Figure shows a series LCR circuit connected to a variable frequency 230 V source. `L = 5.0 H, C = 80muF, 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 LC combination is zero at the resonating frequency.

Explanation

At resonance, potential drop ( p.d.) across R is,
`V_(R ) = I_(rms) R = ( I_(m))/(sqrt(2)) R`
`:. V_(R ) = ( 8.13 xx 40)/( 1.414)`
`:. V_(R ) = 230 V o l t `
For series RLC ac circuit, source voltage is given by,
`V = sqrt((V_(R )^(2) + ( V _(L) - V_(C ))^(2)))`
`:. V^(2) = V_(R )^(2) +(V_(L) - V_(C))^(2)`
`:. ( 230)^(2) = ( 230)^(2) + ( V_(L)- V_(C ))^(2)`
`:. (V_(L) - V_(C ))^(2) = 0`
`:. V_(L) - V_(C ) = 0`
`:.` p.d. across LC combination = 0
OR Net p.d. across LC combination
`= I_(rms) ( X_(L) - X_(C ))`
`= 0 ( :. `At resonance `X_(L ) X_(C)`