Question

LCR circuit has $L=10mH,R=3\Omega$ and $C=1\mu F$ connected in series to a source of $15\cos \omega tV$. Calculate the current amplitude and the average power dissipated per cycle at a frequency that is $10$% lower than the resonance frequency.

Answer 1

As here
${\omega }_0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{{10}^{-2}\times {10}^{-6}}}={10}^{4}rad/sec$
So, $\omega ={\omega }_0-\frac{10}{100}{\omega }_0=\frac{9}{10}{\omega }_0=9\times {10}^{3}rad/sec$
and hence $X_L=\omega L=9\times {10}^3\times {10}^{-2}=90\Omega ;X_C=\omega L=\frac{1}{9\times {10}^3\times {10}^{-6}}=111.11\Omega$
So, $X=X_L-X_C=90-111.11=-21.11\Omega$
And hence $Z=\sqrt{R^2+X^2}=\sqrt{3^2+{(-21.11)}^2}=21.32\Omega$
As here
$E=15\cos \omega T,E_0=15V,I_0=\frac{E_0}{V}=\frac{15}{21.32}=0.704A$
The average power dissipated
$P_{av}=V_{rms}{I}_{rsm}\cos \varphi =(I_{rsm}\times Z)\times I_{rsm}\times \left(R/Z\right)$
ie., $P_{av}=I_{rms}^{2}R=\frac{1}{2}{I}_0^{2}R[I_{rsm}=\frac{I_0}{\sqrt{2}}]$
$P_{av}=0.74W$
Now as $f=\frac{\omega }{2\pi }=\frac{9\times {10}^3}{2\pi }cycle/sec$
So, $\frac{P_{av}}{cycle}=\frac{0.74}{(9\times {10}^3/2\pi )cycle/sec}=5.16\times {10}^{-4}J/cycle$