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`?

Explanation

`R =120 Omega, omega_(r) =4 xx 10^(5) `rad//sec
`V_(R) =60 V, V_(L) =40V`
`V_(R) =iR implies 60 =I (120) implies I = (1)/(2)A`
`V_(L) = iX_(L) implies 40 = (1)/(2) X_(L) implies X_(L) =80Omega =X_(C)`
`X_(L) = omega_(r)L implies 80 = 4 xx 10^(5) L implies L = 2 xx 10^(-4) H`
`X_(C) = (1)/(omega_(r)C) =80 = (1 )/(4 xx 10^(5) C) `
`C = (10^(-6))/(32) = (1)/(32) muF`
Phase difference between voltage and current `=45^(@)`
`tan phi = (X_(L) -X_(C))/(R) implies tan 45^(@) = (X_(L) -X_(C))/(R)`
`X_(L) -X_(C) =R`
`omegaL - (1)/(omegaC) =R`
`omegaL - (1)/(omegaC) =R`
`LComega^(2) - RC omega -1 =0`
`omega = (RC +-sqrt(R^(2) C^(2) + 4LC))/(2LC)`
`=(RC + sqrt(R^(2) C^(2) + 4LC))/(2LC)`
`[since omega+- -ve]`
`= (1)/(2)[ sqrt((R^(2))/(L^(2))+ (4)/(LC))]`
`= (1)/(2) [ 6 xx 10^(5) + sqrt(36 xx 10^(10) + 64 xx10^(10))]`
`=(1)/(2) [ 6 xx 10^(5) + 10 xx 10^(5)] = 8 xx 10^(5) rad//sec`