For very high frequencies, the effective impedance of the circuit shown in the figure
will be: the source feeds R₁=1 Ω in series, then a parallel section whose upper
branch is R₂ in series with an inductor L and whose lower branch is a capacitor C,
and finally R₃=2 Ω in series.

Question

Accepted Answer

(4) 3 Ω. At very high frequency the two reactances go to opposite extremes. X_C=1/(ω C)→0, so the capacitor behaves as a plain wire and short-circuits whatever it is connected across. X_L=ω L→∞, so the inductive branch carries no current and behaves as a break. What is left is the 1 Ω and 2 Ω resistors in series: Z=1+2=3 Ω.

For very high frequencies, the effective impedance of the circuit shown in the figure
will be: the source feeds R₁=1 Ω in series, then a parallel section whose upper
branch is R₂ in series with an inductor L and whose lower branch is a capacitor C,
and finally R₃=2 Ω in series.

Step-by-step solution

Why the other options are wrong

For very high frequencies, the effective impedance of the circuit shown in the figurewill be: the source feeds R₁=1 Ω in series, then a parallel section whose upperbranch is R₂ in series with an inductor L and whose lower branch is a capacitor C,and finally R₃=2 Ω in series.

Step-by-step solution

Why the other options are wrong

For very high frequencies, the effective impedance of the circuit shown in the figure
will be: the source feeds R₁=1 Ω in series, then a parallel section whose upper
branch is R₂ in series with an inductor L and whose lower branch is a capacitor C,
and finally R₃=2 Ω in series.