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Current flows to the midrange speaker (9.0 ohms) in a loudspeaker system through a 2.4-millihenry...

Question:

Current flows to the midrange speaker ({eq}9.0 \, \mathrm{\Omega} {/eq}) in a loudspeaker system through a 2.4-millihenry inductor in series with a capacitor ({eq}13.5 \, \mathrm{\mu F} {/eq}). Although the maximum current flows in the speaker circuit at the resonant frequency, the peak voltage across the capacitor is a maximum at a somewhat lower frequency.

What is that frequency?

If the peak output voltage of the amplifier is {eq}25 \, \mathrm{V} {/eq}, what is the corresponding peak capacitor voltage?

Peak Voltage across the Capacitor in an RLC series Circuit

In an R-L-C series circuit the impedance of the circuit is given by {eq}Z=\sqrt{R^2 + (X_L -X_C)^2} {/eq}.

Where {eq}X_L \ and \ X_C {/eq} are the inductive reactance and capacitive reactance respectively. At resonance impedance of the circuit become minimum and maximum current flows through the circuit. The frequency at which resonance happens is given by the equation {eq}F_r = \dfrac { 1 } { 2 \pi \sqrt { LC } } {/eq}. But at this frequency voltage across the capacitor and inductor are equal. At a lower frequency than the resonance frequency the voltage across the capacitor become maximum. The frequency at which the voltage across the capacitor become maximum is given by the equation {eq}F_c = F_r \sqrt { 1 - \dfrac { R^2 C } { 2 L } } {/eq}. Here R, C, L stands for resistance , cpacitance and inductance in the circuit respectively.

Answer and Explanation:

Given:

A series {eq}RLC {/eq} circuit consists of a

  • {eq}R=9.0 \Omega {/eq} resistor,
  • {eq}L = 2.40mH = 2.40 \times 10^{-3} H {/eq} inductor,
  • {eq}...

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R-L-C Series Circuits

from UExcel Physics: Study Guide & Test Prep

Chapter 13 / Lesson 9
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