Question

An `LCR` circuit has `L = 10 mL, R = 3 Omega`, and `C = 1 mu F` connected in series to a source of `15 cos omega t` volt. The current amplitude at a frequency that is 10% lower then the resonant frequency is
A. `0.5 A`
B. `0.7 A`
C. `0.9 A`
D. `1.1 A`

Answer 1

Correct Answer - 2
`c_(v) = (90)/(100) c_(v_(0)) = (90)/(100) x (1)/(sqrt(LC)) = 9000 "rad"//s`
`i_(0) = (E_(0))/(sqrt(R^(2) + (omega L - (1)/(omega C))^(2)))`