Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 12.0V . To produce a sufficiently loud sound, the capacitor must store 0.0160 J of energy. What values of capacitance and inductance should you choose for your car-alarm circuit?

Short Answer

Expert verified

A) The value of capacitance is222μF

B) The value of inductance is9.31μH

Step by step solution

01

Concept of the energy in the capacitor

The energy in capacitor is Uc=12CV2where, C is the capacitance and V is the voltage

02

Calculate the values of capacitance

From the given value of the stored energy Uc in the capacitor, we can calculate the capacitance C

C=2UcV2=20.016J12V2=222×10-6F=222μF

Therefore, the value of capacitance is222μF

03

Calculate the values of inductance

For the self-inductance, we use the given value of frequency f from equation in the formL=12πf2C.Substitute the values in the given equation we get,

L=12π3500Hz2222μF=9.31μH

Therefore, the value of inductance is9.31μH

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In the circuit shown in Fig. E25.30, the 16.0-V battery is removed and reinserted with the opposite polarity, so that its negative terminal is now next to point a. Find (a) the current in the circuit (magnitude anddirection); (b) the terminal voltage Vbaof the 16.0-V battery; (c) the potential difference Vacof point awith respect to point c. (d) Graph the potential rises and drops in this circuit (see Fig. 25.20).

In the circuit shown in Fig. E26.18,ε=36.V,R1=4.0Ω,R2=6.0Ω,R3=3.0Ω(a) What is the potential difference Vab between points a and b when the switch S is open and when S is closed? (b) For each resistor, calculate the current through the resistor with S open and with S closed. For each resistor, does the current increase or decrease when S is closed?

A 1.50-mcylindrical rod of diameter 0.500cmis connected to

a power supply that maintains a constant potential difference of 15.0Vacross

its ends, while an ammeter measures the current through it. You observe that

at room temperature (20.0C)the ammeter reads 18.5Awhile at 92.0Cit

reads 17.2A. You can ignore any thermal expansion of the rod. Find (a) the

resistivity at and (b) the temperature coefficient of resistivity at for the material of the rod.

An electron at point in figure has a speed v0=1.41×106m/s. Find (a) the magnetic field that will cause the electron to follow the semicircular path from to and (b) The time required for the electron to move fromAtoB.

The power rating of a light bulb (such as a 100-W bulb) is the power it dissipates when connected across a 120-V potential difference. What is the resistance of (a) a 100-W bulb and (b) a 60-W bulb? (c) How much current does each bulb draw in normal use?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free