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⚛️ Physics  ·  Alternating Current  ·  NEET & JEE

An ideal LC circuit starts with maximum charge Q<sub>0</sub> on its capacitor. At time T/6, where T is the oscillation period, what fraction of the total energy is stored in the inductor?

Answer: 3/4.

  • A 1/4
  • B 1/2
  • C 1
  • D 3/4

Correct answer: D. 3/4

Explanation: For an LC oscillator, q = Q<sub>0</sub> cos(&omega;t). At T/6, &omega;t = &pi;/3, so q = Q<sub>0</sub>/2. Capacitor energy is therefore q<sup>2</sup>/(2C) = one-fourth of the total energy. The remaining three-fourths is stored in the inductor.

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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