1.

Series LCR circuit is connected to an AC source of angular frequency omega. Current flowing in the circuit is found to lead the voltage by pi/4. Magnitude of capacitance is

Answer»

`(1)/(OMEGA^(2) (L + R))`
`(1)/(omega (L + R))`
`(1)/(omega^(2) L + omega R)`
`(1)/(omega L + omega^(2) R)`

SOLUTION :(c ): Phase difference of `pi//4 ` or `45^(@)` between current and VOLTAGE means NET reactance of circuit is equal to resistance of circuit . Further it is given that current leads the voltage hence reactance due to capacitance is more than due to inductor . Hence we can write the following equation :
`(1)/(omegaC ) - omegaL = R `
`implies (1)/(omegaC )= R + omegaL`
`implies (1)/(omegaC)=R+omegaL`
`implies C = (1)/(omega^(2)L+omegaR)`
Hence optioin C is correct .


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