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JAMB Physics 2001 Question 49

Question 49

Alternating Current Circuits

A capacitor of 20 × 10−12 F and an inductor are joined in series. The value of the inductance that will give the circuit a resonant frequency of 200 kHz is [π2 = 10]

  1. A.1/16 H
  2. B.1/8 H
  3. C.1/64 H
  4. D.1/32 H
Show the answer and explanation

Answer: D (1/32 H)

Explanation

At resonance f = 1/(2π√(LC)), so L = 1/(4π2f2C) = 1/(4 × 10 × (2 × 105)2 × 20 × 10−12) = 1/32 H.