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A capacitor of 20 x 10\(^{-12}\)F and an inductor are joined in series. The value...

Physics
JAMB 2001

A capacitor of 20 x 10\(^{-12}\)F and an inductor are joined in series. The value of the inductance that will give the circuit a resonance frequency of 200 kHz is

  • A. \(\frac{I}{16}\) H
  • B. \(\frac{I}{8}\) H
  • C. \(\frac{I}{64}\) H
  • D. \(\frac{I}{32}\) H
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Correct Answer: Option D
Explanation

Resonance frequency in an a.c. circuit is given by F = \(\frac{1}{2π \sqrt{LC}}\)
therefore, \(200\times 10^3\) =  \(\frac{1}{2π \sqrt{L\times 20 \times 10^{-12}}}\)
therefore (\(200\times 10^3)^2\)  =\(\frac{1}{4 \times π^2 \times L \times 20 \times 10^{-12}}\)
THEREFORE,  L = \(\frac{1}{4\times10\times (200\times 10^3)^2 \times 20\times 10^{-12}}\),      (π\(^2\)  ≈10)
Therefore L =\(\frac{1}{32}\)H

There is an explanation video available below.


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Explanation Video

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