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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

Expert verified
Answer: The capacitance of the capacitor is approximately 2.53 pF when the circuit is tuned to the given frequency.

Step by step solution

01

Write down the given values

We have a frequency of \(f = 1000\) kHz (which is \(1*10^6\) Hz), and an inductor with an inductance of \(L = 10.0\) mH (which is \(10*10^{-3}\) H).
02

Find the formula for capacitance in terms of frequency and inductance

From the resonant frequency formula, we can rewrite the equation to find the capacitance: \(C = \frac{1}{(2\pi f)^2 L}\)
03

Plug in the given values and solve for capacitance

We can now plug in the given values of \(f\) and \(L\) to find the capacitance: \(C = \frac{1}{(2\pi (1*10^6))^2 (10*10^{-3})} \approx 2.533 * 10^{-12} \mathrm{F}\)
04

Final answer

The capacitance of the capacitor when the circuit is tuned to the given frequency is approximately \(2.53\) pF.

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Key Concepts

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

Resonant Frequency Formula
The resonant frequency formula is essential for understanding how tuning works in an LC (inductor-capacitor) circuit. This formula provides the frequency at which the circuit can oscillate without external influence. The formula is given by
\[ f = \frac{1}{2\pi\sqrt{LC}} \]
where \(f\) is the resonant frequency, \(L\) is the inductance in henrys (H), and \(C\) is the capacitance in farads (F). The relationship shows us that the higher the inductance or capacitance, the lower the resonant frequency, and vice versa. This is the fundamental principle behind tuning a radio, as changing the capacitance will alter the resonant frequency to match the frequency of the desired radio station.
Variable Capacitor
A variable capacitor is a pivotal component in an LC circuit, especially in applications like radio tuning. It consists of two sets of plates: one stationary and one movable. Turning the dial of a radio adjusts the position of the movable plates in relation to the stationary ones, effectively changing the capacitance.

Benefits of a Variable Capacitor

  • Allows precise tuning of resonant frequency in radios and other devices.
  • Enables the circuit to filter out all other frequencies except the one that is being tuned into.
  • Improves user control over device functions.
The ability to fine-tune the separation and surface area of the plates makes the variable capacitor the heart of the tuning process in an LC circuit.
Inductance
Inductance, symbolized as \(L\), is a measure of an inductor's ability to store energy in a magnetic field as it resists changes in current flow. In other words, it's the inductor's form of electrical inertia.

Core Factors Influencing Inductance

  • Core Material: An inductor's core can be air, iron, or another material, which affects its ability to create a magnetic field.
  • Number of Coils: More coils generally mean greater inductance.
  • Coil Shape: The shape and spacing of the coils determine the magnetic field's strength and thus the inductance.
Inductance plays a crucial role in tuning LC circuits, as it defines the amount of opposition to the change in current and, together with capacitance, determines the resonant frequency.
Capacitance Calculation
Capacitance calculation is vital for designing and understanding LC circuits. The capacitance is the measure of a capacitor's ability to store charge per unit voltage. It’s represented by the symbol \(C\) and is measured in farads (F). Calculating the capacitance in an LC circuit involves reorganizing the resonant frequency formula to solve for \(C\), as shown below:
\[ C = \frac{1}{(2\pi f)^2 L} \]
By plugging the values of the frequency and inductance into this formula, one can determine the required capacitance for the circuit to resonate at a specific frequency. This calculation is essential for tuning purposes, such as adjusting a radio to receive the correct signal from the desired station.

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Most popular questions from this chapter

A \(10.0-\mu \mathrm{F}\) capacitor is fully charged by a \(12.0-\mathrm{V}\) battery and is then disconnected from the battery and allowed to discharge through a \(0.200-\mathrm{H}\) inductor. Find the first three times when the charge on the capacitor is \(80.0-\mu C\), taking \(t=0\) as the instant when the capacitor is connected to the inductor.

A series RLC circuit is in resonance when driven by a sinusoidal voltage at its resonant frequency, \(\omega_{0}=(L C)^{-1 / 2} .\) But if the same circuit is driven by a square-wave voltage (which is alternately on and off for equal time intervals), it will exhibit resonance at its resonant frequency and at \(\frac{1}{3}\), \(\frac{1}{5}, \frac{1}{7}, \ldots,\) of this frequency. Explain why.

A circuit contains a \(100 .-\Omega\) resistor, a 0.0500 - H inductor, a 0.400 - \(\mu\) F capacitor, and a source of time-varying emf connected in series. The time-varying emf corresponds to \(V_{\mathrm{rms}}=50.0 \mathrm{~V}\) at a frequency of \(2000 . \mathrm{Hz}\) a) Determine the current in the circuit. b) Determine the voltage drop across each component of the circuit. c) How much power is drawn from the source of emf?

An inductor with inductance \(L\) is connected to an AC power source that supplies \(V_{\mathrm{emf}}=21.5 \mathrm{~V}\) at \(f=797 \mathrm{~Hz}\). If the maximum current in the circuit is to be \(0.1528 \mathrm{~A},\) what should the value of \(L\) be?

A common configuration of wires has twisted pairs as opposed to straight, parallel wires. What is the technical advantage of using twisted pairs of wires versus straight, parallel pairs?

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