Physics, asked by sanjaysaini3850, 10 months ago

In an L-C-R ereis circuit R=150Omega, L=0.0750 H and C=0.0180muF . The source has voltage amplitude V=150V and a frequencey equal to the resonacne frequency of the circuit. (a) What is the power factor ? (b) What is the average power delivered by the source? (c) The capacitor is replaced by one with C=0.0360muF and the source frequency is adjusted to the new resonance value. Then, what is the average power delivered by the source?

Answers

Answered by hiteshvskp1
0

Answer:

averge power delivered by the source=0.12345...

Answered by wajahatkincsem
0

Thus the power delivered by the source is 75 W

Explanation:

Given data:

  • Resistance "R" = 150 Ω
  • Inductance "L" = 0.0750 H
  • Capacitance "C" = 0.0180 muF
  • voltage V = 150 V

Solution:

(a) At resonance frequency,

XL = XC

Z = R and power factor cosϕ = R Z = 1.0

(b) P = V^2rms / R  = (150/√2)^2  / 150 = 75 W

(c) Again P = V^2 rms / R  = 75 W.

Thus the power delivered by the source is 75 W

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