At very high frequencies, the effective impendence of the given circuit will be ________________ Ω.

Solution

Physics sir · 4 steps

At very high frequency, every capacitor becomes a plain wire and the inductor becomes a break.

1. Limit of reactances. X_C = 1/(ω C) → 0 as ω → ∞, so each 0.5 F capacitor acts as a short circuit. X_L = ω L → ∞, so the 20 H inductor acts as an open circuit.
2. Redraw the circuit. The bottom-left branch (the 1 Ω resistor in series with the inductor) is broken, so no current flows through it. The middle capacitor shorts the top and bottom junctions together. The two right-hand capacitors also become wires.
3. What remains. The top-left 1 Ω resistor leads to the joined junction. From there the two 2 Ω resistors run in parallel to the right terminal: (2 × 2)/(2 + 2) = 1 Ω.

   Z = 1 Ω + (2 Ω ∥ 2 Ω) = 1 Ω + 1 Ω = 2 Ω

4. So. The effective impedance at very high frequencies is 2 Ω.

The common slip: 1.5. Keeping the bottom-left 1 Ω in parallel with the top-left 1 Ω gives 0.5 + 1 = 1.5. That branch carries no current, because the inductor in series with it is open.

The idea to keep. For extreme frequencies, swap the parts first: at high ω, C is a wire and L is a break. At ω → 0 it is the other way round. Then solve it like a resistor network.

Answer from the NTA final answer key. Solution by PrepWiser.