In a home stereo system, low sound frequencies are handled by large “woofer” speakers, and high frequencies by smaller “tweeter” speakers. F

Question

In a home stereo system, low sound frequencies are handled by large “woofer” speakers, and high frequencies by smaller “tweeter” speakers. For the best sound reproduction, low-frequency currents from the amplifier should not reach the tweeter. One way to do this is to place a capacitor in series with the 9.0 Ω resistance of the tweeter; one then has an RLC circuit with no inductor L (that is, an RLC circuit with L = 0).
What value of C should be chosen so that the current through the tweeter at 200 Hz is half its value at very high frequencies? Express your answer with the appropriate units.

in progress 0
Kim Chi 4 years 2021-09-01T01:24:15+00:00 1 Answers 33 views 0

Answers ( )

  1. vankhanh
    0
    2021-09-01T01:25:52+00:00
    Reply

    Answer:

    C = 2.9 10⁻⁵ F = 29 μF

    Explanation:

    In this exercise we must use that the voltage is

              V = i X

              i = V/X    

    where X is the impedance of the system

    in this case they ask us to treat the system as an RLC circuit in this case therefore the impedance is

              X = \sqrt{R^2 + ( wL - \frac{1}{wC})^2 }

    tells us to take inductance L = 0.

    The angular velocity is

             w = 2π f

    the current is required to be half the current at high frequency.

    Let’s analyze the situation at high frequency (high angular velocity) the capacitive impedance is very small

             \frac{1}{wC} →0       when w → ∞

    therefore in this frequency regime

             X₀ = \sqrt{R^2 + ( \frac{1}{2\pi 2 10^4 C} )^2 } = R \sqrt{ 1+ \frac{8 \ 10^{-10} }{RC} }

    the very small fraction for which we can despise it

            X₀ = R

    to halve the current at f = 200 H, from equation 1 we obtain

             X = 2X₀

    let’s write the two equations of inductance

              X₀ = R                                    w → ∞

              X= 2X₀ = \sqrt{R^2 +( \frac{1}{wC} )^2 }        w = 2π 200

     

             

    we solve the system

             2R = \sqrt{R^2 +( \frac{1}{wC} )^2 }

             4 R² = R² + 1 / (wC) ²

             1 / (wC) ² = 3 R²

              w C = \frac{1}{\sqrt{3} } \ \frac{1}{R}

              C = \frac{1}{\sqrt{3} } \ \frac{1}{wR}

               

    let’s calculate

               C = \frac{1}{\sqrt{3} } \ \frac{1}{2\pi \ 200 \ 9}

               C = 2.9 10⁻⁵ F

               C = 29 μF

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )