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.

Answer 1

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