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An Wectromagnetic Car Alarm. Your latest invention is a car alarm that produces sound at a particularly annoying frequency of \(3500 \mathrm{~Hz}\). 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 \(12.0 \mathrm{~V}\). To produce a sufticiently 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
The values of capacitance and inductance for the car-alarm circuit should be approximately \(0.222 \mu F\) and \(0.165 H\) respectively.

Step by step solution

01

Resonant Frequency Formula

First, use the resonant frequency formula for an LC circuit which is \(f = \frac{1}{2\pi \sqrt{LC}}\). Substitute the given frequency \(f = 3500 Hz\) into the formula to obtain the relationship between the capacitance \(C\) and inductance \(L\). This forms the equation \(1 = 2\pi f \sqrt{LC}\) or simplified to \(LC = \frac{1}{(2\pi f)^2}\).
02

Capacitor Energy Formula

Second, use the formula for the energy \(U\) stored in a capacitor, which is \(U=\frac{1}{2}CV^2\). Substitute the given energy \(U = 0.0160 J\) and the given voltage \(V=12.0 V\) into the formula to solve for the capacitance \(C\). This forms the equation \(C = \frac{2U}{V^2}\).
03

Solve for Capacitance (C)

Substituting for \(U\) and \(V\), capacitance \(C\) becomes \(C = \frac{2*0.016 J}{(12 V)^2}\), solving for which gives \(C \approx 0.222 \mu F\).
04

Solve for Inductance (L)

Substitute the value of \(C\) derived in step 3 into the relationship we derived in step 1 between \(L\) and \(C\) to solve for the Inductance \(L\). This gives \(L = \frac{1}{(2\pi f)^2 C}\). Substituting the values of \(f\) and \(C\) into this, we get \(L \approx 0.165 H\).

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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 component in many electronic devices, including car alarms. It consists of an inductor (L) and a capacitor (C) connected in series or parallel. The beauty of an LC circuit lies in its ability to oscillate current at a particular frequency, known as the resonant frequency. This resonance is crucial for applications like generating sound at a specific frequency, as in the car alarm example.

The resonant frequency of an LC circuit is determined by the formula:
  • \[ f = \frac{1}{2\pi \sqrt{LC}} \]
This formula establishes a link between the inductance (L) and capacitance (C), dictating the frequency at which the circuit oscillates. At this frequency, energy transfers back and forth between the inductor and the capacitor, resulting in a sustained oscillation. This principle allows your car alarm to emit a tone at the desired frequency, enhancing its effectiveness in alerting of unauthorized access.
Capacitor Energy
Capacitors are electronic components that store and release electrical energy. In the context of the LC circuit for a car alarm, the capacitor must hold a specific amount of energy to produce adequate sound. This energy storage capability is described by the formula for the energy (\(U\)) stored in a capacitor.
  • \[ U = \frac{1}{2} C V^2 \]
Here, \(U\) is the energy stored, \(C\) is the capacitance, and \(V\) is the voltage across the capacitor. For the car alarm to produce a loud enough sound, it needs to store 0.0160 J of energy. Setting this condition allows us to solve for the value of capacitance required. By applying the formula, one can adjust the capacitor's specifications to store the needed energy for the car alarm's operation. This concept is pivotal in ensuring the alarm is not just heard but is also effective in its purpose.
Inductance and Capacitance Calculation
Determining the correct values of inductance (L) and capacitance (C) is a critical step in designing an LC circuit, such as a car alarm. First, using the resonant frequency equation,\(LC = \frac{1}{(2\pi f)^2}\), we derived a relationship between these two variables. This allows us to establish a foundation for further calculations.

Once the capacitance (\(C\)) is calculated using the capacitor energy formula (where \(C \approx 0.222 \mu F\)), the inductance (\(L\)) can then be determined by substituting this value back into the resonant frequency relationship. This calculation gives us the inductance required to achieve resonance at the desired frequency of 3500 Hz, which is approximately \(0.165 H\).

This step-wise approach ensures the car alarm circuit is finely tuned to produce sound at the correct frequency. It also highlights the interdependency between inductance and capacitance, where altering one will affect the necessary value of the other to maintain the system's performance.

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Most popular questions from this chapter

(a) What would have to be the self-inductance of a solenoid for it to store \(10.0 \mathrm{~J}\) of energy when a \(2.00 \mathrm{~A}\) current runs through it? (b) If this solenoid's cross-sectional diameter is \(4.00 \mathrm{~cm}\), and if you could wrap its coils to a density of 10 coils \(/ \mathrm{mm}\), how long would the solenoid be? (See Exercise \(30.11 .)\) Is this a realistic length for ordinary laboratory use?

ln \operatorname{an} L-C\( circuit, \)L=85.0 \mathrm{mH}\( and \)C=3.20 \mu \mathrm{F}\(. During the oscillations the maximum current in the inductor is \)0.850 \mathrm{~mA}\(. (a) What is the maximum charge on the capacitor? (b) What is the magnitude of the charge on the capacitor at an instant when the current in the inductor has magnitude \)0.500 \mathrm{~mA} ?$

If part of the magnet develops resistance and liquid helium boils away, rendering more and more of the magnct nonsuperconducting. how will this quench affect the time for the current to drop to half of its initial value? (a) The time will be shorter bccause the resistance will increase; (b) the time will be longer because the resistance will increase: (c) the time will be the same; (d) not enough information is given.

An air-filled toroidal solcnoid has 300 turns of wire, a mean radius of \(12.0 \mathrm{~cm}\), and a cross-sectional area of \(4.00 \mathrm{~cm}^{2}\). If the current is \(5.00 \mathrm{~A}\), calculate: (a) the magnetic ficld in the solenoid; (b) the selfinductance of the solcnoid: (c) the cnergy stored in the magnctic ficld: (d) the cnergy density in the magnetic ficld. (c) Check your answer for part (d) by dividing your answer to part (c) by the volume of the solenoid.

An air-filled toroidal solenoid has a mean radius of \(15.0 \mathrm{~cm}\) and a cross-sectional area of \(5.00 \mathrm{~cm}^{2}\). When the current is \(12.0 \mathrm{~A}\), the encrgy stored is \(0.390 \mathrm{~J}\). How many turns does the winding have?

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