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Question

LCR circuit has L=10mH,R=3Ω and C=1μF connected in series to a source of 15cosωtV. Calculate the current amplitude and the average power dissipated per cycle at a frequency that is 10% lower than the resonance frequency.

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Solution

As here
ω0=1LC=1102×106=104rad/sec
So, ω=ω010100ω0=910ω0=9×103rad/sec
and hence XL=ωL=9×103×102=90Ω;XC=ωL=19×103×106=111.11Ω
So, X=XLXC=90111.11=21.11Ω
And hence Z=R2+X2=32+(21.11)2=21.32Ω
As here
E=15cosωT,E0=15V,I0=E0V=1521.32=0.704A
The average power dissipated
Pav=VrmsIrsmcosϕ=(Irsm×Z)×Irsm×(R/Z)
ie., Pav=I2rmsR=12I20R[Irsm=I02]
Pav=0.74W
Now as f=ω2π=9×1032πcycle/sec
So, Pavcycle=0.74(9×103/2π)cycle/sec=5.16×104J/cycle

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