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.
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As here,`omega_(0)=(1)/(sqrtLC)=(1)/(sqrt(10^(-2)xx10^(-6)))=10^(4) rad//sec` So,`omega=omega_(0)-(10)/(100)omega_(0)=(9)/(10)omega_(0_=9xx10^(3)red// ec` and hence, `X_(L) = omegaL = 9 xx 10^(3)xx106(-2)=90Omega` `X_(c)=(1)/(omegaC)=(1)/(9xx10^(3)xx10^(-6))=111.11Omega` So, `X = X_(L)-X_(c) = 90 - 111.11 = -21.11Omega` And hence,`Z=sqrt(R^(2)+X^(2))=sqrt(3^(2)+(-21.11)^(2))=21.32Omega` As here `E = 15 cos omegat, E_(0)= 15 V` `I_(0)=(E_(0))/(V)=(15)/(21.32)=0.704A` The average power dissipated,`(P_(av))=V_(rms)|rmscosphi=(I_(rms)xxZ)xxI_(rms)xx(R//Z)` i.e.`P_(av)=I_(rms)^(2)R=(1)/(2)I_(0)^(2)R[asI_(rms)=(I_(0))/(sqrt2)]` `P_(av)=0.74W` Now asf=`(omega)/(2pi)=(9xx10^(3))/(2pi)cycle//sec` `(P_(av)/(cycie)=(0.74)/(9xx10^(3)//2pi)cycie//sec)=5.16xx10^(-4)j//cycle`
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