contestada

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?

Respuesta :

onyebuchinnaji

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) = ?

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

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?

[tex]V_0 =I_o*X_L\\\\X_L = 2\pi F_0L[/tex]

Where XL is the inductive reactance

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

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

[tex]V_0 =I_o*X_C\\\\X_C = \frac{1}{2\pi F_0C}[/tex]

Where XC is the capacitive reactance

[tex]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[/tex]

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

[tex]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[/tex]

Otras preguntas