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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`
`therefore I_(rms) = ((V_(R)_(rms)))/R= 60/120 = 1/2A`
Also, `I_(rms) = V_(L)_(rms))/(omegaL)`
`therefore L=(V_(L)_(rms))/(omegaI_(rms)) = 40/((4x10^(5))(1/2))`
`=2.0 xx 10^(-4)H = 0.2mH`
The resonance frequency is given by, `omega=1/(sqrt(LC))` or `C=1/(omega^(2)L)` Substituting the values, we have
`C= 1/((4xx10^(5))^(2) (2.0 xx 10^(-4))) = 3.125 xx 10^(-8)` F
Current lags the voltage by `45^(@)`, when tan`45^(@)=(omegaL-1/(omegaC))/(R)`
Substituting the values of L,C,R and `tan45^(@)` ltrbgt We get, `omega=8xx10^(5) rad s^(-1)`
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