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 \(A\))

\(=\sqrt{\left(\frac{1}{100 \times 100 \times 10^{-6}}\right)^{2}+100^{2}}\)

\({Z}_{{C}}=\sqrt{(100)^{2}+(100)^{2}}\)

\(=100 \sqrt{2}\)

\({Z}_{{L}}=\sqrt{(\omega {L})^{2}+{R}^{2}}\)

\(\sqrt{(100 \times 0.5)^{2}+50^{2}}\)

\(=50 \sqrt{2}\)

\({i}_{{C}}=\frac{200}{{z}_{{c}}}=\frac{200}{100 \sqrt{2}}=\sqrt{2}\)

\({i}_{{L}}=\frac{200}{{z}_{{L}}}=\frac{200}{50 \sqrt{2}}=2 \sqrt{2}\)

\(\cos \phi_{1}=\frac{100}{10 \sqrt{2}}=\frac{1}{\sqrt{2}} \Rightarrow \phi_{1}=45^{\circ}\)

\(\cos \phi_{2}=\frac{50}{50 \sqrt{2}}=\frac{1}{\sqrt{2}} \Rightarrow \phi_{2}=45^{\circ}\)

\({I}=\sqrt{{I}_{{C}}^{2}+{I}_{{L}}^{2}}\)

\(=\sqrt{2+8}\)

\(=\sqrt{10}\)

\({I}=3.16 \,{A}\)