# NEET Physics Alternating Current Questions Solved

NEET - 2009

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

(a) $\frac{{\epsilon }^{2}R}{\left[{R}^{2}+{\left(L\omega -\frac{1}{C\omega }\right)}^{2}\right]}$

(b) $\frac{{\epsilon }^{2}{\sqrt{{R}^{2}+\left(L\omega -\frac{1}{C\omega }\right)}}^{2}}{R}$

(c) $\frac{{\epsilon }^{2}\left[{R}^{2}+{\left(L\omega -\frac{1}{C\omega }\right)}^{2}\right]}{R}$

(d) $\frac{{\epsilon }^{2}R}{\sqrt{{R}^{2}+{\left(L\omega -\frac{1}{C\omega }\right)}^{2}}}$

Power dissipated in series L-C-R.

$P={I}_{rms}^{2}R=\frac{{\epsilon }_{rms}^{2}R}{{\left|Z\right|}^{2}}=\frac{{\epsilon }^{2}R}{\left[{R}^{2}+{\left(\omega L-\frac{1}{\omega C}\right)}^{2}\right]}$

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