A series resonant LCR circuit has a quality factor (Q-factor)=0.4. If `R=2k Omega, C=0.1 mu F` then the value of inductance is

Options

A. 2H
B. 10 H
C. 0.064 H
D. 0.1 H

Answer

C

Explanation

Quality factor `Q = (omegaL)/R = 1/(sqrt(LC)) xx L/R`
`1/Rsqrt(L/C) [:' omega^(2) = 1/(LC]`
`:. Q^(2) = L/(CR^(2))`
`:. L = Q^(2)CR^(2) = (0.4)^(2) xx 0.1 xx 10^(-6) xx (2 xx 10^(3))^(2)`
`= 16 xx 10^(-2) xx 10^(-7) xx 4 xx 10^(6)`
`=64 xx 10^(-3) = 0.064 H`.
Note: Bandwidth is not necessary to calculate L.