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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^@`? |
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Answer» 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` |
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