Q137 / 200 · NEET 2009 (Screening) · Physics

Power dissipated in an $L-C-R$ series circuit connected to an AC source of emf $\varepsilon$ is

  • Correct answer
    A
    $\frac{\varepsilon^2 R}{\left[R^2+\left(L \omega-\frac{1}{C \omega}\right)^2\right]}$
  • B
    $\frac{\varepsilon^2 \sqrt{\mathrm{R}^2+\left(\mathrm{L} \omega-\frac{1}{\mathrm{C} \omega}\right)^2}}{\mathrm{R}}$
  • C
    $\frac{\varepsilon^2\left[\mathrm{R}^2+\left(\mathrm{L} \omega-\frac{1}{\mathrm{C} \omega}\right)^2\right]}{\mathrm{R}}$
  • D
    $\frac{\varepsilon^2 R}{\sqrt{R^2+\left(L \omega-\frac{1}{C \omega}\right)^2}}$

Solution

The emf of an LCR circuit is \(\varepsilon\). 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 \(\frac{\varepsilon^2 R}{\left[R^2+\left(\omega L-\frac{1}{\omega C}\right)^2\right]}\).

Hence, the correct answer is option: 'A'.

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