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Power dissipated in an LCR series circuit connected to an a.c. source of emf $\varepsilon$ is :
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Verified Answer
The correct answer is:
$\varepsilon^2 \mathrm{R} /\left[\mathrm{R}^2+\left(\mathrm{L} \omega-\frac{1}{\mathrm{C} \omega}\right)^2\right]$
$$
\begin{aligned}
\mathrm{P}_{\mathrm{av}} & =\mathrm{E}_{\mathrm{rms}} \cdot \mathrm{I}_{\mathrm{rms}} \cos \phi \\
& =\varepsilon \cdot \frac{\varepsilon}{\mathrm{z}} \cdot \frac{\mathrm{R}}{\mathrm{Z}}=\frac{\varepsilon^2 \mathrm{R}}{\mathrm{z}^2} \\
& =\frac{\varepsilon^2 \mathrm{R}}{\mathrm{R}^2+\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)^2}
\end{aligned}
$$
\begin{aligned}
\mathrm{P}_{\mathrm{av}} & =\mathrm{E}_{\mathrm{rms}} \cdot \mathrm{I}_{\mathrm{rms}} \cos \phi \\
& =\varepsilon \cdot \frac{\varepsilon}{\mathrm{z}} \cdot \frac{\mathrm{R}}{\mathrm{Z}}=\frac{\varepsilon^2 \mathrm{R}}{\mathrm{z}^2} \\
& =\frac{\varepsilon^2 \mathrm{R}}{\mathrm{R}^2+\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)^2}
\end{aligned}
$$
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