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Capacitors

Q1: With a 500 kHz signal source, what would be the value of a capacitor yielding a capacitive reactance of 1 k omega.gif?

A 318 pF

B 2 nF

C 3.18 F

D 2 F

ANS:A - 318 pF

Given: f = 500 kHz; XC = 1 kΩ; C =?

XC = 1 / (2 * π * f * C).
C = 1 / (2 * π * f * XC).
C = 1 / (2 * 3.14 * 500 * 10^3 * 1 * 10&3) .
C = 1 / 3140 * 10^6.
C = 3.18 * 10^-4 * 10^-6.
C = 3.18 * 10^-10 or 318 * 10^-12.
And we get;
C = 3.18 p F.



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