For a series LCR circuit with R = 100 Ω, L = 0.5 mH and C = 0.1 pF connected across 220V-50 Hz AC supply, the phase angle between current and supplied voltage and the nature of the circuit is :

Solution

Physics sir · 4 steps

Compare X_L and X_C at 50 Hz: a 0.1 pF capacitor has a huge reactance that swamps everything else.

1. Inductive reactance. ω = 2π f = 100π ≈ 314 rad s⁻¹. So X_L = ω L = 314 × 0.5 × 10⁻³ ≈ 0.157 Ω, which is tiny.
2. Capacitive reactance. X_C = 1/(ω C) = 1/(314 × 0.1 × 10⁻¹²) ≈ 3.18 × 10¹⁰ Ω. This is enormous compared with both R = 100 Ω and X_L.

   tan φ = (X_C - X_L)/R ≈ (3.18 × 10¹⁰)/100 = 3.18 × 10⁸

3. Phase angle. When tan φ is this large, φ is essentially 90°. Because X_C ≫ X_L, the current leads the voltage, so the circuit behaves as predominantly capacitive.
4. So. The phase angle is about 90° and the circuit is predominantly capacitive, which is option (d).

The idea to keep. In any series LCR question, find X_L and X_C first. The bigger one decides the nature of the circuit, and tan φ = (X_L - X_C)/R tells you how close the phase is to 90°.

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