Question

In an L-R-C series circuit, 310 ? , 0.399 H , and 6.01×10^?8 F . When the ac source operates at the resonance frequency of the circuit, the current amplitude is 0.508 A

a. What is the voltage amplitude of the source?

b. What is the amplitude of the voltage across the resistor?

c. What is the amplitude of the voltage across the inductor?

d. What is the amplitude of the voltage across the capacitor?

e. What is the average power supplied by the source?

Answer 1

Ans:-

Given data:- R=310? L= 0.399H, C= 6.01*10^-8F, I= 0.508A

Part A

Resonance frequency of the circuit w = 1/sqrt(LC)

= 1/sqrt(0.399*6.01*10^-8)=0.6458*10^4rad/s

Xc = 1/(wC)= 1/(0.6458*10^4*6.01*10^-8 = 0.2577*10^4?=2577?

XL = wL = 0.6458*10^4*0.399=2576.74?

Z= sqrt(R^2 +(XL – Xc)^2) =310?

I= V/Z

V= I*Z = 0.508*310 =157.48V

Partb

Vr = I*R = 0.508 * 310 = 157.48 V

Partc
Vl = I*XL = 1308.98 V

Partd
Vc = I*XC = 1309.12 V

Parte

Power delivered = V*IPowerfactor = 157.48*0.508*1 = 80W