An AC circuit has R= 100 Ω, C = 2 μF and L = 80 mH, connected in series. The quality factor of the circuit is :

Solution

Physics sir · 4 steps

Q is the reactance at resonance divided by R, so find the resonant frequency first and then compare.

1. What Q means. For a series LCR circuit, Q = (ω₀ L)/R, where ω₀ = 1/√LC is the resonant angular frequency. Substituting ω₀ gives a form that needs no frequency at all.

   Q = 1/R√(L/C)

2. Put in numbers. In SI units, L = 0.08 H and C = 2 × 10⁻⁶ F. Then L/C = 0.08/(2 × 10⁻⁶) = 4 × 10⁴ Ω², so √(L/C) = 200 Ω.
3. Check by frequency. ω₀ = 1/√(0.08 × 2 × 10⁻⁶) = 2500 rad s⁻¹, so ω₀ L = 2500 × 0.08 = 200 Ω. The two methods agree.
4. So. Q = 200/100 = 2, which is option (b).

The idea to keep. For a series LCR circuit, remember Q = 1/R√(L/C). Convert mH and μF to H and F before you take the square root.

Answer from the NTA final answer key. Solution by PrepWiser.