Q3: For the circuit shown below use phasor method (frequency domain) to find the current $i_s(t)$, $i_L(t)$ and $i_C(t)$
$V_s = 10\angle20^\circ$
$j\omega L = j(4)(2) = j8\Omega$
$\frac{1}{j\omega C} = \frac{1}{j(4)(0.2)} = -j1.25\Omega$
$2 + j8 - j1.25 = 2 + j6.75\Omega$
$\sqrt{(2)^2 + (6.75)^2} = 7.04\Omega$
$\theta = \tan^{-1}(\frac{6.75}{2}) = 73.3^\circ$
$Z_{eq} = 7.04\angle73.3^\circ\Omega$
$V_C(t) = 10\cos(4t + 20^\circ)$
$\frac{10\angle20^\circ}{7.04\angle73.3^\circ} = 1.42\angle-53.3^\circ A$
$i(t) = 1.42\cos(4t - 53.3^\circ)A$