Question

A series L-C-R circuit containing a resistance of 120Omega has resonance frequency 4xx10^5rad//s. At resonance the voltages across resistance and inductance are 60V and 40V, respectively. Find the values of L and C.At what angular frequency the current in the circuit lags the voltage by pi//4?

Answer 1

The value of angular frequency is ω = 8 × 10^5 rad/s.

Explanation:

At resonance XL − XC = 0

and Z = R = 120 Ω

irms = (VR)rms / R = 60 / 4120 = 12 A

Also, irms = (VL) rms ωL

L  =(VL)rms / ω(irms) = 40 / (4×10^5(1 / 2)

L = 2.0 × 10^−4 H

L = 0.2 mH

The resonance frequency is given by

ω = 1  / √LC  

or C = 1 / ωL

Substituting the values, we have

C = 1 / (4 × 10^5)^2 (2.0 × 10^−4)

C = 3.125 × 10^−8 F

Current lags the voltage by 45∘ when

tan (45∘) = ωL− 1 / ωC ÷ R

Substituting the values of L,C,R and tan(45∘) we get

ω = 8 × 10^5 rad/s.

Hence the value of angular frequency is ω = 8 × 10^5 rad/s.