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
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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
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