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Chapter 21: Problem 62

In the crossover network of Problem \(61,\) the inductance \(L\) is $1.20 \mathrm{mH}$. The capacitor is variable; its capacitance can be adjusted to set the crossover point according to the frequency response of the woofer and tweeter. What should the capacitance be set to for a crossover point of $180 \mathrm{Hz} ?[\text {Hint}:$ At the crossover point, the currents are equal in amplitude. \(]\)

Short Answer

Expert verified
Question: Determine the capacitance of a capacitor in a crossover network, given that the crossover point is at 180 Hz and the inductor has an inductance of 1.20 mH. Answer: The capacitance should be set to approximately 6.15 µF for a crossover point of 180 Hz.

Step by step solution

01

Write down the given values

Inductance \(L = 1.20 \times 10^{-3} \mathrm{H}\) Crossover frequency \(f = 180 \mathrm{Hz}\)
02

Calculate the angular frequency

\(\omega = 2 \pi f\) \(\omega = 2 \pi (180)\) \(\omega = 1130.97 \mathrm{rad/s}\)
03

Write the equation of equal magnitudes of inductive and capacitive impedance

At the crossover point, we have: \(|Z_L| = |Z_C|\) \(|j\omega L| = |\frac{1}{j\omega C}|\)
04

Solve the equation for capacitance

Remove the j from both sides: \(\omega L = \frac{1}{\omega C}\) Now, solve for C: \(C = \frac{1}{\omega^2 L}\)
05

Plug in the values and calculate the capacitance

\(C = \frac{1}{(1130.97)^2 (1.20 \times 10^{-3})}\) \(C = 6.15489 \times 10^{-6} \mathrm{F}\) So, the capacitance should be set to approximately \(6.15 \mu \mathrm{F}\) for a crossover point of 180 Hz.

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Most popular questions from this chapter

A 4.00 -mH inductor is connected to an ac voltage source of \(151.0 \mathrm{V}\) rms. If the rms current in the circuit is \(0.820 \mathrm{A},\) what is the frequency of the source?
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