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
Verified Correct Answer: D.
Physical Principles & Numerical Solution
Resonance frequency in an a.c. circuit is given by F = \(\frac{1}{2\pi \sqrt{LC}}\)
therefore, \(200\times 10^3\) = \(\frac{1}{2\pi \sqrt{L\times 20 \times 10^{-12}}}\)
therefore (\(200\times 10^3)^2\) =\(\frac{1}{4 \times \pi^2 \times L \times 20 \times 10^{-12}}\)
THEREFORE, L = \(\frac{1}{4\times10\times (200\times 10^3)^2 \times 20\times 10^{-12}}\), (\pi\(^2\) ≈10)
Therefore L =\(\frac{1}{32}\)H
Source: Myschool Past Questions Archive (Record #9217)