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
Explanation
Correct Option
dVideo Explanation
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CRIMHAZ
3 years ago
Myschool pls change d ans to option A.
Because. (where c=capacitor
L= 1/F^2*4*π^2*C. L=inductor
1/2*10^10 * 4 * 10 * 20*10^-12.] F=freq)
1/80*10^10 * 20*10^-12. ...............
1/8*10^11 *2*10^-11
1/8*2
:. L=1/16
