1.

For a series LCR circuit the power loss at resonance is

Answer»

`(V^(2))/([OMEGA L-(1)/(omega C)])`
`I^(2)L omega`
`I^(2)R`
`(V^(2))/(C omega)`

Solution :The impedance Z of a series LCR circuit is GIVEN by, `Z=sqrt(R^(2)+(X_(L)-X_(C ))^(2))`
where `X_(L)=omega L` and `X_(C )=(1)/(omega C), omega` is ANGULAR frequency.
At resonance, `X_(L)=X_(C )`, hence Z = R :
`therefore V_(R )= V` (supply voltage)
`therefore` R.M.S. current, `I=(V_(R ))/(R )=(V)/(R )`
Power loss `I^(2)R=V^(2)//R`.


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