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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 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 1.65×10−2 J .

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Answer:

Step-by-step explanation:

There is a "?" mark in the amount of energy stored.I am taking it as 1.65*10^-2 J. I am also mentioning the formulae used to solve the problem.If the energy value is different from the one i used then plug into the formula correct energy value and get solution

(a)Energy stored by a capacitor =
(0.5)Cv^2 = 1.65*10^(-2)\\\\</p><p>(0.5)C* 12^2 = 1.65*10^(-2)</p><p>\\\\C = 229*10^(-6) F

(b)
f = \frac{1}{(2\pi √(LC))\\\\</p><p>L =(1)/((2\pi f)^(2)C))\\\\</p><p>L =(1)/((2\pi * 3500)^(2)* 229* 10^(-6))\\\\L = 9.03* 10^(-6) H

User Mathias Nielsen
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