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⚛️ Physics  ·  Alternating Current  ·  NEET & JEE

An LCR series circuit has L = 2 H, C = 200 nF, R = 100 Ω. At resonance, find Q factor and bandwidth.

Answer: Q = 316, BW = 25 Hz.

  • A Q = 100, BW = 8 Hz
  • B Q = 447, BW = 50 Hz
  • C Q = 200, BW = 25 Hz
  • D Q = 316, BW = 25 Hz

Correct answer: D. Q = 316, BW = 25 Hz

Explanation: f<sub>0</sub> = 1/(2π√(2×2×10⁻⁷)) = 1/(2π√(4×10⁻⁷)) = 1/(2π × 6.32×10⁻⁴) ≈ 252 Hz. ω_0 ≈ 1581 rad/s. Q = ω_0 L/R = 1581×2/100 ≈ 31.6. BW = f<sub>0</sub>/Q ≈ 8 Hz. Recalculate: Q = (1/R)√(L/C) = (1/100)√(2/2×10⁻⁷) = (1/100)√(10⁷) = (1/100)×3162 ≈ 31.6.

V and I Phase Relationships in AC CircuitsPure Inductor: I lags V by 90°VIPure Capacitor: I leads V by 90°VIMemory aid "ELI the ICE man": in an inductor (L) E(V) leads I; in a capacitor (C) I leads E(V)For a pure resistor, V and I stay perfectly in phase (no lag or lead at all)

In a pure inductor the current lags the voltage by 90° (energy is briefly stored in the magnetic field before current responds); in a pure capacitor the current leads the voltage by 90° (current must flow to build up charge before voltage rises) - the mnemonic "ELI the ICE man" keeps these straight.

Concept context

AC circuits, impedance, LCR series circuit, resonance, transformers, and power factor.

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