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An Electromagnetic Car Alarm. Your latest invention is a car alarm that produces sound at a particularly annoying frequency of 3500 Hz. To do this, the car-alarm circuitry must produce an alternating electric current of the same frequency. That's why your design includes an inductor and a capacitor in series. The maximum voltage across the capacitor is to be 12.0 \(\mathrm{V}\) (the same voltage as the car battery. To produce a sufficiently loud sound, the capacitor must store 0.0160 \(\mathrm{J}\) of energy. What values of capacitance and inductance should you choose for your car-alarm circuit?

Short Answer

Expert verified
C ≈ 222 μF, L ≈ 9.18 mH.

Step by step solution

01

Find the Capacitance

First, let's find the capacitance using the formula for energy stored in a capacitor: \[ U = \frac{1}{2} C V^2 \]where \( U = 0.0160 \ \mathrm{J} \) is the energy, and \( V = 12.0 \ \mathrm{V} \) is the voltage. Rearrange the formula to solve for capacitance \( C \):\[ C = \frac{2U}{V^2} \]Substitute the values: \[ C = \frac{2 \times 0.0160}{12.0^2} = \frac{0.0320}{144} = 0.000222 \ \mathrm{F} \]Thus, \( C \) is approximately \( 222 \; \mu \mathrm{F} \).
02

Find the Inductance

To find the inductance, use the resonant frequency formula for an LC circuit:\[ \omega = 2\pi f = \frac{1}{\sqrt{LC}} \]where \( f = 3500 \ \mathrm{Hz} \) is the frequency and \( C = 222 \times 10^{-6} \ \mathrm{F} \). Rearrange the formula to solve for the inductance \( L \):\[ L = \frac{1}{(2\pi f)^2 C} \]Substitute the values:\[ L = \frac{1}{(2\pi \times 3500)^2 \times 222 \times 10^{-6}} = \frac{1}{(21991)^2 \times 222 \times 10^{-6}} \approx 9.18 \times 10^{-3} \ \mathrm{H} \]Thus, \( L \) is approximately \( 9.18 \ \mathrm{mH} \).

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Key Concepts

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

Capacitance
Capacitance is a measure of a capacitor's ability to store charge. In simpler terms, it's about how much electric charge a capacitor can hold at a certain voltage. The formula to find this is based on the amount of energy stored in the capacitor, given by:\[ U = \frac{1}{2} C V^2 \]where \( U \) is the energy in joules, \( C \) is the capacitance in farads, and \( V \) is the voltage across the capacitor in volts. This tells us how the energy stored depends on the capacitance and the voltage squared. In the context of a car alarm, selecting a capacitor with the right capacitance ensures that it can store enough energy to produce a loud noise.
  • Capacitance is essential for controlling how much energy is stored.
  • In LC circuits, capacitance affects the timing of oscillations.
  • It is crucial for determining the response of the circuit to electrical changes.
By figuring out the specific capacitance value, you can tailor the circuit to meet specific electrical requirements.
Inductance
Inductance is the property of a circuit or component that allows it to oppose a change in the current flowing through it. This naturally arises in components called inductors, which are often just coils of wire. Think of inductance as a kind of electrical inertia; it resists changes to the current, much like how mass resists changes in motion.The formula involving an inductor in an LC circuit is:\[ L = \frac{1}{(2\pi f)^2 C} \]where \( L \) represents the inductance in henrys, \( f \) is the frequency, and \( C \) is the capacitance. Understanding inductance involves:
  • Knowing that it affects the rate at which current can ramp up or down.
  • Recognizing its influence on the resonant frequency of LC circuits.
  • Using it to control how responsive circuits are to changes in voltage.
In the car alarm scenario, choosing the right inductor ensures that the circuit oscillates at the desired frequency, producing the needed sound.
Resonant Frequency
The resonant frequency is the special frequency at which inductors and capacitors in an LC circuit naturally prefer to oscillate. It's like the sweet spot where the energy just seamlessly transfers between the inductor's magnetic field and the capacitor's electric field.The mathematical expression for resonant frequency \( f \) is:\[ \omega = 2\pi f = \frac{1}{\sqrt{LC}} \]
  • This relationship links capacitor capacitance \( C \), inductor inductance \( L \), and the resonant frequency \( f \).
  • Resonant frequency defines how fast the circuit naturally oscillates.
  • It's key to designing circuits that need to respond to or filter specific frequencies, such as radio or audio circuits.
In the car alarm case, calculating this frequency correctly ensures the alarm sounds at a very specific and annoying pitch, which is perfect for deterring car thieves.

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Most popular questions from this chapter

Two toroidal solenoids are wound around the same form so that the magnetic field of one passes through the turns of the other. Solenoid 1 has 700 turns, and solenoid 2 has 400 turns. When the current in solenoid 1 is 6.52 \(\mathrm{A}\) , the average flux through each turn of solenoid 2 is 0.0320 \(\mathrm{Wb}\) . (a) What is the mutual inductance of the pair of solenoids? (b) When the current in solenoid 2 is 2.54 \(\mathrm{A}\) , what is the average flux through each turn of solenoid 1 1 ?

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?

A 10.0 -cm-long solenoid of diameter 0.400 \(\mathrm{cm}\) is wound uniformly with 800 turns. A second coil with 50 turns is wound around the solenoid at its center. What is the mutual inductance of the combination of the two coils?

An \(L-C\) circuit containing an \(80.0-\mathrm{mH}\) inductor and a \(1.25-\mathrm{nF}\) capacitor oscillates with a maximum current of 0.750 \(\mathrm{A}\) . Calculate: (a) the maximum charge on the capacitor and (b) the oscillation frequency of the circuit. (c) Assuming the capacitor had its maximum charge at time \(t=0\) , calculate the energy stored in the inductor after 2.50 \(\mathrm{ms}\) of oscillate the energy stored in.

One solenoid is centered inside another. The outer one has a length of 50.0 \(\mathrm{cm}\) and contains 6750 coils, while the coaxial inner solenoid is 3.0 \(\mathrm{cm}\) long and 0.120 \(\mathrm{cm}\) in diameter and contains 15 coils. The current in the outer solenoid is changing at 49.2 A/s. (a) What is the mutual inductance of these solenoids? (b) Find the emf induced in the innner solenoid.

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