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

An LCR circuit has L = 10 mH, `R = 3 Omega` and `C = 1 mu F` and is connected in series to a source of `(20 sin omega t)` volt. Calculate the current amplitude at a frequency 20 % lower than the resonance frequency of the circuit.

Text Solution

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Here, `L = 10 mH = 10^(2) H, R = 3 Omega`
`C = 1 mu F = 10^(-6) F`,
`E = 20 sin omega t = E_(0) sin omega t`
`:. E_(0) = 20 V, I_(0) = ?`
Resonant frequency, `v_(0) = (1)/(2 pi sqrt(LC))`
`= (1)/(2 pi sqrt(10^(-2) xx 10^(-6))) = (10^(4))/(2 pi)`
Frequency at which current amplitude is
required `= v = 80% v_(0) = (80)/(100) xx (10^(4))/(2 pi) = (8 xx 10^(-3))/(2 pi) Hz`
`X_(L) = omega L = 2 pi v L`
`= 2 pi xx (8 xx 10^(3))/(2 pi) xx 10^(-2) = 80 ohm`
`X_(C ) = (1)/(omega C) = (1)/(2 pi v C)`
`= (2 pi)/(2 pi xx (8 xx 10^(3)) xx 10^(-6))`
`= (10^(3))/(8) = 125 ohm`
`Z = sqrt(R^(2) + (X_(L) - X_(C ))^(2))`
`= sqrt(3^(2) + (80 - 125)^(2)) = 45.1 ohm`
`I_(0) = (E_(0))/(Z) = (20)/(45.1) = 0.443 A`
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