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Q.
A value of $\omega$ for which the current amplitude is $1 / \sqrt{2}$ times its maximum value. At this value, the power dissipated by the circuit becomes
Alternating Current
Solution:
The amplitude of the current in the series $L-C-R$ circuit is given by
$i_{m}=\frac{V_{m}}{\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}}$
$\Rightarrow i_{\max }=\frac{V_{m}}{R} \text { now } \frac{1}{\sqrt{2}} i_{\max }=i_{m}$
$\Rightarrow \frac{1}{2} \frac{V_{m}^{2}}{R^{2}}=\frac{V_{m^{2}}}{R^{2}+(\omega L-1 / \omega C)^{2}}$
$\Rightarrow R^{2}=(\omega L-1 / \omega C)^{2}$
$\Rightarrow R=(\omega L-1 / \omega C)^{2}$
$\Rightarrow \tan \phi=\frac{\left(\omega L-1 / \omega C^{2}\right)}{R}=1$ $\Rightarrow \phi=45^{\circ}$
$P=\frac{I_{ \text{rms }}}{\sqrt{R^{2}+(\omega L-1 / \omega C)^{2}}}=\frac{I_{ rms }^{2}}{\sqrt{2} R} \times \frac{1}{\sqrt{2}}$
$=\frac{I_{ rms }}{2 R}=\frac{P_{ max }}{2}$