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An LCR circuit has L = 10 mL, R = 3 Omeg...

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`

Text Solution

Verified by Experts

The correct Answer is:
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)))`
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