A capacitor of 20 x 10-12F and an inductor are joined in series. The value of the inductance that will give the circuit a resonance frequency of 200kHz is
- I/16 H
- I/8 H
- I/64 H
-
I/32 H
The correct answer is: 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 .