in a lcr circuit L=10H,c=160uf and R =80 ohm find the value of angular frequencies for which the power is half of its maximum value are nearly
Answers
Answer:
(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
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