A series LCR circuit containing a resistance of `120Omega` has angular resonance frequency `4times10^5rad//s` .At resonance, the voltages across resistance and inductancr are 60V and 40V, respectively. Find the values of L and C. At what frequency , does not current in the circuit lag behind the voltage by `45^@`?

Explanation

At resonance, `X_L=X_C`
`thereforeI=V_R/R`
`because`voltage acorrs the resistance, `V_R=60V`)
=0.5A
Now, voltage across the indicator,
`V_L=IX_L=IomegaL`
or,`L=V_L/(Iomega)=40/(0.5times4times10^5)`
[`because` angular frequency, `omega=4times10^5rad//s`]
`=2times10^-4H`
We know that at resonance,
`X_L=X_C`or,`omegaL=1/(omegaC)`
or,`C=1/(omega^2L)=1/((4times10^5)^2times0.2times10^-3)=3.125times10^-8F`
Let the angular frequency be `omega_1` when the current lags behind the voltage by `45^@`.
`thereforetan45^@=(omega_1L-1/(omega_1C))/R`
`[C=3.125times10^-8F=1/32times10^6F]`
or,`1times120=omega_1times2times10^-4-1/(omega_1(1/32)times10^-6)`
or,`omega_1^2-6times10^5omega_1-16times10^10=0`
The physically meaningful solution of the above equation is,
`omega_1=(6times10^5times10times10^5)/2=8times10^5rad//s`