A resonance circuit having inductance and resistance 2 × 10^-4 H and 6.28 Ω respectively oscillates at 10⁷ Hz frequency. The value of quality factor of this resonator is _________. [π = 3.14]

Explanation: 

The ratio of the initial energy stored in the resonator to the energy lost in one radian of the oscillation cycle is known as the quality factor of the resonator.

The quality factor Q = \(\frac{\omega_{0}L}{R} = \frac{2\pi L f}{R}\)    ----(1)

Calculation:

Given: 

Inductance L = 2 × 10-4 H 

 Resistance R = 6.28 Ω

 frequency f = 107 Hz

 Quality factor =?

From equation (1) we get:

 Q = \(\frac{2\pi L f}{R}\)    where L = 2 × 10-4 H , R = 6.28 Ω, f = 107 Hz

  putting values we get: \(Q = \frac{2\pi\times 10^{7}\times 2\times 10^{-4}}{6.28} = 2 \times 10^{3}\)

 Hence the answer is = 2000 = value of quality factor.