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

Solution

Physics sir · 4 steps

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

1. High-frequency limit. X_L = ω L grows without limit, so each inductor acts as an open circuit. X_C = 1/(ω C) drops to zero, so each capacitor acts as a short circuit.
2. Remove the inductors. The top-left branch (8 Ω with 20 mH) and the top-right branch (50 mH with 8 Ω) carry no current. The two vertical 0.5 μF capacitors become wires, so each top junction joins the junction directly below it.
3. Redraw the circuit. What remains is 1 Ω in series with a middle section, then 2 Ω. The middle section is two 4 Ω resistors in parallel, the top one and the bottom one, which give 2 Ω.

   R_(eq) = 1 + (4 × 4)/(4 + 4) + 2 = 5 Ω

4. So. I = 220/5 = 44 A.

The idea to keep. Very high frequency: open every inductor, short every capacitor. Very low frequency (DC): do the opposite. Then redraw the circuit and solve it with plain resistors.

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