Figure here, shows a series L-C-R circuit connected to a variable frequency 230 V source. L = 5.0H, C = `80 muF` and r =`40 Omega`
(a) Determine the source frequency which drives the circuit in resonance.
(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
(c) Determine the rms potential drops across the three elements of the circuit. show that the potential drop across the L-C combination is zero at the resonating frequency.

Explanation

Resonant frequency (source frequency ) is given by
`omega_(0) = (1)/(sqrt(LC)) = (1)/(sqrt(5.0 xx 80 xx 10^(-6))`
at resonance, impedance of he circuit is
Z = R = 40 `Omega`
Peak value of current is
` I_(0) = (E_(0))/(Z) = (E_(0))/(R)` (at resonance )
Here, `E_(0)` is the peak value of voltage = `sqrt(2) xx 230 ` V
`therefore " " I_(0) = (sqrt(2) xx 230)/(40) ` = 8.13 A
(c) rms value of current is
` I_(rms) = (E_(rms))/(Z) = (230)/(40) = 5.75` A
Across inductor , rms potenial difference is
`V_(L) = I_(rms) xx omega_(0) L `
`rArr " " = 5.75 xx 50 xx 5 ` = 1437.5 V
Across capacitor, rms potential difference is
` V_(C ) = I_(rms) xx (1)/( omega_(0)C )`
` = 5.75 xx (1)/( 50 xx 80 xx 10^(-6)) = ` 1437.5 V
Total potential drop across L and C
` I_(rms) xx (omega L - (1)/(omega_(0)C) )`
1437.5 - 1437.5 = 0
Across R, `V_(R) = 5.75 xx 40 ` = 230 V