JAMB - Physics (2001 - No. 36)
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
\(\frac{I}{16}\) H
\(\frac{I}{8}\) H
\(\frac{I}{64}\) H
\(\frac{I}{32}\) H
설명
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
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