Physics - EMI/AC Question with Solution | TestHub

For$\omega \to \infty$
$\frac{1}{\omega C}\to 0\Rightarrow$capacitance acts as short circuit
&$\omega \mathrm{L}\to \infty$Inductance acts as open circuit

$Z=1+\frac{2\times 2}{2+2}=2\mathrm{\Omega }$