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For the RLC parallel resonant circuit when R = 8kΩ, L = 40 mH and C = 0.25 μF, the quality factor Q is
  • a)
    40
  • b)
    20
  • c)
    30
  • d)
    10
Correct answer is option 'B'. Can you explain this answer?
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Calculation of Quality Factor Q:

Given values:
- Resistance R = 8kΩ = 8000Ω
- Inductance L = 40 mH = 0.04 H
- Capacitance C = 0.25 μF = 0.25 x 10^-6 F

Formula for Quality Factor Q:
Q = 1 / R * sqrt(L / C)

Substitute the given values:
Q = 1 / 8000 * sqrt(0.04 / 0.25 x 10^-6)
Q = 1 / 8000 * sqrt(160000)
Q = 1 / 8000 * 400
Q = 0.05 * 400
Q = 20
Therefore, the quality factor Q for the RLC parallel resonant circuit is 20. Hence, the correct answer is option B.
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