Physics - EMI/AC Question with Solution | TestHub

For $\omega \to \infty$

$\frac{1}{\omega \mathrm{C}}\to 0\Rightarrow$ capacitance acts as short circuit

& $\omega L\to \infty$ Inductance acts as open circuit

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