Let us examine the behavior of capacitors and inductors for both DC and high-frequency AC:

• For DC (frequency $f = 0$):
  • Capacitive reactance $X_C = \frac{1}{2\pi f C} \to \infty$ (capacitors act as OPEN circuits).

• Inductive reactance $X_L = 2\pi f L = 0$ (inductors act as SHORT circuits / ideal zero-resistance wires).
• Looking at the circuit: Capacitors $C_1$ and $C_2$ act as open circuits, blocking current through the branches containing $R_2$ and $R_3$. The inductors $L_1$ and $L_2$ act as short circuits. Thus, the only conducting path connected across the source has resistance $R_1$.

• With DC applied ($V = 100\,\text{V}, I_{\text{dc}} = 1.0\,\text{A}$):
  In DC, an inductor offers zero inductive reactance, so only the ohmic resistance $R$ limits the current:
  $$R = \frac{V_{\text{dc}}}{I_{\text{dc}}} = \frac{100}{1.0} = 100\,\Omega$$
• With AC applied ($V = 100\,\text{V}, I_{\text{ac}} = 0.5\,\text{A}, f = 50\,\text{Hz}$):
  The total impedance $Z$ is:
  $$Z = \frac{V_{\text{ac}}}{I_{\text{ac}}} = \frac{100}{0.5} = 200\,\Omega$$
• Calculating the inductive reactance $X_L$:
  $$Z = \sqrt{R^2 + X_L^2} \implies X_L = \sqrt{Z^2 - R^2} = \sqrt{200^2 - 100^2} = \sqrt{40000 - 10000} = \sqrt{30000} = 100\sqrt{3} \approx 173.2\,\Omega$$
• Inductance $L$:
  $$X_L = 2\pi f L \implies L = \frac{X_L}{2\pi f} = \frac{173.2}{2 \times 3.1416 \times 50} = \frac{173.2}{314.16} \approx 0.551\,\text{H} \approx 0.55\,\text{H}$$