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

 

As here,
$\begin{array}{ll}{\omega }_0& =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{{10}^{-2}\times {10}^{-6}}}={10}^4\mathrm{rad}/\mathrm{s}\\ \text{ so }& \omega ={\omega }_0-\frac{10}{100}{\omega }_0=\frac{9}{10}{\omega }_0=9\times {10}^3\mathrm{rad}/\mathrm{s}\\ \text{ and hence, }& X_L=\omega L=9\times {10}^3\times {10}^{-2}=90\Omega ,\\ \text{ So, }& X_C=\frac{1}{\omega C}=\frac{1}{9\times {10}^3\times {10}^{-6}}=111.11\Omega \\ \text{ and hence, }& Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}=\sqrt{3^2+(-21.11{)}^2},\\ \text{ i.e., }& Z=\sqrt{9+445.63}=21.32\Omega \\ \text{ and as here }& v=15\cos \omega t,\text{ i.e., }{V}_M=15 \mathrm{V}\\ & I_M=\frac{V_M}{Z}=\frac{15}{21.32}=0.704 \mathrm{A} \end{array}$

The average power dissipated,
$P_{av.}=V_{rms}{I}_{rms}\cos \varphi =\left(I_{rms}\times Z\right)\times I_{rms}\times \frac{R}{Z}$

i.e., $P_{av.}=I_{rms}^{2}R=\frac{1}{2}{I}_0^{2}R$ as  $I_{rms}=\frac{I_0}{\sqrt{2}}$
so,
$P_{avD}=\frac{1}{2}\times (0.704{)}^2\times 3=0.743 \mathrm{W}$