Power dissipated in an `L-C-R` series circuit connected to an `AC` source of emf `epsilon` is

Options

A. `(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)])`

Answer

D

Explanation

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)])`