The effective current $$I$$ in the given circuit at very high frequencies will be ______ A.

We need to determine the effective current $$I$$ in the circuit at a very high frequency ($$f \to \infty$$).

1. Circuit Behavior at Very High Frequencies

The reactances of inductors ($$X_L$$) and capacitors ($$X_C$$) depend on the frequency ($$\omega = 2\pi f$$) as follows:

• Inductors: $$X_L = \omega L \to \infty$$ (acts as an open circuit / broken wire)
• Capacitors: $$X_C = \frac{1}{\omega C} \to 0$$ (acts as a short circuit / plain wire)

2. Simplify the Circuit

Applying these conditions to the circuit diagram 

• Any branch containing an inductor ($$20\text{ mH}$$ and $$50\text{ mH}$$) carries no current and is completely removed.
• Any capacitor is replaced by a straight wire.

Following the continuous path through the remaining active branches from the voltage source:

• The top-left path ($8\ \Omega$) is blocked by the open $$20\text{ mH}$$ inductor.
• The current flows cleanly through the bottom paths via the shorted capacitors. The only resistors that remain part of a completed closed loop with the source are the two $$4\ \Omega$$ resistors connected in parallel across the network.

The equivalent resistance ($$R_{\text{eq}}$$) of these two parallel $$4\ \Omega$$ resistors is:

$$R_{\text{eq}} = \frac{4 \times 4}{4 + 4} = 2\ \Omega$$

3. Calculate the Effective Current ($$I$$)

Given the source voltage $$V = 220\text{ V}$$, we apply Ohm's law to find the current:

$$I = \frac{V}{R_{\text{eq}}} = \frac{220}{2} = 110\text{ A}$$