Question

The AM radio band covers the frequency range from \(520 \mathrm{kHz}\) to \(1610 \mathrm{kHz}\). Assuming a fixed inductance in a simple \(\mathrm{LC}\) circuit, what ratio of capacitance is necessary to cover this frequency range? That is, what is the value of \(C_{\mathrm{h}} / C_{\mathrm{p}}\), where \(C_{\mathrm{h}}\) is the capacitance for the highest frequency and \(C_{1}\) is the capacitance for the lowest frequency? a) 9.59 b) 0.104 c) 0.568 d) 1.76

Short answer

Answer: (b) 0.104

Step-by-step solution

Write down the resonance frequency formula

Given the LC circuit, its resonance frequency can be calculated with the formula: \(f = \frac{1}{2\pi\sqrt{LC}}\)

Solve for capacitance

We want to find the capacitance necessary to cover the frequencies given. Thus, we need to solve for \(C\) in the resonance frequency formula. Rearrange the equation to get: \(C = \frac{1}{(2\pi f)^2 L}\)

Calculate capacitance for the highest and lowest frequencies

Use the given frequencies to calculate the required capacitance. The highest frequency is \(1610\,\mathrm{kHz}\), and the lowest is \(520\,\mathrm{kHz}\). Plug these values into the rearranged equation and find the capacitance for both: \(C_{\mathrm{h}} = \frac{1}{(2\pi \cdot 1610\times10^3)^2 L}\) \(C_{\mathrm{p}} = \frac{1}{(2\pi \cdot 520\times10^3)^2 L}\)

Calculate the capacitance ratio

To find the ratio \(C_{\mathrm{h}} / C_{\mathrm{p}}\), divide the capacitance at highest frequency by the capacitance at lowest frequency: \(\frac{C_{\mathrm{h}}}{C_{\mathrm{p}}} = \frac{\frac{1}{(2\pi \cdot 1610\times10^3)^2 L}}{\frac{1}{(2\pi \cdot 520\times10^3)^2 L}}\)

Simplify and find the answer

Simplify the ratio: \(\frac{C_{\mathrm{h}}}{C_{\mathrm{p}}} = \frac{(2\pi \cdot 520\times10^3)^2 L}{(2\pi \cdot 1610\times10^3)^2 L}\) Notice that \(L\) cancels out, and we can further simplify to get: \(\frac{C_{\mathrm{h}}}{C_{\mathrm{p}}} = \frac{(520)^2}{(1610)^2}\) Calculate the value: \(\frac{C_{\mathrm{h}}}{C_{\mathrm{p}}} = 0.104\) Comparing the result to the given options, the answer is (b) 0.104.

Key concepts

Capacitive Reactance

In an LC circuit, capacitive reactance is a critical concept, much like resistance in a traditional electrical circuit. However, instead of impeding current in a direct manner like resistance, capacitive reactance impedes alternating current (AC) due to the capacitor's charge and discharge cycle. The formula for capacitive reactance (\(X_C\)) is given by:\[\begin{equation}X_C = \frac{1}{2\text{π}fC}\text{\end{equation}\]}where

• \(f\) is the frequency of the alternating current,
• \(C\) is the capacitance of the capacitor.

The key takeaway is that capacitive reactance decreases with increasing frequency or capacitance. This behavior is pivotal when tuning an LC circuit to resonate at different frequencies, such as those found in the AM radio frequency range.

When calculating how to cover the AM band frequencies using a fixed inductance LC circuit, an understanding of capacitive reactance allows us to establish that a higher frequency will require a lower capacitance and vice versa, to maintain resonance. The selection of the correct capacitance values is thus essential to ensure the circuit can effectively resonate across the desired frequency spectrum.

LC Circuit

An LC circuit is an essential type of resonant circuit composed of an inductor (represented by the symbol \(L\)) and a capacitor (\(C\)). It is also known as a tank circuit, a tuned circuit, or an RLC circuit when resistance (\(R\)) is considered. The behavior of an LC circuit is largely characterized by its ability to resonate at a specific frequency, known as the resonance frequency.\[\begin{equation}f_{\text{res}} = \frac{1}{2\text{π}\sqrt{LC}}\text{\end{equation}\]}In the context of our initial exercise,

• The resonance frequency formula allows the calculation of the specific capacitance needed to tune the LC circuit to a desired frequency.
• This demonstrates how changing the value of the capacitor can adjust the resonance frequency, expanding or narrowing the range of frequencies the circuit can tune into.

In the AM radio band example, an LC circuit must be capable of tuning across a range of frequencies. The range is determined by the inverse of the square root of the product of inductance and the capacitance. By modifying the capacitance alone, while keeping the inductance constant, the circuit can be made to resonate across the intended frequency spectrum. In practical terms, this is achieved by using variable capacitors that can change their capacitance to tune to different broadcast frequencies within the AM band.

Radio Frequency

Radio frequencies (RF) encompass a large segment of the electromagnetic spectrum, and they are utilized in a wide array of applications from broadcasting to wireless communications. In our exercise example, we focused on the AM radio band, which lies typically within the frequency range of 520 kHz to 1610 kHz.

• The term kHz stands for kilohertz, which is a unit equalling 1,000 hertz. Hertz (Hz) is the measurement unit for frequency, indicating the number of cycles per second.
• AM radio utilizes these radio frequencies to transmit audio signals through modulation of the amplitude of the carrier wave (hence AM - amplitude modulation).

The resonance of an LC circuit at specific radio frequencies enables the selection and reception of a desired radio station. The capacitance ratio needed to cover the range from 520 kHz to 1610 kHz, as calculated in the exercise, is a measure of how the capacitive values need to be altered to resonate across the entire range of the AM radio band. In practice, this tuning is often done with the assistance of a dial or slider that adjusts the capacitance to match the intended frequency, thus allowing listeners to tune into their selected AM radio station with clarity and precision.