Problem 57
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 12.0 \(\mathrm{V}\) (the same voltage as the car battery. To produce a sufficiently 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?
Step-by-Step Solution
Verified Answer
C ≈ 222 μF, L ≈ 9.18 mH.
1Step 1: Find the Capacitance
First, let's find the capacitance using the formula for energy stored in a capacitor: \[ U = \frac{1}{2} C V^2 \]where \( U = 0.0160 \ \mathrm{J} \) is the energy, and \( V = 12.0 \ \mathrm{V} \) is the voltage. Rearrange the formula to solve for capacitance \( C \):\[ C = \frac{2U}{V^2} \]Substitute the values: \[ C = \frac{2 \times 0.0160}{12.0^2} = \frac{0.0320}{144} = 0.000222 \ \mathrm{F} \]Thus, \( C \) is approximately \( 222 \; \mu \mathrm{F} \).
2Step 2: Find the Inductance
To find the inductance, use the resonant frequency formula for an LC circuit:\[ \omega = 2\pi f = \frac{1}{\sqrt{LC}} \]where \( f = 3500 \ \mathrm{Hz} \) is the frequency and \( C = 222 \times 10^{-6} \ \mathrm{F} \). Rearrange the formula to solve for the inductance \( L \):\[ L = \frac{1}{(2\pi f)^2 C} \]Substitute the values:\[ L = \frac{1}{(2\pi \times 3500)^2 \times 222 \times 10^{-6}} = \frac{1}{(21991)^2 \times 222 \times 10^{-6}} \approx 9.18 \times 10^{-3} \ \mathrm{H} \]Thus, \( L \) is approximately \( 9.18 \ \mathrm{mH} \).
Key Concepts
CapacitanceInductanceResonant Frequency
Capacitance
Capacitance is a measure of a capacitor's ability to store charge. In simpler terms, it's about how much electric charge a capacitor can hold at a certain voltage. The formula to find this is based on the amount of energy stored in the capacitor, given by:\[ U = \frac{1}{2} C V^2 \]where \( U \) is the energy in joules, \( C \) is the capacitance in farads, and \( V \) is the voltage across the capacitor in volts. This tells us how the energy stored depends on the capacitance and the voltage squared. In the context of a car alarm, selecting a capacitor with the right capacitance ensures that it can store enough energy to produce a loud noise.
- Capacitance is essential for controlling how much energy is stored.
- In LC circuits, capacitance affects the timing of oscillations.
- It is crucial for determining the response of the circuit to electrical changes.
Inductance
Inductance is the property of a circuit or component that allows it to oppose a change in the current flowing through it. This naturally arises in components called inductors, which are often just coils of wire. Think of inductance as a kind of electrical inertia; it resists changes to the current, much like how mass resists changes in motion.The formula involving an inductor in an LC circuit is:\[ L = \frac{1}{(2\pi f)^2 C} \]where \( L \) represents the inductance in henrys, \( f \) is the frequency, and \( C \) is the capacitance. Understanding inductance involves:
- Knowing that it affects the rate at which current can ramp up or down.
- Recognizing its influence on the resonant frequency of LC circuits.
- Using it to control how responsive circuits are to changes in voltage.
Resonant Frequency
The resonant frequency is the special frequency at which inductors and capacitors in an LC circuit naturally prefer to oscillate. It's like the sweet spot where the energy just seamlessly transfers between the inductor's magnetic field and the capacitor's electric field.The mathematical expression for resonant frequency \( f \) is:\[ \omega = 2\pi f = \frac{1}{\sqrt{LC}} \]
- This relationship links capacitor capacitance \( C \), inductor inductance \( L \), and the resonant frequency \( f \).
- Resonant frequency defines how fast the circuit naturally oscillates.
- It's key to designing circuits that need to respond to or filter specific frequencies, such as radio or audio circuits.
Other exercises in this chapter
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