Question

An $LCR$ circuit has $L=10mH,R=150\Omega$ and $C=1\mu F$ connected in series to a source of $150\sqrt{2}\cos \omega t$ volt. At a frequency that is $50\%$ of the resonant frequency, calculate the average power (in $watt$) dissipated per cycle.

Answer 1

${\omega }_0=$ resonance frequency.

${\omega }_0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{10\times {10}^{-3}\times 1\times {10}^{-6}}}=\frac{1}{{10}^{-4}}={10}^{4}Hz$

For, $\omega =\frac{{\omega }_0}{2}=5\times {10}^{3}Hz$

$Z=\sqrt{R^2+({\omega L-\frac{1}{C\omega })}^2}=\sqrt{{\left(150\right)}^2+(50-{200)}^2}$$=150\sqrt{2}$

$P$ decipated$=\frac{Vrm{s}^2}{R}cos\varphi =\frac{{150}^2}{150\sqrt{2}}\times cos\frac{\pi }{4}$

$=\frac{150\times 150}{150\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{150}{2}=75W$