Draw the effective equivalent circuit of the circuit shown in the figure at very high frequencies and find the effective impedance.

(N/A) The inductive reactance is given by $X_{L} = 2 \pi \nu L$ and the capacitive reactance is given by $X_{C} = \frac{1}{2 \pi \nu C}$.
For very high frequencies,$\nu \rightarrow \infty$.
As $\nu \rightarrow \infty$,the inductive reactance $X_{L} \rightarrow \infty$,which acts as an open circuit. Conversely,the capacitive reactance $X_{C} \rightarrow 0$,which acts as a short circuit.
Therefore,in the given circuit,the capacitors $C_{1}$ and $C_{2}$ behave as short circuits (wires),and the inductors $L_{1}$ and $L_{2}$ behave as open circuits (breaks in the path).
By replacing the components with their high-frequency equivalents,the circuit simplifies to a series combination of resistors $R_{1}$ and $R_{3}$.
Thus,the effective impedance is $Z_{eq} = R_{1} + R_{3}$.