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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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The correct Answer is:
`0.2 mH, 1/32 muF, 8xx10^(5) rad//s`
At resonance `V_(L)=V_(C )=40V`
`V=V_(R)=60V` and `I_(max)=60/120=0.5 Amp`.
`I_(max)X_(L)=40 implies X_(L)=80 Omega`
`omegaL=80 implies L=0.2 mH`
`X_(C )=80 implies 1/(4xx10^(5))C=80`
`C=1/32xx10^(-6)=1/32 muF`
Let at frequency `omega` current lags the voltage by `45^(@)`
`tanphi=x/R implies X=R implies X_(L)-X_(C )=R`
`omegaL=1/(omegaC)=R`
Putting the value of `X_(L), X_(C )` and `R` we get `omega=8xx10^(5) rad//s`
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