Physics - EMI/AC Question with Solution | TestHub

The emf of an LCR circuit is $\epsilon$. The impedance of a series LCR circuit is given as, $Z=\sqrt{\left[R^2+{\left(\omega L-\frac{1}{\omega C}\right)}^2\right]}$ The power factor in a series LCR circuit is given as, $\cos \varphi =\frac{\mathrm{R}}{|\mathrm{Z}|}$ The power dissipated in the circuit is given as, $\begin{array}{rl}& \mathrm{P}={\mathrm{V}}_{\mathrm{rms}}{\mathrm{I}}_{\mathrm{rms}}\cos \varphi \\ & =\epsilon \times \frac{\epsilon }{|\mathrm{Z}|}\times \frac{\mathrm{R}}{|\mathrm{Z}|}\\ & =\frac{{\epsilon }^2\mathrm{R}}{\left[{\mathrm{R}}^2+{\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)}^2\right]}\end{array}$ Thus, the power dissipated in the series $LCR$ circuit is $\frac{{\epsilon }^{2}R}{\left[R^2+{\left(\omega L-\frac{1}{\omega C}\right)}^2\right]}$.