Question

When you turn the dial on a radio to tune it, you are adjusting a variable capacitor in an LC circuit. Suppose you tune to an AM station broadcasting at a frequency of \(1000 . \mathrm{kHz},\) and there is a \(10.0-\mathrm{mH}\) inductor in the tuning circuit. When you have tuned in the station, what is the capacitance of the capacitor?

Short answer

Answer: The capacitance needed for tuning the radio to the given frequency is approximately \(2.53 \times 10^{-12}\,\text{F}\).

Step-by-step solution

Write down the given information

We are given the following information: - Frequency of the radio station: \(f = 1000 \,\text{kHz}\) - Inductance of the inductor: \(L = 10.0\, \text{mH}\)

Convert the frequency and inductance to SI units

Before we proceed with the calculations, let's convert the frequency and inductance to the appropriate SI units: - Frequency: \(f = 1000 \,\text{kHz} = 1000 \times 10^{3} \,\text{Hz} = 10^6\, \text{Hz}\) - Inductance: \(L = 10.0 \,\text{mH} = 10.0 \times 10^{-3}\, \text{H}\)

Apply the resonant frequency formula

We can use the formula for the resonance frequency of an LC circuit: $$ f = \dfrac{1}{2\pi\sqrt{LC}} $$ We need to rearrange this formula to get the capacitance, \(C\): $$ C = \dfrac{1}{(2\pi f)^2 L} $$

Substitute the values and calculate the capacitance

Now, let's substitute the values of \(f\) and \(L\) that we found earlier into the formula: $$ C =\dfrac{1}{(2\pi (10^6\,\text{Hz}))^2 (10.0 \times 10^{-3}\,\text{H})} $$ Calculate the value of the capacitance: $$ C \approx 2.53 \times 10^{-12}\,\text{F} $$

Write down the final answer

The capacitance of the capacitor needed to tune the radio to the given frequency is approximately \(2.53 \times 10^{-12}\,\text{F}\).

Key concepts

Resonant Frequency

In an LC circuit, the resonant frequency is the frequency at which the circuit naturally oscillates. This happens when the inductive reactance and capacitative reactance are equal in magnitude, but opposite in phase. At this frequency, the oscillation is at its maximum amplitude. The formula to find the resonant frequency \( f \) of an LC circuit is: \[ f = \frac{1}{2\pi\sqrt{LC}} \] In this formula, \( L \) is the inductance, and \( C \) is the capacitance. Finding the resonant frequency is crucial when tuning devices like radios, as it ensures that the LC circuit is perfectly resonant with the frequency of the incoming signal. This results in clearer sound or stronger signal reception, as it minimizes losses and maximizes efficiency. When dealing with real-life circuits, detuning and losses often occur, which deviate the system from its ideal resonant frequency. Nevertheless, understanding the basic concept helps in setting the foundation for optimizing the performance of LC circuits.

Capacitance Calculation

Calculating capacitance in an LC circuit is a crucial skill for tasks like tuning a radio. In the given problem, we needed to find the capacitance that tunes the radio to a specific station. By knowing the resonant frequency, which in this case is the radio station frequency, we can rearrange the resonant frequency formula to solve for capacitance \( C \): \[C = \frac{1}{(2\pi f)^2 L} \] Here, \( f \) is the frequency of the radio station, and \( L \) is the inductance of the coil used in the circuit. It is important to ensure that all units are in the standard international (SI) units during the calculation. Frequencies should be in Hertz (Hz) and inductance in Henrys (H). By substituting the correct values into the formula, we obtain the desired capacitance value, which is critical for achieving resonance. This process underscores the importance of using correct units and understanding the relation between frequency, inductance, and capacitance in LC circuits.

Variable Capacitor

A variable capacitor is a type of capacitor whose capacitance value can be adjusted or varied. These capacitors are widely used in radio tuning circuits to change the resonant frequency of the circuit, allowing us to tune into different radio stations. In an LC circuit like the one discussed, rotating the dial alters the plate separation or the overlapping area of the capacitor plates, changing its capacitance. Variable capacitors are practical because they allow precise control over the tuning. Unlike fixed capacitors, they enable dynamic changes without having to swap out components. These components typically have a limited range of capacitance and are designed mainly for low-frequency applications like radios. Understanding how to manipulate them helps us grasp how traditional radios functioned before digital technology took over.

Inductance

Inductance is a property of an electrical circuit or component that opposes changes in current. It is an essential aspect of LC circuits and is due to the magnetic field created by the current flowing through a coil or inductor. It is measured in Henrys (H). In our problem, an inductance of \(10.0 \, \text{mH}\) is used to determine the tuning capacitance for the radio. Inductors are often used in combination with capacitors to form LC circuits, which serve as basic filters or tuners in electronic systems. The inductance value plays a crucial role in determining the resonant frequency of the circuit, making it possible to selectively filter and enhance specific frequencies. Some key features of inductors include:

• They work based on Faraday's law of induction.
• In an AC circuit, they cause a phase shift between voltage and current.
• Their ability to store energy in a magnetic field.

A solid understanding of inductance and its behaviors is crucial in designing circuits like radios, which rely heavily on accurate tuning.