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

V At resonance as`X=0,I=(V)/(R)=(60)/(120)=(1)/(2)A`
and`V_(L)=IX_(L)=IomegaL,L=(V_(L))/(Iomega)`
So`L=(40)/(0.5xx4xx10^(5))=0.25mH`
`i.e.,C=(1)/(Lomega_(@)^(2))=(1)/0.2xx10^(-3)xx(4xx10^(5))^(2)=(1)/(32)muF`
Now in case of series LCR circuit
`tanphi=(X_(L)-X_(C))/(R)`
So current will lag the applied voltage by `45^(@) if, tan 45^(@)=(omega-(1)/(omegaC))/(R)`
i.e.`1xx120=omegaxx2xx10^(-4)-(1)/(omega(1//32)xx10^(-6)`
i.e.,`omega-6xx10^(5)omega-16xx10^(10)=0`
i.e.,`omega=(6xx10^(5)pmsqrt((6xx10^(5))^(2)+64xx10^(10)))/(2)`
i.e.,`omega=(6xx10^(5)+10xx10^(5))/(2)=8xx10^(5)rad//S`