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?

A. I/16 H
B. I/8 H
C. I/64 H
D. I/32 H

Correct Answer: Option D
Explanation
Resonant, frequency in an a.c. circuit is given by F = I/2π√LC
therefore 200 * 103 = I/2 * π√l * 20 * 10-12
therefore (2.0 * 105)2 = I/4 * π2 * L * 20 * 10-12
THEREFORE L = I/4.0 * 1010 * 4 * 10 * 20 * 10-12
(if π2 = 10)
= I/16 * 1011 * 2.0 * 10-11
Therefore L = I/32 H