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An LCR circuit has L = 10 mH. R = 3 ohm ...

An `LCR` circuit has `L = 10 mH`. `R = 3` ohm and `C = 1 mu F` connected in series to a source is `15 cos omega t` volt. What is average power dissipated per cycle at a frequency that is `10%` lower than the resonant requency?

Text Solution

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Here, `L = 10^(-2) H, R = 3 Omega, C = 10^(-6) F`
Resonant frequency,
`omega_(0) = (1)/(sqrt(LC)) = (1)/(sqrt(10^(-2) xx 10^(-6))) = 10^(4) "rad"//s`
Actual frequency, `omega = (90%) omega_(0)`
`= 9 xx 10^(3) "rad"//s`
`X_(L) = omega L = 9 xx 10^(3) xx 10^(-2) = 90 Omega`
`X_(C ) = (1)/(omega C) = (1)/(9 xx 10^(3) xx 10^(-6)) = (1000)/(9) Omega`
`Z = sqrt(R^(2) + (X_(C ) - X_(L))^(2))`
`= sqrt(3^(2) + ((100)/(9) - 90)^(2)) = 21.3 Omega`
Power dissipated`//` cycle `=E_(v)I_(v) cos phi`
`=E_(0) (E_(v)/Z) R/Z=(E_(v)/Z)^(2)xxR`
`=(15/(sqrt(2)xx21.3))^(3)xx3=0.744 W`
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