Exercise 31.29 - Enhanced - with Solution In an L-R-C series circuit, 310 , 0.40

Question

Exercise 31.29 - Enhanced - with Solution In an L-R-C series circuit, 310 , 0.404 H , and 6.04×108 F . When the ac source operates at the resonance frequency of the circuit, the current amplitude is 0.504 A . You may want to review ( pages 1036 - 1038) .

Part A What is the voltage amplitude of the source? V = V SubmitMy AnswersGive Up

Part B What is the amplitude of the voltage across the resistor? VR = V SubmitMy AnswersGive Up

Part C What is the amplitude of the voltage across the inductor? VL = V SubmitMy AnswersGive Up

Part D What is the amplitude of the voltage across the capacitor? VC = V SubmitMy AnswersGive Up

Part E What is the average power supplied by the source? P = W SubmitMy AnswersGive Up

Explanation / Answer

given

R = 310 ohms

L = 0.404 H

C = 6.04*10^-8 F

A) at resosnat frequency, z = R

Vmax = Imax*z

= Imax*R

= 0.504*310

= 156 V

B) Voltage across resistor, VR = Imax*R

= 0.504*310

= 156 V

C)
resonnat frequency, fo = 1/(2*pi*sqrt(L*C))

= 1/(2*pi*sqrt(0.404*6.04*10^-8))

= 1019 Hz

Xc = XL = 2*pi*f*L

= 2*pi*1019*0.404

= 2587 ohms

VL = XL*Imax

= 2587*0.504

= 1304 V

d) VC = XC*Imax

= 2587*0.504

= 1304 V

f) the average power supplied by the source,

Pavg = Imax^2*R/2

= 0.504^2*310/2

= 39.4 W

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