Question

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

A) The value of capacitance is$222\mathrm{\mu F}$

B) The value of inductance is$9.31\mathrm{\mu H}$

Step-by-step solution

Concept of the energy in the capacitor

The energy in capacitor is ${\mathbf{U}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{CV}}^{\mathbf{2}}$where, C is the capacitance and V is the voltage

Calculate the values of capacitance

From the given value of the stored energy ${\mathrm{U}}_{\mathrm{c}}$ in the capacitor, we can calculate the capacitance C

$$\begin{gathered} \mathrm{C}=\frac{2{\mathrm{U}}_{\mathrm{c}}}{{\mathrm{V}}^2} \\ =\frac{2\left(0.016\mathrm{J}\right)}{{\left(12\mathrm{V}\right)}^2} \\ =222\times {10}^{-6}\mathrm{F} \\ =222\mathrm{\mu F} \end{gathered}$$

Therefore, the value of capacitance is$222\mathrm{\mu F}$

Calculate the values of inductance

For the self-inductance, we use the given value of frequency f from equation in the form$\mathrm{L}=\frac{1}{{\left(2\mathrm{\pi f}\right)}^2\mathrm{C}}$.Substitute the values in the given equation we get,

$$\begin{gathered} \mathrm{L}=\frac{1}{2\mathrm{\pi }{\left(3500\mathrm{Hz}\right)}^2\left(222\mathrm{\mu F}\right)} \\ =9.31\mathrm{\mu H} \end{gathered}$$

Therefore, the value of inductance is$9.31\mathrm{\mu H}$