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A series L-C-R circuit containing a resi...

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`?

Text Solution

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At resonance `X_L-X_C=0`
and `Z=R=120Omega`
`:. i_("rms")=((V_R)_("rms"))/R=60/4120=1/2A`
Also, `i_("rms")=((V_L)_("rms"))/(omegaL)`
`:. L=((V_L)_("rms"))/(omegai_("rms"))=40/((4xx10^5(1/2)`
`=2.0xx10^-4H`
`=0.2mH`
The resonance frequency is given by
`omega=1/(sqrt(LC))` or `C=1/(omegaL)`
Substituting the values, we have
`C=1/((4xx10^5)^2(2.0xx10^-4))`
`=3.125 xx 10^(-8) F`
Current lags the voltage by `45^@` when
`tan45^@=(omegaL-1/(omegaC))/R`
Substituting the values of `L,C,R` and `tan45^@` we get
`omega=8xx10^5rad//s`.
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