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⚛️ Physics  ·  Alternating Current  ·  NEET & JEE

The resonant frequency of a circuit with L = 4 mH and C = 100 nF is:

Answer: 7958 Hz.

  • A 7958 Hz
  • B 2500 Hz
  • C 1000 Hz
  • D 500 Hz

Correct answer: A. 7958 Hz

Explanation: f<sub>0</sub> = 1/(2π√(LC)) = 1/(2π√(4×10⁻³ × 10⁻⁷)) = 1/(2π√(4×10⁻¹⁰)) = 1/(2π × 2×10⁻⁵) = 1/(4π×10⁻⁵) ≈ 7958 Hz.

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.

Read the full Alternating Current notes →