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Chapter 30: Problem 36

A Radio Tuning Circuit. The minimum capacitance of a variable capacitor in a radio is 4.18 \(\mathrm{pF}\) . (a) What is the inductance of a coil connected to this capacitor if the oscillation frequency of the \(L\) circuit is \(1600 \times 10^{3} \mathrm{Hz}\) , corresponding to one end of the AM radio broadcast band, when the capacitor is set to its minimum capacitance? (b) The frequency at the other end of the broadcast band is \(540 \times 10^{3}\) Hz. What is the maximum capacitance of the capacitor if the oscillation frequency is adjustable over the range of the broadcast band?

Short Answer

Expert verified
(a) Find \( L \), then (b) find max \( C \) using new \( f \).

Step by step solution

01

Understand the Formula

The resonant frequency of an "L-C" circuit is given by the formula \( f = \frac{1}{2\pi\sqrt{LC}} \), where \( f \) is the frequency, \( L \) is the inductance, and \( C \) is the capacitance. We'll use this formula to find \( L \) for part (a) and the maximum capacitance for part (b).
02

Solve for Inductance - Setup Equation

For part (a), rearrange the formula to solve for \( L \):\[ L = \frac{1}{(2\pi f)^2 C} \] Substitute the given values: \( f = 1600 \times 10^3 \) Hz and \( C = 4.18 \times 10^{-12} \) F.
03

Solve for Inductance - Calculate

Plugging in the values:\[ L = \frac{1}{(2\pi \cdot 1600 \times 10^3)^2 \times 4.18 \times 10^{-12}} \]Calculate to find \( L \).
04

Solve for Maximum Capacitance - Setup Equation

For part (b), use the same formula rearranged to solve for \( C \):\[ C = \frac{1}{(2\pi f)^2 L} \]Now, substitute \( f = 540 \times 10^3 \) Hz and the value of \( L \) obtained from part (a).
05

Solve for Maximum Capacitance - Calculate

Substitute into the equation:\[ C = \frac{1}{(2\pi \cdot 540 \times 10^3)^2 \times L} \]Compute to find the maximum capacitance \( C \).
06

Conclusion

Based on the calculations:- For part (a), the inductance \( L \) is approximately found from the given frequency and capacitance.- For part (b), substituting the obtained inductance into the frequency equation gives the maximum capacitance.

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Key Concepts

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

Resonant Frequency
The resonant frequency is an essential aspect of radio tuning circuits. It represents the frequency at which the circuit naturally oscillates with the greatest amplitude. This is the key frequency where maximum energy transfer occurs between the inductor and the capacitor in the circuit. Mathematically, it is given by the formula:\[ f = \frac{1}{2\pi\sqrt{LC}} \]where:- \(f\) is the resonant frequency,- \(L\) is the inductance in henrys (H),- \(C\) is the capacitance in farads (F).This formula is crucial in determining how a radio tuning circuit can select different frequencies. By adjusting either the inductance or the capacitance, the circuit can tune into different stations on the AM band.
Inductance
Inductance is a critical component in radio tuning circuits. It refers to the property of a coil, which resists changes in electric current through it. Inductance is measured in henrys (H) and affects how the circuit interacts with the magnetic fields generated by alternating current.- The larger the inductance, the more the coil opposes the change in current, making the circuit resonate at a lower frequency if the capacitance remains constant.- The coil, often called an inductor, stores energy in the form of a magnetic field when electric current flows through it.In our exercise, finding the inductance was essential for setting the resonant frequency at the higher end of the AM band. By rearranging the resonant frequency formula to solve for \(L\):\[ L = \frac{1}{(2\pi f)^2 C} \]students learn how to calculate the inductance needed to achieve a specific resonant frequency.
Capacitance
Capacitance is another vital component in radio tuning circuits, defined as the ability of a system to store an electric charge. Capacitance is measured in farads (F), and in radio circuits, it plays a crucial role in determining resonant frequency alongside inductance.- A larger capacitance means more electric charge can be stored, which will lower the resonant frequency if the inductance remains constant.- Variable capacitors are used in tuning circuits to allow adjustment across a desired frequency range, such as the AM broadcast band.For example, the exercise required calculating the maximum capacitance of a capacitor to cover the entire range of the AM band. By using the formula rearranged for capacitance:\[ C = \frac{1}{(2\pi f)^2 L} \]students can understand how capacitance affects the ability of a circuit to select specific frequencies.

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Most popular questions from this chapter

A 7.50 -nF capacitor is charged up to 12.0 \(\mathrm{V}\) , then disconnected from the power supply and connected in series through a coil. The period of oscillation of the circuit is then measured to be \(8.60 \times 10^{-5}\) s. Calculate: (a) the inductance of the coil; (b) the maximum charge on the capacitor; (c) the total energy of the circuit; \((\mathrm{d})\) the maximum current in the circuit.

Two coils are wound around the same cylindrical form, like the coils in Example \(30.1 .\) When the current in the first coil is decreasing at a rate of \(-0.242 \mathrm{A} / \mathrm{s}\) , the induced emf in the second coil has magnitude 1.65 \(\times 10^{-3} \mathrm{V}\) . (a) What is the mutual inductance of the pair of coils? (b) If the second coil has 25 turns, what is the flux through each turn when the current in the first coil equals 1.20 \(\mathrm{A} ?\) (c) If the current in the second coil increases at a rate of \(0.360 \mathrm{A} / \mathrm{s},\) what is the magnitude of the induced emf in the first coil?

A solenoidal coil with 25 turns of wire is wound tightly around another coil with 300 turns (see Example 30.1\() .\) The inner solenoid is 25.0 \(\mathrm{cm}\) long and has a diameter of 2.00 \(\mathrm{cm} .\) At a certain time, the current in the inner solenoid is 0.120 \(\mathrm{A}\) and is increasing at a rate of \(1.75 \times 10^{3} \mathrm{A} / \mathrm{s}\) . For this time, calculate: (a)the average magnetic flux through each turn of the inner solenoid; (b) the mutual inductance of the two solenoids; (c) the emf induced in the outer solenoid by the changing current in the inner solenoid.

In a proton accelerator used in elementary particle physics experiments, the trajectories of protons are controlled by bending magnets that produce a magnetic field of 4.80 T. What is the magnetic-field energy in a 10.0 -cm' volume of space where \(B=4.80 \mathrm{T} ?\)

A toroidal solenoid with mean radius \(r\) and cross-sectional area \(A\) is wound uniformly with \(N_{1}\) turns. A second toroidal solenoid with \(N_{2}\) turns is wound uniformly on top of the first, so that the two solenoids have the same cross-sectional area and mean radius. (a) What is the mutual inductance of the two solenoids? Assume that the magnetic field of the first solenoid is uniform across the cross section of the two solenoids. (b) If \(N_{1}=500\) turns, \(N_{2}=300\) turns, \(r=10.0 \mathrm{cm},\) and \(A=0.800 \mathrm{cm}^{2},\) what is the value of the mutual inductance?
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