An LCR circuit has L = 10 mH, `R = 3 Omega` and `C = 1 mu F` connected in series to a source of `15cos omega t`.volt. Calculate the current ampliuted and the average power dissipated per cycle at a frequency 10% lower than the resonance frequency.

Answer

5

Explanation

`omega_(o)=(1)/(LC)=(1)/sqrt(10^(-2)xx10^(-6))=10^(4) rad//s`
`w=w_(o)-(10)/(100)omega_(o)=9xx10^(3) rad//s`
Hence `X_(L)=omegaL=9xx10^(3)xx10^(-2)=90Omega`
`X_(C)=(1)/(omega C)=(1)/(9xx10^(3)xx10^(-6))=111.11Omega`
`X=X_(C)-X_(L)=111.11-90=21.11Omega`
Impedance `Z=sqrt(R^(2)+X^(2))=sqrt(3^(2)+(21.11)^(2))`
`Z=21.32Omega`
`l_(0)=(15)/(21.32)=0.704 A, P_(av)=i_(ms)^(2)R=(i_(o)^(2))/(2)R`
`=(1)/(2)xx(0.704)^(2)xx3=0.74` watt
`f=(W)/(2pi)=(9xx10^(3))/(2pi)` cycle`//`s So energy per cycle
`=(0.74)/(((9xx10^(3))/(2pi)))=5.16xx10^(-4) cong 5xx10^(-4) "Joules"//"cycle"`