Draw the effective equivalent circuit of the circuit show in Fig. at very high frequencies and find the effective impedance.

Explanation

At very high frequencies, `X_(C ) = (1)/(omega C) = (1)/(2 pi v C) rarr` Zero
`X_(L) = omega L = 2 pi v L rarr` extremely high
i.e., capacitor is short circuited and inductor blocks the currents (serving as open key). The effective equivalent circuits are show in Fig. (a) and (b).
Effective impedance, `Z = R_(1) + R_(3)`.