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A resonant circuit in a radio receiver is tuned to a certain station when the inductor has a value of \(3.00 \mu \mathrm{H}\) and the capacitor has a value of \(2.50 \mathrm{pF}\). The resistance of the circuit is \(12 \Omega\). (a) Find the frequency of the radio station. (b) Is there any information given in the problem that is not needed to solve it? Explain.

Short Answer

Expert verified
The frequency of the radio station is approximately \(23.0 MHz\). The resistance of \(12 \Omega\) does not affect the tuning of the radio and is therefore not needed to solve this problem.

Step by step solution

01

Convert Units

Inductance is given as \(3.00 \mu H\) which equals \(3.00 \times 10^{-6} H\) and the capacitance is given as \(2.50 pF\) which equals \(2.50 \times 10^{-12} F\). Always remember to work with the standard metric units.
02

Calculate the Resonant Frequency

Use the formula for resonant frequency in an LC circuit: \( f = \frac{1}{2 \pi \sqrt{LC}} \). Insert the converted values of L and C into the formula to find the frequency: \( f = \frac{1}{2 \pi \sqrt{3.00 \times 10^{-6} H \times 2.50 \times 10^{-12} F}}\).
03

Evaluate the Importance of Resistance

Understand that in an ideal LC circuit, the resistance does not affect the resonance frequency. Therefore, the given resistance value of \(12 \Omega\) is not necessary to solve the problem.

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Key Concepts

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

LC Circuit
An LC circuit is a fundamental electrical circuit that consists of an inductor, represented by the symbol 'L', and a capacitor, represented by 'C'. Think of it like a swinging pendulum, but with electric signals. The inductor and capacitor interact with each other by exchanging energy back and forth. The inductor stores energy in a magnetic field when electrical current flows through it, while the capacitor stores energy in an electric field as an electric charge.

This back-and-forth energy exchange between the magnetic field of the inductor and the electric field of the capacitor creates a phenomenon known as 'resonance'. During resonance, the circuit naturally oscillates at a particular frequency called the 'resonant frequency'. This frequency is particularly special because at this point, the energy losses within the circuit are minimal, and the current and voltage are at their maximum values—a bit like pushing a swing at just the right moment to make it go higher.

In the context of our textbook exercise, the LC circuit is being used to select a specific radio frequency. To do this, we alter the circuit until its resonant frequency matches the desired radio signal frequency. Then, the radio signal is at its strongest, and the station comes in clear.
Radio Station Tuning
Radio station tuning with an LC circuit is the process of adjusting the resonant frequency until it matches the frequency of the radio station you'd like to listen to. Imagine turning a radio dial and finding your favorite song—it's that, but on a circuit level.

The resonant frequency of an LC circuit, which determines what station you're tuned into, can be found using the formula: \[ f = \frac{1}{2 \pi \sqrt{LC}} \]. In practice, by varying either the inductance 'L' or the capacitance 'C', you can 'dial in' to different stations. These adjustments are often made with variable capacitors or inductors.

When you tune a radio, what you're typically doing is changing the capacitance, which directly affects the resonant frequency. As you move the tuning knob, the resonant frequency slides up and down the radio spectrum, allowing you to 'lock in' on your desired station, filtering out all other signals thanks to the physics of resonance.
Inductance and Capacitance
Inductance (L) and capacitance (C) are fundamental concepts in electronics that describe how an inductor and a capacitor, respectively, behave in a circuit.

Inductance is the property of an inductor to resist changes in the electric current flowing through it. It does this by creating a magnetic field around itself. The strength of an inductor is measured in henrys (H), and higher inductance means a stronger magnetic field and more stored energy.

Capacitance, on the other hand, is the ability of a capacitor to hold an electrical charge. It's measured in farads (F), and a higher capacitance means the capacitor can store more charge at a given voltage. Just like a small bucket can hold only so much water, a small capacitor can hold only so much charge.

In the resonance formula \[ f = \frac{1}{2 \pi \sqrt{LC}} \], 'L' and 'C' play crucial roles. A larger inductance or capacitance will result in a lower resonant frequency, while smaller values will give a higher frequency. It's a delicate balance, like adjusting the length of a pendulum to make it swing faster or slower. This concept is at the heart of tuning electronic circuits to specific frequencies, be it in radios, TVs, or even your wireless router!

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Most popular questions from this chapter

An AC power generator produces \(50 \mathrm{~A}\) (rms) at \(3600 \mathrm{~V}\). The voltage is stepped up to \(100000 \mathrm{~V}\) by an ideal transformer, and the energy is transmitted through a longdistance power line that has a resistance of \(100 \Omega\). What percentage of the power delivered by the generator is dissipated as heat in the power line?

Assume the solar radiation incident on Earth is \(1340 \mathrm{~W} / \mathrm{m}^{2}\) (at the top of Earth's atmosphere). Calculate the total power radiated by the Sun, taking the average separation between Earth and the Sun to be \(1.49 \times 10^{11} \mathrm{~m}\).

35\. An inductor and a resistor are connected in series. When connected to a \(60-\mathrm{Hz}, 90-\mathrm{V}\) (rms) source, the voltage drop across the resistor is found to be \(50 \mathrm{~V}\) (rms) and the power delivered to the circuit is \(14 \mathrm{~W}\). Find (a) the value of the resistance and (b) the value of the inductance.

Operation of the pulse oximeter (see previous problem). The transmission of light energy as it passes through a solution of light-absorbing molecules is described by the Beer-Lambert law $$ I=I_{0} 10^{-c a} \text { or } \log _{10}\left(\frac{l}{I_{0}}\right)=-\epsilon C L $$ which gives the decrease in intensity \(I\) in terms of the distance \(L\) the light has traveled through a fluid with a concentration \(C\) of the light- absorbing molecule. The quantity \(\epsilon\) is called the extinction coefficient, and its value depends on the frequency of the light. (It has units of \(\mathrm{m}^{2} / \mathrm{mol}\).) Assume the extinction coefficient for \(660-\mathrm{nm}\) light passing through a solution of oxygenated hemoglobin is identical to the coefficient for \(940-\mathrm{nm}\) light passing through deoxygenated hemoglobin. Also assume \(940-\mathrm{nm}\) light has zero absorption \((\epsilon=0)\) in oxygenated hemoglobin and \(660-\mathrm{nm}\) light has zero absorption in deoxygenated hemoglobin. If \(33 \%\) of the energy of the red source and \(76 \%\) of the infrared energy is transmitted through the blood, what is the fraction of hemoglobin that is oxygenated?

Consider a series RLC circuit with \(R=25 \Omega, L=\) \(6.0 \mathrm{mH}\), and \(C=25 \mu \mathrm{F}\). The circuit is connected to \(10-\mathrm{V}(\mathrm{rms}), 600-\mathrm{Hz} \mathrm{AC}\) source. (a) Is the sum of the voltage drops across \(R, L\), and \(C\) equal to \(10 \mathrm{~V}(\mathrm{rms}) ?\) (b) Which is greatest, the power delivered to the resistor, to the capaci tor, or to the inductor? (c) Find the average power deliv ered to the circuit.

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