A resonance circuit having inductance and resistance 2 × 10⁻⁴ H and 6.28Ω respectively oscillates at 10 MHz frequency. The value of quality factor of this resonator is ___________. [π = 3.14]

Solution

Physics sir · 4 steps

At resonance, Q is just the inductor's reactance divided by the resistance, so find ω L first.

1. Quality factor. For a series resonant circuit, the quality factor compares the inductive reactance at resonance with the resistance.

   Q = (ω₀ L)/R = (2π f₀ L)/R

2. Put in numbers. f₀ = 10 MHz = 10⁷ Hz, so ω₀ L = 2 × 3.14 × 10⁷ × 2 × 10⁻⁴ = 6.28 × 2 × 10³ Ω.
3. Divide by R. Q = (6.28 × 2 × 10³)/6.28 = 2 × 10³. The 6.28 cancels neatly, which is why R was chosen as 2π.
4. So. The quality factor of the resonator is 2000.

The common slip: 318. Using f in place of ω drops the factor 2π and gives about 318 instead of 2000.

The idea to keep. In a series LCR circuit, Q = ω₀ L / R = 1/(ω₀ C R). Always use angular frequency ω = 2π f, never f alone.

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