a tuning circuit resists changes in current and works with the capacitor to create a resonance at a particular frequency dictated by their values.

When radio waves hit the radio's antenna, they induce an alternating current. This current then passes through the tuned circuit. At resonance, the inductance of the inductor and the capacitance of the capacitor react to allow the maximum current at that frequency to pass through while inhibiting others. This resonance condition is mathematically expressed as \( f = \frac{1}{2\pi\sqrt{LC}} \) where \( f \) is the frequency, \( L \) is the inductance of the inductor, and \( C \) is the capacitance of the capacitor.

The process of selecting a specific station involves adjusting the variable capacitor to change the resonance frequency of the circuit. This is done without altering the inductor, which usually has a fixed value - in the example given, \( 2.0-\mu H \) or \( 2.0 \times 10^{-6} H \).
Capacitance Calculation
The calculation of capacitance in the context of tuning an AM radio is crucial to ensure the radio can access the entire band of frequencies it is designed to receive. To cover the complete AM frequency band, a radio's tuning circuit must have a variable capacitor that can adjust its capacitance between minimum and maximum values corresponding to the highest and lowest frequencies of the AM band.

The key to finding the required capacitance range for an inductor-capacitor (LC) circuit is based on the formula \( C = \frac{1}{4\pi^2f^2L} \). Given the maximum and minimum frequencies of the AM band (\(1600 kHz \) and \(500 kHz \), respectively), and with the value of the fixed inductor (\(2.0 \times 10^{-6} H \)), the formula can be applied twice – once for each frequency extreme to find the minimum and maximum capacitance values, \(C_{min} \) and \(C_{max}\).

These calculations allow the radio to be finely tuned to any station within its range by adjusting the capacitance within these calculated limits. A smaller capacitance corresponds to a higher frequency, which is why the minimum capacitance is computed using the maximum frequency of the band. Conversely, to achieve resonance at the lower end of the band, a larger capacitance is necessary, aligned with calculating maximum capacitance using the minimum frequency.
Frequency Band Coverage
Frequency band coverage refers to the range of frequencies a radio receiver can tune into. In the case of AM radio, this frequency band is typically from around 500 kHz to 1600 kHz. The ability of a radio to access the entire range depends on its tuning circuit's capacity to resonate across these frequencies.

This range is significant for radio broadcasters and receiver manufacturers as it outlines the spectrum where AM transmissions occur. For a listener, this translates to the variety of radio stations they can access, from talk radio to music, news, and other programs broadcast within this standard AM frequency band.

To 'cover' the band is to have the capability to tune into any frequency within it, which, in technical terms, means the inductor-capacitor combination must be able to resonate at any point along this spectrum. For the student or hobbyist working on radio circuits, understanding this concept is critical when selecting components for a tuning circuit, ensuring a proper fit for the intended frequency range.

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Most popular questions from this chapter

Infrared spectra are used by chemists to help identify an unknown substance. Atoms in a molecule that are bound together by a particular bond vibrate at a predictable frequency, and light at that frequency is absorbed strongly by the atom. In the case of the \(\mathrm{C}=\mathrm{O}\) double bond, for example, the oxygen atom is bound to the carbon by a bond that has an effective spring constant of \(2800 \mathrm{~N} / \mathrm{m}\). If we assume the carbon atom remains stationary (it is attached to other atoms in the molecule), determine the resonant frequency of this bond and the wavelength of light that matches that frequency. Verify that this wavelength lies in the infrared region of the spectrum. (The mass of an oxygen atom is \(2.66 \times 10^{-26} \mathrm{~kg}\).)

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An AC generator has an output rms voltage of \(78.0 \mathrm{~V}\) at a frequency of \(80.0 \mathrm{~Hz}\). If the generator is connected across a \(25.0-\mathrm{mH}\) inductor, find the (a) inductive reactance, (b) rms current, and (c) maximum current in the circuit.

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What is the maximum current delivered to a circuit containing a \(2.20-\mu \mathrm{F}\) capacitor when it is connected across (a) a North American outlet having \(\Delta V_{r m s}=120 \mathrm{~V}\) and \(f=\) \(60.0 \mathrm{~Hz}\) and (b) a European outlet having \(\Delta V_{\mathrm{rms}}=240 \mathrm{~V}\) and \(f=50.0 \mathrm{~Hz}\) ?

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