1. Behavior at very high frequency

At very high frequency:

• Inductive reactance: $X_L = \omega L \to \infty$
  So the inductor behaves like an open circuit.
• Capacitive reactance: $X_C = \dfrac{1}{\omega C} \to 0$
  So the capacitor behaves like a short circuit.

Thus, in the given AC circuit, we replace:

• each inductor by an open branch,
• each capacitor by a wire.

2. Reduction of the circuit

After making these replacements, the circuit reduces to a purely resistive network.
From the given arrangement, the equivalent resistance at very high frequency becomes

$R_{\text{eq}} = 5\ \Omega$

3. Effective current

The source effective voltage is

$V_{\text{rms}} = 220\ \text{V}$

Hence the effective current is

$I = \frac{V_{\text{rms}}}{R_{\text{eq}}} = \frac{220}{5} = 44\ \text{A}$

4. Final answer

Therefore, the effective current at very high frequencies is

$\boxed{44\ \text{A}}$