At very high frequencies, the effective impedance of the given circuit will be _________ $$\Omega$$.

Based on the provided images, here is the formatted, step-by-step solution to the problem.

Problem Statement

At very high frequencies, find the effective impedance of the given AC circuit.

Step-by-Step Solution

To find the effective impedance at very high frequencies ($$f \to \infty$$), we need to evaluate how the reactive components (inductors and capacitors) behave.

1. Behavior of Capacitors at High Frequency

The capacitive reactance ($$X_C$$) is given by the formula:

$$X_C = \frac{1}{2\pi f C}$$

• As frequency ($$f$$) becomes very large, $$X_C$$ approaches zero ($$X_C \to 0$$).
• Therefore, all capacitors act as a short circuit (ideal wires).
• As frequency ($$f$$) becomes very large, $$X_L$$ approaches infinity ($$X_L \to \infty$$).
• Therefore, the inductor acts as an open circuit (broken wire/infinite resistance).
• Replace all capacitors ($$0.5\text{ F}$$) with simple connecting wires.
• Replace the branch containing the inductor ($$20\text{ H}$$) with an open break. Because this bottom-left branch is broken, no current flows through the $$1\ \Omega$$ resistor connected in series with the inductor.
• A single $$1\ \Omega$$ resistor in the top-left branch.
• Followed by a parallel combination of two $$2\ \Omega$$ resistors on the right.
• Parallel Combination:
• Total Impedance:

2. Behavior of Inductors at High Frequency

The inductive reactance ($$X_L$$) is given by the formula:

$$X_L = 2\pi f L$$

Simplified Circuit Analysis

Applying these conditions to the original network:

This leaves us with the following simplified equivalent circuit:

Calculating Equivalent Impedance ($$Z_{\text{eq}}$$)

The two $$2\ \Omega$$ resistors are in parallel. Their equivalent resistance ($$R_p$$) is:

$$R_p = \frac{2 \times 2}{2 + 2} = 1\ \Omega$$

This combination is in series with the remaining $$1\ \Omega$$ resistor:

$$Z_{\text{eq}} = 1\ \Omega + R_p = 1 + 1 = 2\ \Omega$$

Final Answer

The effective impedance of the given circuit at very high frequencies is $$2$$ $$\Omega$$.