InterviewSolution
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InterviewSolution
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A series LCR circuit with `L= 0.12H, C=480 nF,` and R=23 Omega` is connected to a 230-V variable frequency supply. (a) What is the source frequency for which current amplitude is maximum? Find this maximum value. (b) What is the source frequency for which average power absorbed by the circuit is maximum? Obtain the value of maximum power. (c ) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? (d) What is the Q-factor of the circuit? |
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Answer» (a) Current amplitude is maximum at resonace and source frequency at resonance is given by `f_(0)=(1)/(2 pi sqrt(LC))=(1)/(2 pi sqrt(0.12xx480xx10^(-9))) = 663.14 Hz` and `omega_(0) 2pi f_(0)=4166.66 rad s^(-1)` The maximum current amplitude is given by `I_(0(max))=(E_0)/(R ) =(230 sqrt(2))/(23)= 14.14A` (b) Average power absorbaed by the circuit is maximum at resonance for wich frequency is calculate above as 663.14 Hz. Maximum power absorbed is given by `P_(max)=(E_(v)^(2))/(R ) =((230)^(2))/(23) =2300W` (c ) `P=E_(v) I_(v) cos phi =E_(v) I_(v) (R )/(Z) =(E_(v)^(2))/(Z^2) R and P_(max)=(E_(v)^(2))/(R )` Given `P=P_(max)//2` `implies (E_(v)^(2)R)/(Z^2)=1/2 (E_(v)^(2))/(R ) implies Z^(2)=2 R^(2)` `implies R^(2) + (X_(C )-X_(L))^(2)=2R^(2)` `implies X_(C )-X_(L) = +_ R` Solve to get `omega_(1)=(-RC+sqrt(R^(2)C^(2)+4LC))/(2LC)=4263.6 rad s^(-1)` (d) Q-factor is given by `Q=1/R sqrt((L)/(C )) =1/23 sqrt(0.12)(480xx10^(-9))=21.74`. |
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