A series combination of resistor of resistance 100 Ω, inductor of inductance 1 H and capacitor of capacitance 6.25 μ F is connected to an ac source. The quality factor of the circuit will be __________

Solution

Physics sir · 3 steps

Find the resonant frequency first, then Q is just how big the inductive reactance is compared with R.

1. Resonant frequency. At resonance ω₀ = 1/√LC. With L = 1 H and C = 6.25 × 10⁻⁶ F, LC = 6.25 × 10⁻⁶ s², so √LC = 2.5 × 10⁻³ s and ω₀ = 400 rad s⁻¹.

   Q = (ω₀ L)/R = 1/R√(L/C)

2. Put in numbers. ω₀ L = 400 × 1 = 400 Ω. Divide by R = 100 Ω to get Q = 400/100.
3. So. The quality factor of the circuit is 4.

The common slip: 0.004. Using C = 6.25 instead of 6.25 × 10⁻⁶ F gives √(L/C) = 0.4 and Q = 0.004, not 4.

The idea to keep. For a series LCR circuit, Q = 1/R√(L/C). Always convert μF to farads before taking the square root.

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