InterviewSolution
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InterviewSolution
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A series LCR circuit containing a resistance of `120 Omega` has angular resonance frequency `4 xx 10^(5) rad s^(-1)`. At resonance the vlotage across resistance and inductance are 60V and 40 V, repectively, the value of inductance L isA. `0.1 mH`B. `0.2 mH`C. `0.35 mH`D. `0.4 mH` |
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Answer» Correct Answer - B At resonance as `X=0, I+V/R =60/120 = 1/2 A` as `V_(L) = IX_(L) = I omega L, L+(V_L)/(I omega)` so, ` L+(40)/((1//2)xx4xx(10^5))=0.2 mH` i.e. `C=(1)/(0.2 xx 10^(-3) xx (4 xx 10^(5))^(2))=(1)/(32 (mu)F`. Now in case of series LCR circuit. ` tan(phi)=(X_(L)-X_(C ))/(R )` So current will lag the applied voltage by `45^(@)` if , `tan 45^(@) =(omega L -(1)/(omega L))/(R )` ` 1 xx 120 = omega xx 2 xx 10^(-4)-(1)/(omega (1//32)xx10^(-6))` `omega^(2) - 6 xx 10^(5) omega - 16 xx 10^(10)=0` i.e. `omega = (6 xx(10^5) +- sqrt((6 xx 10^(5))^(2)+64 xx 10^(10)))/(2)` i.e., `omega = (6 xx (10^5) + 10 xx (10^5))/(2) = 8 xx 10^(5) rad s ^(-1)`. |
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