Question

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

• A. double
• B. one-fourth
• C. one-third
• D. half

Answer 1

Answer: (D)

The amplitude of the current in the series $L-C-R$ circuit is given by
$i_m=\frac{V_m}{\sqrt{R^2+{\left(\omega L-\frac{1}{\omega C}\right)}^2}}$
$\Rightarrow i_{\max }=\frac{V_m}{R}\text{ now }\frac{1}{\sqrt{2}}{i}_{\max }=i_m$
$\Rightarrow \frac{1}{2}\frac{V_m^2}{R^2}=\frac{V_{m^2}}{R^2+(\omega L-1/\omega C{)}^2}$
$\Rightarrow R^2=(\omega L-1/\omega C{)}^2$
$\Rightarrow R=(\omega L-1/\omega C{)}^2$
$\Rightarrow \tan \varphi =\frac{\left(\omega L-1/\omega C^2\right)}{R}=1$ $\Rightarrow \varphi =45^{\circ }$
$P=\frac{I_{\text{rms }}}{\sqrt{R^2+(\omega L-1/\omega C{)}^2}}=\frac{I_{rms}^2}{\sqrt{2}R}\times \frac{1}{\sqrt{2}}$
$=\frac{I_{rms}}{2R}=\frac{P_{max}}{2}$