Power dissipated in an LCR series circuit connected to an a.c. source of emf $\varepsilon$ is :

Power dissipated in an LCR series circuit connected to an a.c. source of emf $\varepsilon$ is :
  1. $\varepsilon^2 R / \sqrt{R^2+\left(L \omega-\frac{1}{C \omega}\right)^2}$
  2. $\varepsilon^2 \mathrm{R} /\left[\mathrm{R}^2+\left(\mathrm{L} \omega-\frac{1}{\mathrm{C} \omega}\right)^2\right]$
  3. $\varepsilon^2 \sqrt{\left[\mathrm{R}^2+\left(\mathrm{L} \omega-\frac{1}{\mathrm{C} \omega}\right)^2\right]} / \mathrm{R}$
  4. $\frac{\varepsilon^2\left[\mathrm{R}^2+\left(\mathrm{L} \omega-\frac{1}{\mathrm{C} \omega}\right)^2\right]}{\mathrm{R}}$

Solution

$\begin{aligned} \mathrm{P}_{\mathrm{av}} & =\mathrm{E}_{\mathrm{rms}} \cdot \mathrm{I}_{\mathrm{rms}} \cos \phi \\ & =\varepsilon \cdot \frac{\varepsilon}{\mathrm{z}} \cdot \frac{\mathrm{R}}{\mathrm{Z}}=\frac{\varepsilon^2 \mathrm{R}}{\mathrm{z}^2} \\ & =\frac{\varepsilon^2 \mathrm{R}}{\mathrm{R}^2+\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)^2} \end{aligned}$

Asked in: NEET 2009 (Mains)

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