Question

Two circuits are shown in the figure (a) & (b). At a frequency of ${\text{rad s}}^{-1}$, the average power dissipated in one cycle will be the same in both the circuits.

Answer 1

For circuit (a)
$P_a=\frac{V_{rms}^2}{R}$

For circuit (b)
$P_b=\frac{v_{rms}^2}{Z^2}\times R$

Given that, $P_a=P_b$

$\Rightarrow \frac{V_{rms}^2}{R}=\frac{v_{rms}^2}{Z^2}\times R$

$\Rightarrow R^2=Z^2$

$\Rightarrow 25={\left(\sqrt{(X_C-X_L{)}^2+5^2}\right)}^2$

$\Rightarrow 25={(X_C-X_L)}^2+25$

$\Rightarrow X_C=X_L$

$\Rightarrow \frac{1}{\omega C}=\omega L$

$\Rightarrow {\omega }^2=\frac{1}{LC}=\frac{{10}^6}{0.1\times 40}$

$\Rightarrow \omega =500{\text{rad s}}^{-1}$

Hence, $500$ is the correct answer.