Question

A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 W is connected to a 230 V variable frequency supply. (a) What is the source frequency for which current amplitude is maximum. Obtain this maximum value. (b) What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power. (c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies? (d) What is the Q-factor of the given circuit?

Answer 1

(a) at resonance frequency, the current amplitude is maximum.

so, source frequency, f = 1/{2π√(LC)}

here, L = 0.12H, C = 480nF = 480 × 10^-9F

= 1/{2π√(0.12 × 480 × 10^-9)}

= 663Hz

maximum value of current, I =√2 × virtual current in the coil (Iv)

= √2(E/R)

= √2 × 230V/(23Ω)

= 14.14 A

( b) maximum power loss at resonant frequency, P = Iv. Ev cos∅

= Ev.(Ev/R)cos0°

= (Ev)²/R , [ Ev is virtually emf of circuit]

= (230)²/23 = 2300W

(c) The power transferred to the circuit is half the power at resonant frequency.

frequency at which power transferred to the circuit is half = w ± ∆w

where ∆w = R/2L

= (23 ohm)/(2 × 0.12H)

= 95.83 rad/s

so, band frequency , ∆f = 95.83/2π

= 15.26 Hz

Hence two frequencies for half power

f1 = f - ∆f

= 663 - 15.26 = 647.74Hz

and f2 = f + ∆f

= 663 + 15.26 = 678.26 Hz

at theses frequencies the current amplitude is I' = I/√2 = 14.14/√2 = 10A

(d) Q - factor = 1/R√{L/C}

= 1/(23) × √{0.12/480 × 10^-9}

= 21.7