In an L-R-C series circuit, 305 , 0.401 , and 5.98×10-8 . When the ac source operates at the resonance frequency of the circuit, the current

Question

In an L-R-C series circuit, 305 , 0.401 , and 5.98×10-8 . When the ac source operates at the resonance frequency of the circuit, the current amplitude is 0.491A
What is the voltage amplitude of the source?
What is the amplitude of the voltage across the resistor?
What is the amplitude of the voltage across the inductor?
What is the amplitude of the voltage across the capacitor?
What is the average power supplied by the source?

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Khánh Gia 3 years 2021-08-31T18:34:37+00:00 1 Answers 14 views 0

Answers ( )

    0
    2021-08-31T18:35:56+00:00

    Answer:

    • the amplitude voltage of the source is 0.197 V
    • the amplitude of the voltage across the resistor is 0.197 V
    • the amplitude of the voltage across the inductor is 35,063.91 V
    • the amplitude of the voltage across the capacitor is 35,067.22 V
    • the average power supplied by the source is 0.0967 V

    Explanation:

    Given;

    inductance, L = 305 H

    Resistance , R = 0.401 Ω

    Capacitance, C = 5.98 x 10⁻⁸ F

    Amplitude current, I₀ = 0.491A

    resonance frequency (F₀) = ?

    F_o = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{305*5.98*10^{-8}}} = 37.26 \ Hz

    Part (a) What is the voltage amplitude of the source?

    V₀ = I₀ Z

    Z is the impedance, since the circuit is in resonance Z = R

    V₀ = I₀ Z = I₀ R = 0.491 x 0.401 = 0.197 V

    Part (b) What is the amplitude of the voltage across the resistor?

    V₀ =  I₀ R = 0.491 x 0.401 = 0.197 V

    Part (c) What is the amplitude of the voltage across the inductor?

    V_0 =I_o*X_L\\\\X_L = 2\pi F_0L

    Where XL is the inductive reactance

    V_0 =I_o*2\pi F_0L\\\\V_o =0.491*2\pi*37.26*305 = 35,063.91 \ V

    Part (d) What is the amplitude of the voltage across the capacitor?

    V_0 =I_o*X_C\\\\X_C = \frac{1}{2\pi F_0C}

    Where XC is the capacitive reactance

    V_o = I_o*\frac{1}{2\pi F_0C} = \frac{0.491}{2\pi *37.26*5.98*10^{-8}} = 35,067.22 \ V

    Part (e) What is the average power supplied by the source?

    Average \ power, P = I_o\sqrt{V_R^2 +(V_L-V_C)^2} \\\\ P = I_o\sqrt{V_R^2 } = I_oV_R\\\\P = 0.491*0.197 =0.0967 \ V

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