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Power dissipated in an L-C-R series circ...

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

A

`(epsion^(2) sqrt(R^(2) + (L omega - (1)/(C omega) )^(2)))/(R)`

B

`(epsilon [R^(2) + (L omega - (1)/(C omega) )^(2)])/(R)`

C

`(epsilon^(2) R)/(sqrt(R^(2) + (L omega - (1)/(C omega) )^(2)))`

D

`(epsilon^(2) R)/([R^(2) + (L omega - (1)/(C omega) )^(2)])`

Text Solution

Verified by Experts

The correct Answer is:
D
(d) Power is consumed by resistance R only . Average power consumed in L and C is zero .So we have
`P =I_("rms")^(2)R=(epsilon^(2)R)/((sqrt((X_(L)-X_(C))^(2)+(R^(2))))^(2))`
`= (epsilon^(2)R)/([(omegaL-(1)/(omegaC))^(2)+R^(2)])`
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