In the given circuit,the AC source has omega | vedclass

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(D) The circuit consists of two parallel branches connected to a $200 \, V$ $AC$ source.For the upper branch ($RC$ circuit):$Z_{C} = \sqrt{R_{1}^{2} +

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$3.16$

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(D) The circuit consists of two parallel branches connected to a $200 \, V$ $AC$ source.For the upper branch ($RC$ circuit):$Z_{C} = \sqrt{R_{1}^{2} + \left(\frac{1}{\omega C}ight)^{2}} = \sqrt{100^{2} + \left(\frac{1}{100 	imes 100 	imes 10^{-6}}ight)^{2}} = \sqrt{100^{2} + 100^{2}} = 100\sqrt{2} \, \Omega$.The current in the capacitor branch is $I_{C} = \frac{V}{Z_{C}} = \frac{200}{100\sqrt{2}} = \sqrt{2} \, A$.For the lower branch ($RL$ circuit):$Z_{L} = \sqrt{R_{2}^{2} + (\omega L)^{2}} = \sqrt{50^{2} + (100 	imes 0.5)^{2}} = \sqrt{50^{2} + 50^{2}} = 50\sqrt{2} \, \Omega$.The current in the inductor branch is $I_{L} = \frac{V}{Z_{L}} = \frac{200}{50\sqrt{2}} = 2\sqrt{2} \, A$.In the $RC$ branch,current leads voltage by $\phi_{1} = 	an^{-1}\left(\frac{1/\omega C}{R_{1}}ight) = 	an^{-1}(1) = 45^{\circ}$.In the $RL$ branch,current lags voltage by $\phi_{2} = 	an^{-1}\left(\frac{\omega L}{R_{2}}ight) = 	an^{-1}(1) = 45^{\circ}$.The total current $I$ is the vector sum of $I_{C}$ and $I_{L}$. The angle between $I_{C}$ and $I_{L}$ is $45^{\circ} + 45^{\circ} = 90^{\circ}$.$I = \sqrt{I_{C}^{2} + I_{L}^{2} + 2I_{C}I_{L}\cos(90^{\circ})} = \sqrt{I_{C}^{2} + I_{L}^{2}} = \sqrt{(\sqrt{2})^{2} + (2\sqrt{2})^{2}} = \sqrt{2 + 8} = \sqrt{10} \approx 3.16 \, A$.

In the given circuit,the $AC$ source has $\omega = 100 \, rad \, s^{-1}$. Considering the inductor and capacitor to be ideal,what will be the current $I$ flowing through the circuit? (in $A$)

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