For very high frequencies, the effective impedance of the circuit (shown in the figure) will be [2023]

Question

Smart Achievers · Accepted Answer

(4)

As frequency is very high. Here, reactance due to capacitance varies inversely whereas reactance due to inductor varies directly.

Then, $X_{C} \approx 0$

$X_{L} \to \infty$

Effective circuit will be

Thus, capacitor acts as a short circuit while inductor as an open circuit. Effective impedance will be $1 Ω + 2 Ω = 3 Ω$

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