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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`? |
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Answer» Correct Answer - `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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