InterviewSolution
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InterviewSolution
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Power dissipated in an `L-C-R` series circuit connected to an `AC` source of emf `epsilon` isA. `(epsilon^(2)sqrt(R^(2)+(omega L-(1)/(omega C))^(2)))/(R )`B. `(epsilon^(2)[R^(2)+(omega L-(1)/(omega C))^(2)])/(R )`C. `(epsilon^(2)R)/(sqrt(R^(2)+(omega L-(1)/(omega C))^(2)))`D. `(epsilon^(2)R)/([R^(2)+(omega L-(1)/(omega C))^(2)])` |
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Answer» Correct Answer - D Average power, `P = V_("rms")I_("rms")cos phi` Here, `Z = sqrt(R^(2)+(X_(L)-X_(C ))^(2)), cos phi = (R )/(Z)` But `I_("rms")=(V_("rms"))/(Z). Therefore P= V_("rms")^(2)(R )/(Z^(2))` `therefore P = V_("rms")^(2) (R )/({R^(2)+(X_(L)-X_(C ))^(2)})=(epsilon^(2)R)/([R^(2)+(omega L-(1)/(omega C))^(2)])` |
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