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Your latest invention is a car alarm that produces sound at a particularly annoying frequency of 3600 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 capacitor in series. The maximum voltage across the capacitor is to be 12.0 V . To produce a sufficiently loud sound, the capacitor must store an amount of energy equal to 1.55×10−2 J . What values of capacitance and inductance should you choose for your car-alarm circuit?

Respuesta :

davidlcastiblanco

Answer:

The capacitance and the inductance can choose for a car-alarm circuit are

C = 215.27 μF

L = 9.078 μH

Explanation:

[tex]V =12.0 V[/tex], [tex]E = 1.55*10^2 J[/tex], [tex]f = 3600 Hz[/tex]

To determine the capacitance can use the equation

[tex]U_c= \frac{1}{2}*C*V^2[/tex]

Solve to C'

[tex]C = \frac{U_c*2}{V^2}=\frac{1.55x10^2J*2}{12.0^2V}[/tex]

[tex]C=215.27 uF[/tex]

To find the inductance can use the frequency of the circuit

[tex]f = \frac{1}{2\pi* \sqrt{C*L} }[/tex]

Solve to L'

[tex]L = \frac{1}{4\pi^2*f^2*C}=\frac{1}{4\pi^2*3600^2*215.27 uF}}[/tex]

[tex]L = 9.078 uH[/tex]

jayilych4real

The capacitance and the inductance which can be chosen for a car-alarm circuit are

  • C = 215.27 μF
  • L = 9.078 μH

Calculations and Parameters:

Hence, we need to lay the parameters:

V= 120v

E= 1.55 x 10^2J

f= 3600Hz

Then we solve for capacitance which is Uc= 0.5 x CV^2

=>1.55 x 10^2J2/12^2V

=> 215. 27uF

Then the inductance would be:

f= 1/(2[tex]\pi[/tex][tex]\sqrt{C x L}[/tex]

=> 9.078uH

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