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

1. Component Behavior at Very High Frequencies ($$f \to \infty$$)

In alternating current (AC) circuits, the reactances of inductors and capacitors are highly frequency-dependent:

• Inductors ($$20 \,\, \text{mH}$$ and $$50 \,\, \text{mH}$$): The inductive reactance formula is $$X_L = \omega L$$. As $$\omega \to \infty$$, $$X_L \to \infty$$. This means inductors act as open circuits (broken wires), completely blocking current from entering the outer top branches.
• Capacitors ($$6 \,\, \mu\text{F}$$, $$0.5 \,\, \mu\text{F}$$, and $$0.8 \,\, \mu\text{F}$$): The capacitive reactance formula is $$X_C = \frac{1}{\omega C}$$. As $$\omega \to \infty$$, $$X_C \to 0$$. This means capacitors offer zero reactance and act as short circuits (ideal connecting wires).

2. Tracing the Simplified Network Paths

By substituting these high-frequency conditions into the schematic, we can track the continuous path from the left terminal of the $$220 \,\, \text{V}$$ source to the right terminal:

• The current first passes through the initial series resistor on the left:

  $$R_{\text{left}} = 1 \,\, \Omega$$

• Next, the current reaches a split junction (let's call it Node A). Because the vertical capacitors are now shorted into clean wires, they link the top and bottom paths directly across the same electric potentials:
  1. Path 1: Directs current horizontally through the lower-middle $$4 \,\, \Omega$$ resistor.
  2. Path 2: Directs current vertically upward through the shorted capacitor wire, horizontally across the upper-middle $$4 \,\, \Omega$$ resistor, and back down through the second shorted capacitor wire.
  Both paths re-converge perfectly at the same downstream junction (Node B). Because these two $$4 \,\, \Omega$$ resistors are bounded by the exact same nodes, they form a parallel combination.
• Finally, after exiting Node B, the current must pass through the last series resistor connected to the right terminal:

  $$R_{\text{right}} = 2 \,\, \Omega$$

3. Calculate the Total Equivalent Resistance ($$R_{\text{eq}}$$)

First, evaluate the parallel network containing the two middle $$4 \,\, \Omega$$ resistors:

$$R_{\text{parallel}} = \frac{4 \times 4}{4 + 4} = \frac{16}{8} = 2 \,\, \Omega$$

Now, combine the entire remaining pathway, since $$R_{\text{left}}$$, $$R_{\text{parallel}}$$, and $$R_{\text{right}}$$ sit sequentially in series with each other:

$$R_{\text{eq}} = R_{\text{left}} + R_{\text{parallel}} + R_{\text{right}}$$

$$R_{\text{eq}} = 1 \,\, \Omega + 2 \,\, \Omega + 2 \,\, \Omega = 5 \,\, \Omega$$

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

Applying Ohm's Law with the source root-mean-square voltage ($$V = 220 \,\, \text{V}$$) and our corrected load impedance configuration:

$$I = \frac{V}{R_{\text{eq}}} = \frac{220 \,\, \text{V}}{5 \,\, \Omega} = 44 \,\, \text{A}$$

Correct Numerical Answer: 44 A