For very high frequencies, the effective impedance of the circuit (shown in the figure) will be

For very high frequencies, the effective impedance of the circuit (shown in the figure) will be

1. $1 \Omega$
2. $3 \Omega$
3. $4 \Omega$
4. $6 \Omega$

Solution

$X_L=\omega L$ $x_C=\frac{1}{\omega C}$ at very high frequency $X_C=0, X_L=\infty$
$Z=(1+2) \Omega=3 \Omega$

Asked in: NEET 2023 (Manipur)

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