Power dissipated in an $L-C-R$ series circuit connected to an AC source of emf $\varepsilon$ is
- $\frac{\varepsilon^2 R}{\left[R^2+\left(L \omega-\frac{1}{C \omega}\right)^2\right]}$
- $\frac{\varepsilon^2 \sqrt{\mathrm{R}^2+\left(\mathrm{L} \omega-\frac{1}{\mathrm{C} \omega}\right)^2}}{\mathrm{R}}$
- $\frac{\varepsilon^2\left[\mathrm{R}^2+\left(\mathrm{L} \omega-\frac{1}{\mathrm{C} \omega}\right)^2\right]}{\mathrm{R}}$
- $\frac{\varepsilon^2 R}{\sqrt{R^2+\left(L \omega-\frac{1}{C \omega}\right)^2}}$
Solution
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,
\(\cos \phi=\frac{\mathrm{R}}{|\mathrm{Z}|}\)
The power dissipated in the circuit is given as,
\(\begin{aligned}
& \mathrm{P}=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \phi \\
& =\varepsilon \times \frac{\varepsilon}{|\mathrm{Z}|} \times \frac{\mathrm{R}}{|\mathrm{Z}|} \\
& =\frac{\varepsilon^2 \mathrm{R}}{\left[\mathrm{R}^2+\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)^2\right]}
\end{aligned}\)
Thus, the power dissipated in the series \(L C R\) circuit is \(\frac{\varepsilon^2 R}{\left[R^2+\left(\omega L-\frac{1}{\omega C}\right)^2\right]}\).
Asked in: NEET 2009 (Screening)