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⚛️ Physics  ·  Alternating Current  ·  NEET & JEE

A 100 Ω resistor, 10 mH inductor, and 100 μF capacitor are connected in series to 200 V, 50 Hz AC. Find X<sub>L</sub> and X<sub>C</sub>.

Answer: X L = 31.4 Ω, X C = 31.8 Ω.

  • A X<sub>L</sub> = 3.14 Ω, X<sub>C</sub> = 31.8 Ω
  • B X<sub>L</sub> = 31.4 Ω, X<sub>C</sub> = 31.8 Ω
  • C X<sub>L</sub> = 314 Ω, X<sub>C</sub> = 3.18 Ω
  • D X<sub>L</sub> = 3.14 Ω, X<sub>C</sub> = 318 Ω

Correct answer: B. X<sub>L</sub> = 31.4 Ω, X<sub>C</sub> = 31.8 Ω

Explanation: X<sub>L</sub> = 2πfL = 2π × 50 × 0.01 = π ≈ 3.14 Ω. Wait: 2 × 3.14 × 50 × 0.01 = 3.14 Ω. X<sub>C</sub> = 1/(2πfC) = 1/(2π × 50 × 10⁻⁴) = 1/(0.03142) = 31.8 Ω. So option A is correct.

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 →